OP Malhotra Class 11 Maths Solutions Chapter 7 Properties of Triangle Exercise 7

S Chand Class 11 ICSE Maths Solutions Chapter 7 Properties Of Triangle Ex 7

Question 1. In \( \triangle ABC \),
(i) If \( a = 2, b = 3, c = 4 \), prove that \( \cos A = \frac { 7 }{ 8 } \).
(ii) If the sides are \( 7, 4\sqrt{3} \) and \( \sqrt{13} \) cm, prove that the smallest angle is \( 30^\circ \).
(iii) If \( a = 9, b = 8, c = 4 \), prove that \( 6 \cos C = 4 + 3 \cos B \).
(iv) The sines of the angles of a triangle are in the ratio of \( 4 : 5 : 6 \); prove that the cosines are in the ratio \( 12 : 9 : 2 \).
(v) If the two angles of a triangle are \( 30^\circ \) and \( 45^\circ \) and the included side is \( (\sqrt{3} + 1) \) cm, find the area of the triangle.
(vi) If in a \( \triangle ABC \), \( a = 6, b = 3 \) and \( \cos (A – B) = \frac { 4 }{ 5 } \), find its area.
(vii) In a triangle ABC, \( \angle C = 60^\circ \) and \( \angle A = 75^\circ \). If D is a point on AC such that the area of the \( \triangle BAD \) is \( \sqrt{3} \) times the area of the \( \triangle BCD \), find the \( \angle ABD \).
Answer:
(i) Given the sides \( a = 2, b = 3, c = 4 \). To find \( \cos A \), we use the cosine formula:
\( \cos A = \frac{b^2+c^2-a^2}{2bc} \)
Now, substitute the given values into the formula:
\( \cos A = \frac{3^2+4^2-2^2}{2 \times 3 \times 4} \)
\( \cos A = \frac{9+16-4}{24} \)
\( \cos A = \frac{21}{24} \)
\( \cos A = \frac{7}{8} \)
Thus, it is proven that \( \cos A = \frac{7}{8} \). The cosine rule is especially useful when we know all three sides of a triangle and want to find an angle.
In simple words: The cosine rule helps us find an angle when we know all three sides of a triangle. We put the side lengths into the formula, calculate, and get the value for cos A, which matches 7/8.

🎯 Exam Tip: Remember to correctly identify sides opposite to angles (e.g., side 'a' is opposite angle 'A') when using the cosine formula.

(ii) Let the sides be \( a = 7 \), \( b = 4\sqrt{3} \), and \( c = \sqrt{13} \). To find the smallest angle, we first identify the smallest side. The approximate values are:
\( a = 7 \)
\( b = 4\sqrt{3} \approx 4 \times 1.732 = 6.928 \)
\( c = \sqrt{13} \approx 3.606 \)
Since \( c \) is the smallest side, the smallest angle is \( \angle C \). We use the cosine formula for \( \angle C \):
\( \cos C = \frac{a^2+b^2-c^2}{2ab} \)
Substitute the values:
\( \cos C = \frac{7^2+(4\sqrt{3})^2-(\sqrt{13})^2}{2 \times 7 \times 4\sqrt{3}} \)
\( \cos C = \frac{49+48-13}{56\sqrt{3}} \)
\( \cos C = \frac{84}{56\sqrt{3}} \)
\( \cos C = \frac{3}{2\sqrt{3}} \)
\( \cos C = \frac{3\sqrt{3}}{2\sqrt{3}\sqrt{3}} \)
\( \cos C = \frac{3\sqrt{3}}{6} \)
\( \cos C = \frac{\sqrt{3}}{2} \)
Since \( \cos C = \frac{\sqrt{3}}{2} \), the angle \( C = 30^\circ \) or \( \frac{\pi}{6} \) radians. In any triangle, the smallest angle will always be found opposite the shortest side, and the largest angle opposite the longest side.
In simple words: First, we find which side is the shortest. The smallest angle is always across from the shortest side. We then use the cosine rule for that angle and find its value, which turns out to be 30 degrees.

🎯 Exam Tip: When finding the smallest angle, always identify the smallest side first, as the angle opposite to it will be the smallest.

(iii) Given sides \( a = 9, b = 8, c = 4 \). We need to prove \( 6 \cos C = 4 + 3 \cos B \).
First, calculate \( \cos B \) using the cosine formula:
\( \cos B = \frac{a^2+c^2-b^2}{2ac} \)
\( \cos B = \frac{9^2+4^2-8^2}{2 \times 9 \times 4} \)
\( \cos B = \frac{81+16-64}{72} \)
\( \cos B = \frac{33}{72} = \frac{11}{24} \)
Next, calculate \( \cos C \) using the cosine formula:
\( \cos C = \frac{a^2+b^2-c^2}{2ab} \)
\( \cos C = \frac{9^2+8^2-4^2}{2 \times 9 \times 8} \)
\( \cos C = \frac{81+64-16}{144} \)
\( \cos C = \frac{129}{144} = \frac{43}{48} \)
Now, substitute these values into the left-hand side (LHS) of the equation:
LHS \( = 6 \cos C = 6 \times \frac{43}{48} = \frac{43}{8} \)
Now, substitute the values into the right-hand side (RHS) of the equation:
RHS \( = 4 + 3 \cos B = 4 + 3 \left(\frac{11}{24}\right) \)
RHS \( = 4 + \frac{11}{8} \)
RHS \( = \frac{32+11}{8} = \frac{43}{8} \)
Since LHS = RHS, the statement \( 6 \cos C = 4 + 3 \cos B \) is proven. This problem demonstrates how we can use the cosine rule to verify relationships between angles and sides within a triangle.
In simple words: We calculate `cos B` and `cos C` using the cosine rule for the given side lengths. Then, we put these values into the left and right sides of the equation `6 cos C = 4 + 3 cos B` to show that both sides are equal.

🎯 Exam Tip: Clearly calculate each cosine value before substituting them into the main equation to avoid errors.

(iv) Given that the sines of the angles of a triangle are in the ratio \( \sin A : \sin B : \sin C = 4 : 5 : 6 \).
According to the sine formula, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \) (a constant).
This implies that the ratio of the sides is \( a : b : c = \sin A : \sin B : \sin C = 4 : 5 : 6 \).
So, we can write \( a = 4k' \), \( b = 5k' \), and \( c = 6k' \) for some constant \( k' \).
Now, we calculate the cosines of the angles using the cosine formula:
\( \cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(5k')^2+(6k')^2-(4k')^2}{2(5k')(6k')} \)
\( \cos A = \frac{25k'^2+36k'^2-16k'^2}{60k'^2} = \frac{45k'^2}{60k'^2} = \frac{3}{4} \)
\( \cos B = \frac{c^2+a^2-b^2}{2ac} = \frac{(6k')^2+(4k')^2-(5k')^2}{2(4k')(6k')} \)
\( \cos B = \frac{36k'^2+16k'^2-25k'^2}{48k'^2} = \frac{27k'^2}{48k'^2} = \frac{9}{16} \)
\( \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{(4k')^2+(5k')^2-(6k')^2}{2(4k')(5k')} \)
\( \cos C = \frac{16k'^2+25k'^2-36k'^2}{40k'^2} = \frac{5k'^2}{40k'^2} = \frac{1}{8} \)
Now, we find the ratio \( \cos A : \cos B : \cos C \):
\( \cos A : \cos B : \cos C = \frac{3}{4} : \frac{9}{16} : \frac{1}{8} \)
To express this ratio in whole numbers, we multiply each term by the least common multiple (LCM) of the denominators (4, 16, 8), which is 16:
\( \left(\frac{3}{4} \times 16\right) : \left(\frac{9}{16} \times 16\right) : \left(\frac{1}{8} \times 16\right) \)
\( 12 : 9 : 2 \)
Thus, it is proven that the cosines are in the ratio \( 12 : 9 : 2 \). The sine rule links the ratio of sides to the ratio of the sines of the opposite angles, allowing us to find side ratios from angle sine ratios.
In simple words: First, we use the sine rule to understand that if the sines of angles are in a certain ratio, then the sides opposite those angles are in the same ratio. We use this to set up the side lengths with a common factor. Then, we use the cosine rule for each angle to find the values of `cos A`, `cos B`, and `cos C`. Finally, we simplify these fractions to show they are in the ratio 12:9:2.

🎯 Exam Tip: Remember that the ratio of sines of angles is equal to the ratio of the opposite sides of the triangle.

(v) Given two angles of a triangle are \( 30^\circ \) and \( 45^\circ \), and the included side is \( a = (\sqrt{3} + 1) \) cm. We need to find the area of the triangle.
Let \( \angle B = 30^\circ \) and \( \angle C = 45^\circ \).
The sum of angles in a triangle is \( 180^\circ \). So, \( \angle A = 180^\circ - (\angle B + \angle C) \):
\( \angle A = 180^\circ - (30^\circ + 45^\circ) \)
\( \angle A = 180^\circ - 75^\circ = 105^\circ \)
Next, we use the sine formula \( \frac{a}{\sin A} = \frac{b}{\sin B} \) to find side \( b \):
\( \frac{\sqrt{3}+1}{\sin 105^\circ} = \frac{b}{\sin 30^\circ} \)
First, calculate \( \sin 105^\circ \):
\( \sin 105^\circ = \sin (60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
\( \sin 105^\circ = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}} \)
Now, substitute back into the sine formula:
\( \frac{\sqrt{3}+1}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \frac{b}{\frac{1}{2}} \)
\( 2\sqrt{2} = 2b \)
\( b = \sqrt{2} \) cm.
Finally, we calculate the area of \( \triangle ABC \) using the formula \( \text{Area} = \frac{1}{2}ab\sin C \):
\( \text{Area} = \frac{1}{2}(\sqrt{3}+1)(\sqrt{2})\sin 45^\circ \)
\( \text{Area} = \frac{1}{2}(\sqrt{3}+1)(\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) \)
\( \text{Area} = \frac{1}{2}(\sqrt{3}+1) \) sq. units.
Knowing two angles and the included side is enough to uniquely determine a triangle and calculate its area.
In simple words: First, we find the third angle using the fact that angles in a triangle add up to 180 degrees. Then, we use the sine rule to find the length of side 'b'. Finally, we use the formula `\frac{1}{2}ab\sin C` to calculate the area of the triangle, using the values we found.

🎯 Exam Tip: Remember to calculate \( \sin 105^\circ \) correctly using the sum of angles formula, as this is a common point of error.

(vi) Given \( a = 6, b = 3 \) and \( \cos (A – B) = \frac { 4 }{ 5 } \). We need to find the area of the triangle.
We use the half-angle formula for cosine: \( \cos(A-B) = \frac{1-\tan^2\left(\frac{A-B}{2}\right)}{1+\tan^2\left(\frac{A-B}{2}\right)} \).
Substitute the given value:
\( \frac{4}{5} = \frac{1-\tan^2\left(\frac{A-B}{2}\right)}{1+\tan^2\left(\frac{A-B}{2}\right)} \)
Let \( X = \tan^2\left(\frac{A-B}{2}\right) \). So, \( \frac{4}{5} = \frac{1-X}{1+X} \).
Cross-multiply: \( 4(1+X) = 5(1-X) \)
\( 4 + 4X = 5 - 5X \)
\( 9X = 1 \implies X = \frac{1}{9} \)
So, \( \tan^2\left(\frac{A-B}{2}\right) = \frac{1}{9} \), which implies \( \tan\left(\frac{A-B}{2}\right) = \frac{1}{3} \) (assuming \( A > B \) since \( a > b \)).
Now, we use Napier's analogy (Tangent rule): \( \tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}\cot\left(\frac{C}{2}\right) \).
Substitute the known values:
\( \frac{1}{3} = \frac{6-3}{6+3}\cot\left(\frac{C}{2}\right) \)
\( \frac{1}{3} = \frac{3}{9}\cot\left(\frac{C}{2}\right) \)
\( \frac{1}{3} = \frac{1}{3}\cot\left(\frac{C}{2}\right) \)
\( \implies \cot\left(\frac{C}{2}\right) = 1 \)
Since \( \cot\left(\frac{C}{2}\right) = 1 \), the angle \( \frac{C}{2} = 45^\circ \), which means \( C = 90^\circ \).
Finally, calculate the area of \( \triangle ABC \) using the formula \( \text{Area} = \frac{1}{2}ab\sin C \):
\( \text{Area} = \frac{1}{2}(6)(3)\sin 90^\circ \)
\( \text{Area} = \frac{1}{2}(18)(1) \)
\( \text{Area} = 9 \) sq. units.
Napier's analogies provide a direct way to relate sides and half-angles, often simplifying calculations in triangle problems.
In simple words: We use a trigonometric identity for `cos(A-B)` to find the value of `tan((A-B)/2)`. Then, we apply Napier's rule, which connects the sides and half-angles of a triangle, to find `cot(C/2)`. From `cot(C/2)`, we find angle `C`. With angle `C` and sides 'a' and 'b', we can calculate the area of the triangle using the formula `\frac{1}{2}ab\sin C`.

🎯 Exam Tip: Napier's analogies are very helpful for solving triangles when you have a mix of side and angle information; ensure you use the correct form.

(vii) Given \( \angle C = 60^\circ \) and \( \angle A = 75^\circ \). D is a point on AC such that Area(\( \triangle BAD \)) \( = \sqrt{3} \times \) Area(\( \triangle BCD \)). We need to find \( \angle ABD \).
First, find \( \angle B \) in \( \triangle ABC \):
\( \angle B = 180^\circ - (\angle A + \angle C) \)
\( \angle B = 180^\circ - (75^\circ + 60^\circ) = 180^\circ - 135^\circ = 45^\circ \)
Let \( \angle ABD = \theta \). Then \( \angle CBD = \angle B - \theta = 45^\circ - \theta \).
The area of a triangle can be given by \( \frac{1}{2} \text{side}_1 \text{side}_2 \sin(\text{included angle}) \).
Area(\( \triangle BAD \)) \( = \frac{1}{2} AB \cdot BD \sin \theta \)
Area(\( \triangle BCD \)) \( = \frac{1}{2} BC \cdot BD \sin (45^\circ - \theta) \)
Given Area(\( \triangle BAD \)) \( = \sqrt{3} \times \) Area(\( \triangle BCD \)):
\( \frac{1}{2} AB \cdot BD \sin \theta = \sqrt{3} \times \frac{1}{2} BC \cdot BD \sin (45^\circ - \theta) \)
\( AB \sin \theta = \sqrt{3} BC \sin (45^\circ - \theta) \)
Using the sine formula in \( \triangle ABC \), \( \frac{AB}{\sin C} = \frac{BC}{\sin A} \). So, \( AB = BC \frac{\sin C}{\sin A} \).
Substitute \( AB \):
\( \left(BC \frac{\sin C}{\sin A}\right) \sin \theta = \sqrt{3} BC \sin (45^\circ - \theta) \)
\( \frac{\sin C}{\sin A} \sin \theta = \sqrt{3} \sin (45^\circ - \theta) \)
Substitute \( \sin C = \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Calculate \( \sin A = \sin 75^\circ = \sin (45^\circ + 30^\circ) \):
\( \sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
\( \sin 75^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}} \)
Now substitute these values back into the equation:
\( \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} \sin \theta = \sqrt{3} (\sin 45^\circ \cos \theta - \cos 45^\circ \sin \theta) \)
\( \frac{\sqrt{3}}{\sqrt{3}+1} \times \sqrt{2} \sin \theta = \sqrt{3} \left(\frac{1}{\sqrt{2}}\cos \theta - \frac{1}{\sqrt{2}}\sin \theta\right) \)
Divide both sides by \( \sqrt{3} \) and multiply by \( \sqrt{2} \):
\( \frac{\sqrt{2}}{\sqrt{3}+1} \sin \theta = \frac{1}{\sqrt{2}} (\cos \theta - \sin \theta) \)
\( 2 \sin \theta = (\sqrt{3}+1)(\cos \theta - \sin \theta) \)
\( 2 \sin \theta = (\sqrt{3}+1)\cos \theta - (\sqrt{3}+1)\sin \theta \)
Gather \( \sin \theta \) terms on one side:
\( 2 \sin \theta + (\sqrt{3}+1)\sin \theta = (\sqrt{3}+1)\cos \theta \)
\( (2 + \sqrt{3}+1) \sin \theta = (\sqrt{3}+1)\cos \theta \)
\( (3 + \sqrt{3}) \sin \theta = (\sqrt{3}+1)\cos \theta \)
Factor out \( \sqrt{3} \) from \( (3+\sqrt{3}) \):
\( \sqrt{3}(\sqrt{3}+1) \sin \theta = (\sqrt{3}+1)\cos \theta \)
Divide both sides by \( (\sqrt{3}+1) \):
\( \sqrt{3} \sin \theta = \cos \theta \)
Divide both sides by \( \cos \theta \):
\( \tan \theta = \frac{1}{\sqrt{3}} \)
Therefore, \( \theta = 30^\circ \). So, \( \angle ABD = 30^\circ \). The ratio of areas of two triangles sharing a common altitude is equal to the ratio of their bases.
In simple words: First, we find angle `B` in the main triangle. We then split angle `B` into two parts using `\theta`. We write down the area formula for `\triangle BAD` and `\triangle BCD`. Using the given ratio of areas, we form an equation. By applying the sine rule to `\triangle ABC` and simplifying, we get an equation involving `\sin \theta` and `\cos \theta`, which helps us find `\tan \theta`. This gives us the value of `\theta`, which is `\angle ABD`.

🎯 Exam Tip: When a point on a side divides a triangle into two smaller triangles, their areas can be related to the ratio of their bases if they share a common vertex.

Question 2. In any \( \triangle ABC \), Prove that \( \frac{\sin A}{\sin (A+B)}=\frac{a}{c} \).
Answer:
We start with the sine formula for a triangle, which states:
\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \)
From this, we know that \( a = k \sin A \) and \( c = k \sin C \).
The sum of the angles in any triangle \( ABC \) is \( 180^\circ \) (or \( \pi \) radians). So, \( A + B + C = \pi \).
This means \( A + B = \pi - C \).
Therefore, \( \sin (A+B) = \sin (\pi - C) \).
Since \( \sin (\pi - x) = \sin x \), we have \( \sin (A+B) = \sin C \).
Now, let's look at the left-hand side (LHS) of the equation we need to prove:
LHS \( = \frac{\sin A}{\sin (A+B)} \)
Substitute \( \sin (A+B) \) with \( \sin C \):
LHS \( = \frac{\sin A}{\sin C} \)
Now, substitute \( \sin A = \frac{a}{k} \) and \( \sin C = \frac{c}{k} \) from the sine formula:
LHS \( = \frac{\frac{a}{k}}{\frac{c}{k}} \)
LHS \( = \frac{a}{c} \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. The sum of angles in any triangle is always 180 degrees or \( \pi \) radians, a fundamental property used here.
In simple words: We start with the sine rule, which connects the sides of a triangle to the sines of its opposite angles. Since `A`, `B`, and `C` are angles of a triangle, their sum is 180 degrees. This means `sin(A+B)` is the same as `sin(C)`. By using this and the sine rule, we can easily show that the left side of the equation equals the right side, `a/c`.

🎯 Exam Tip: Always remember the angle sum property of a triangle \( (A+B+C = \pi) \) as it's frequently used in proving trigonometric identities related to triangles.

Question 3. In any \( \triangle ABC \), Prove that \( \frac{a-b}{a+b}=\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)} \).
Answer:
We begin with the left-hand side (LHS) of the equation. Using the sine formula, we know that \( a = k \sin A \) and \( b = k \sin B \) for some constant \( k \).
LHS \( = \frac{a-b}{a+b} \)
Substitute \( a \) and \( b \) in terms of \( k \) and sines:
LHS \( = \frac{k \sin A - k \sin B}{k \sin A + k \sin B} \)
Factor out \( k \) from the numerator and denominator and cancel it:
LHS \( = \frac{\sin A - \sin B}{\sin A + \sin B} \)
Now, apply the sum-to-product trigonometric formulas:
\( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \)
\( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \)
Substitute these into the LHS:
LHS \( = \frac{2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}{2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)} \)
Cancel the common factor of 2 and rearrange the terms:
LHS \( = \left(\frac{\sin\left(\frac{A-B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}\right) \times \left(\frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)}\right) \)
Recognize that \( \frac{\sin x}{\cos x} = \tan x \) and \( \frac{\cos x}{\sin x} = \cot x \):
LHS \( = \tan\left(\frac{A-B}{2}\right) \cot\left(\frac{A+B}{2}\right) \)
Since \( \cot x = \frac{1}{\tan x} \):
LHS \( = \frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)} \)
This matches the right-hand side (RHS) of the given equation. So, LHS = RHS, and the identity is proven. This identity, known as the Law of Tangents, provides another relationship between the sides and angles of a triangle, which is very useful in trigonometry.
In simple words: We use the sine rule to replace sides 'a' and 'b' with `sin A` and `sin B`. Then, we apply sum-to-product trigonometric formulas to simplify the expression. After canceling common terms and rewriting `cot` as `1/tan`, we arrive at the right side of the equation.

🎯 Exam Tip: \( \frac{a-b}{a+b}=\frac{\tan \frac{A-B}{2}}{\tan \frac{A+B}{2}} \) is a direct formula (Law of Tangents) for triangles; be familiar with its application.

Question 4. In any \( \triangle ABC \), Prove that \( ac \cos B - bc \cos A = a^2 - b^2 \).
Answer:
We start with the left-hand side (LHS) of the equation: \( ac \cos B - bc \cos A \).
Using the cosine formula, we know that:
\( \cos B = \frac{a^2+c^2-b^2}{2ac} \)
\( \cos A = \frac{b^2+c^2-a^2}{2bc} \)
Substitute these expressions for \( \cos B \) and \( \cos A \) into the LHS:
LHS \( = ac \left(\frac{a^2+c^2-b^2}{2ac}\right) - bc \left(\frac{b^2+c^2-a^2}{2bc}\right) \)
Cancel out the common terms \( ac \) and \( bc \) in the numerators and denominators:
LHS \( = \frac{a^2+c^2-b^2}{2} - \frac{b^2+c^2-a^2}{2} \)
Since both terms have a common denominator of 2, we can combine them:
LHS \( = \frac{(a^2+c^2-b^2) - (b^2+c^2-a^2)}{2} \)
Distribute the negative sign in the second part:
LHS \( = \frac{a^2+c^2-b^2 - b^2-c^2+a^2}{2} \)
Combine like terms (note that \( c^2 \) and \( -c^2 \) cancel out):
LHS \( = \frac{2a^2 - 2b^2}{2} \)
Factor out 2 from the numerator and cancel it with the denominator:
LHS \( = a^2 - b^2 \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. This equation is an example of a projection formula in trigonometry, showing how side lengths relate to each other through the cosine of angles.
In simple words: We replace `cos B` and `cos A` in the left side of the equation with their cosine rule formulas, which involve the side lengths. After substitution, we simplify the fractions and combine the terms. This calculation will directly lead to `a^2 - b^2`, proving the equality.

🎯 Exam Tip: When proving trigonometric identities in triangles, the cosine rule is often the key to transforming angle terms into side terms.

Question 5. In any \( \triangle ABC \), Prove that \( \frac{\sin (A-B)}{\sin (A+B)}=\frac{a^2-b^2}{c^2} \).
Answer:
We will start by simplifying the right-hand side (RHS) of the equation. Using the sine formula, we know that \( a = k \sin A \), \( b = k \sin B \), and \( c = k \sin C \) for some constant \( k \).
RHS \( = \frac{a^2-b^2}{c^2} \)
Substitute \( a, b, c \) in terms of \( k \) and sines:
RHS \( = \frac{(k \sin A)^2 - (k \sin B)^2}{(k \sin C)^2} \)
RHS \( = \frac{k^2 \sin^2 A - k^2 \sin^2 B}{k^2 \sin^2 C} \)
Factor out \( k^2 \) from the numerator and denominator and cancel it:
RHS \( = \frac{\sin^2 A - \sin^2 B}{\sin^2 C} \)
Now, use the trigonometric identity: \( \sin^2 X - \sin^2 Y = \sin(X+Y)\sin(X-Y) \).
RHS \( = \frac{\sin(A+B)\sin(A-B)}{\sin^2 C} \)
For a triangle \( ABC \), the sum of angles is \( A+B+C = \pi \). So, \( A+B = \pi - C \).
Therefore, \( \sin (A+B) = \sin (\pi - C) = \sin C \).
Substitute \( \sin(A+B) = \sin C \) into the expression for RHS:
RHS \( = \frac{\sin C \sin(A-B)}{\sin^2 C} \)
Cancel one \( \sin C \) term from the numerator and denominator:
RHS \( = \frac{\sin(A-B)}{\sin C} \)
Since \( \sin(A+B) = \sin C \), we can substitute \( \sin C \) with \( \sin(A+B) \):
RHS \( = \frac{\sin(A-B)}{\sin(A+B)} \)
This matches the left-hand side (LHS) of the given equation. So, LHS = RHS, and the identity is proven. The identity \( \sin^2 X - \sin^2 Y = \sin(X+Y)\sin(X-Y) \) is a powerful tool for simplifying trigonometric expressions involving squares of sines.
In simple words: We use the sine rule to express the sides 'a', 'b', and 'c' in terms of sines of angles. We then substitute these into the right side of the equation and simplify using a known trigonometric identity `(\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B))`. Because `A+B+C = 180^\circ`, `sin(A+B)` is the same as `sin C`. This helps us show that the right side matches the left side.

🎯 Exam Tip: Always remember the identity \( \sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B) \) as it simplifies many problems related to sines of angles in triangles.

Question 6. In any \( \triangle ABC \), Prove that \( a (\sin B - \sin C) + b (\sin C – \sin A) + c (\sin A - \sin B) = 0 \).
Answer:
We start with the left-hand side (LHS) of the equation: \( a (\sin B - \sin C) + b (\sin C – \sin A) + c (\sin A - \sin B) \).
Using the sine formula, we know that \( a = k \sin A \), \( b = k \sin B \), and \( c = k \sin C \) for some constant \( k \).
Substitute these expressions for \( a, b, c \) into the LHS:
LHS \( = k \sin A (\sin B - \sin C) + k \sin B (\sin C – \sin A) + k \sin C (\sin A - \sin B) \)
Factor out the common term \( k \):
LHS \( = k [\sin A (\sin B - \sin C) + \sin B (\sin C – \sin A) + \sin C (\sin A - \sin B)] \)
Expand each term inside the square brackets:
LHS \( = k [\sin A \sin B - \sin A \sin C + \sin B \sin C - \sin B \sin A + \sin C \sin A - \sin C \sin B] \)
Now, observe the terms. Many terms cancel each other out:
\( \sin A \sin B \) cancels with \( - \sin B \sin A \).
\( - \sin A \sin C \) cancels with \( + \sin C \sin A \).
\( + \sin B \sin C \) cancels with \( - \sin C \sin B \).
So, all terms inside the square brackets sum to zero:
LHS \( = k [0] \)
LHS \( = 0 \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. This identity shows a balanced relationship between the sides and sines of the angles in a cyclic sum, often indicating a property of triangles.
In simple words: We use the sine rule to replace the sides 'a', 'b', and 'c' with terms involving `sin A`, `sin B`, and `sin C`. When we expand the expression, we notice that each term `k \sin X \sin Y` has a positive and a negative counterpart. They all cancel each other out, resulting in zero.

🎯 Exam Tip: When proving identities involving cyclic sums (like \( a(x-y) + b(y-z) + c(z-x) \)), substituting using the sine rule \( a=k \sin A \) often leads to straightforward cancellation.

Question 7. In any \( \triangle ABC \), Prove that \( a \cos (A + B + C) - b \cos (B + A) - c \cos (A + C) = 0 \).
Answer:
We start with the left-hand side (LHS) of the equation: \( a \cos (A + B + C) - b \cos (B + A) - c \cos (A + C) \).
For any triangle \( ABC \), the sum of its internal angles is \( A + B + C = \pi \) radians (or \( 180^\circ \)).
Using this property, we can make the following substitutions:
\( A + B + C = \pi \)
\( B + A = \pi - C \)
\( A + C = \pi - B \)
Now, substitute these into the LHS expression:
LHS \( = a \cos(\pi) - b \cos(\pi - C) - c \cos(\pi - B) \)
We know the following trigonometric identities:
\( \cos(\pi) = -1 \)
\( \cos(\pi - x) = -\cos x \)
Apply these identities:
LHS \( = a(-1) - b(-\cos C) - c(-\cos B) \)
LHS \( = -a + b \cos C + c \cos B \)
Now, recall the projection formula for a triangle, which states: \( a = b \cos C + c \cos B \).
Substitute this into the LHS:
LHS \( = -a + a \)
LHS \( = 0 \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. The projection formulas are fundamental relationships that connect a side of a triangle to the other two sides and their respective cosines.
In simple words: We know that the sum of angles in a triangle is 180 degrees `(\pi)`. Using this, we replace `(A+B+C)` with `\pi`, `(B+A)` with `(\pi-C)`, and `(A+C)` with `(\pi-B)`. Then, we use the properties of cosine (`cos(\pi) = -1`, `cos(\pi-X) = -cos X`) to simplify the expression. Finally, we use the projection formula `a = b cos C + c cos B` to show the entire expression equals zero.

🎯 Exam Tip: \( a = b \cos C + c \cos B \) (and its cyclic permutations) is a standard projection formula; recognizing it can simplify proofs in triangle trigonometry.

Question 8. In any \( \triangle ABC \), Prove that \( a (\cos C - \cos B) = 2 (b - c) \cos^2 \frac { A }{ 2 } \).
Answer:
We will prove this identity by simplifying both the left-hand side (LHS) and the right-hand side (RHS) to a common expression.
First, consider the LHS: \( a (\cos C - \cos B) \).
Using the sine formula, \( a = k \sin A \).
LHS \( = k \sin A (\cos C - \cos B) \).
Apply sum-to-product formula: \( \cos C - \cos B = -2 \sin\left(\frac{C+B}{2}\right)\sin\left(\frac{C-B}{2}\right) \).
Since \( A+B+C = \pi \), we have \( \frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2} \).
So, \( \sin\left(\frac{B+C}{2}\right) = \cos\left(\frac{A}{2}\right) \).
Also, \( \sin\left(\frac{C-B}{2}\right) = -\sin\left(\frac{B-C}{2}\right) \).
LHS \( = k \sin A \left(-2 \cos\left(\frac{A}{2}\right) \left(-\sin\left(\frac{B-C}{2}\right)\right)\right) \)
LHS \( = k \sin A \left(2 \cos\left(\frac{A}{2}\right) \sin\left(\frac{B-C}{2}\right)\right) \)
Now use \( \sin A = 2 \sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) \):
LHS \( = k \left(2 \sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\right) \left(2 \cos\left(\frac{A}{2}\right) \sin\left(\frac{B-C}{2}\right)\right) \)
LHS \( = 4k \sin\left(\frac{A}{2}\right)\cos^2\left(\frac{A}{2}\right)\sin\left(\frac{B-C}{2}\right) \).
Now, consider the RHS: \( 2 (b - c) \cos^2 \frac{A}{2} \).
Using the sine formula, \( b = k \sin B \) and \( c = k \sin C \).
RHS \( = 2 (k \sin B - k \sin C) \cos^2 \frac{A}{2} \)
RHS \( = 2k (\sin B - \sin C) \cos^2 \frac{A}{2} \).
Apply sum-to-product formula: \( \sin B - \sin C = 2 \cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right) \).
Again, \( \cos\left(\frac{B+C}{2}\right) = \sin\left(\frac{A}{2}\right) \).
RHS \( = 2k \left(2 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B-C}{2}\right)\right) \cos^2 \frac{A}{2} \)
RHS \( = 4k \sin\left(\frac{A}{2}\right)\sin\left(\frac{B-C}{2}\right)\cos^2\left(\frac{A}{2}\right) \).
Since LHS = RHS, the identity is proven. This identity relates the difference of cosines of two angles to the difference of their opposite sides, scaled by the cosine squared of the third angle's half-value.
In simple words: We prove this by expressing both sides of the equation in terms of `k` (from the sine rule) and half-angles. We use the sum-to-product formulas for sines and cosines, and the property that angles in a triangle sum to 180 degrees. After substituting and simplifying, both the left side and the right side of the equation become equal.

🎯 Exam Tip: Identities involving \( \cos^2(A/2) \) often require converting full angles to half angles using identities like \( \sin A = 2\sin(A/2)\cos(A/2) \) and \( \cos(B+C)/2 = \sin(A/2) \).

Question 9. In any \( \triangle ABC \), Prove that \( a \sin \frac { 1 }{ 2 } (B – C) = (b - c) \cos \frac { 1 }{ 2 }A \).
Answer:
We start by simplifying the right-hand side (RHS) of the equation: \( (b - c) \cos \frac{A}{2} \).
Using the sine formula, we know that \( b = k \sin B \) and \( c = k \sin C \) for some constant \( k \).
RHS \( = (k \sin B - k \sin C) \cos \frac{A}{2} \)
RHS \( = k (\sin B - \sin C) \cos \frac{A}{2} \)
Apply the sum-to-product trigonometric formula: \( \sin B - \sin C = 2 \cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right) \).
For a triangle \( ABC \), the sum of angles is \( A+B+C = \pi \). So, \( \frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2} \).
Therefore, \( \cos\left(\frac{B+C}{2}\right) = \sin\left(\frac{A}{2}\right) \).
Substitute these into the RHS expression:
RHS \( = k \left(2 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B-C}{2}\right)\right) \cos \frac{A}{2} \)
Rearrange the terms:
RHS \( = k \sin\left(\frac{B-C}{2}\right) \left(2 \sin\frac{A}{2}\cos\frac{A}{2}\right) \)
Recall the double angle identity for sine: \( 2 \sin x \cos x = \sin 2x \). Here, \( 2 \sin\frac{A}{2}\cos\frac{A}{2} = \sin A \).
RHS \( = k \sin\left(\frac{B-C}{2}\right) \sin A \)
From the sine formula, we know \( k \sin A = a \).
RHS \( = a \sin\left(\frac{B-C}{2}\right) \)
This matches the left-hand side (LHS) of the given equation. So, LHS = RHS, and the identity is proven. This specific identity is useful in solving triangles when information about differences in angles and sides is given.
In simple words: We start by expressing sides 'b' and 'c' using the sine rule. Then, we use a trigonometric sum-to-product formula for `sin B - sin C`. We also use the angle sum property of a triangle to relate `(B+C)/2` to `A/2`. After simplifying and recognizing that `k \sin A` is equal to side 'a', we show that the right side of the equation is identical to the left side.

🎯 Exam Tip: Mastering sum-to-product and product-to-sum formulas, along with half-angle relations, is crucial for proving triangle identities.

Question 10. In any \( \triangle ABC \), Prove that \( a \sin (\frac { A }{ 2 } + B) = (b + c) \sin \frac { A }{ 2 } \).
Answer:
We will simplify both the left-hand side (LHS) and the right-hand side (RHS) to a common form.
First, consider the RHS: \( (b + c) \sin \frac{A}{2} \).
Using the sine formula, \( b = k \sin B \) and \( c = k \sin C \).
RHS \( = (k \sin B + k \sin C) \sin \frac{A}{2} \)
RHS \( = k (\sin B + \sin C) \sin \frac{A}{2} \)
Apply the sum-to-product trigonometric formula: \( \sin B + \sin C = 2 \sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right) \).
For a triangle \( ABC \), the sum of angles is \( A+B+C = \pi \). So, \( \frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2} \).
Therefore, \( \sin\left(\frac{B+C}{2}\right) = \cos\left(\frac{A}{2}\right) \).
Substitute these into the RHS expression:
RHS \( = k \left(2 \cos\left(\frac{A}{2}\right)\cos\left(\frac{B-C}{2}\right)\right) \sin \frac{A}{2} \)
Rearrange the terms:
RHS \( = k \left(2 \sin\frac{A}{2}\cos\frac{A}{2}\right) \cos\left(\frac{B-C}{2}\right) \)
Recall the double angle identity for sine: \( 2 \sin x \cos x = \sin 2x \). Here, \( 2 \sin\frac{A}{2}\cos\frac{A}{2} = \sin A \).
RHS \( = k \sin A \cos\left(\frac{B-C}{2}\right) \)
From the sine formula, \( k \sin A = a \).
So, RHS \( = a \cos\left(\frac{B-C}{2}\right) \).
Now, let's simplify the LHS: \( a \sin (\frac { A }{ 2 } + B) \). We need to show this is equal to \( a \cos\left(\frac{B-C}{2}\right) \). This means we need to show \( \sin (\frac { A }{ 2 } + B) = \cos\left(\frac{B-C}{2}\right) \).
From \( A+B+C = \pi \), we have \( A = \pi - (B+C) \). So \( \frac{A}{2} = \frac{\pi}{2} - \frac{B+C}{2} \).
Substitute this into \( \frac{A}{2} + B \):
\( \frac{A}{2} + B = \left(\frac{\pi}{2} - \frac{B+C}{2}\right) + B \)
\( = \frac{\pi}{2} - \frac{B}{2} - \frac{C}{2} + B \)
\( = \frac{\pi}{2} + \frac{B}{2} - \frac{C}{2} \)
\( = \frac{\pi}{2} + \frac{B-C}{2} \)
Now, take the sine of this expression:
\( \sin\left(\frac{A}{2} + B\right) = \sin\left(\frac{\pi}{2} + \frac{B-C}{2}\right) \)
Using the identity \( \sin(\frac{\pi}{2} + x) = \cos x \):
\( \sin\left(\frac{A}{2} + B\right) = \cos\left(\frac{B-C}{2}\right) \)
Thus, LHS \( = a \cos\left(\frac{B-C}{2}\right) \).
Since LHS = RHS, the identity is proven. This identity elegantly combines angle sum properties with half-angle and sum-to-product formulas, showcasing the interconnectedness of triangle trigonometry.
In simple words: We start with the right side, replacing 'b' and 'c' using the sine rule. We then use the sum-to-product formula for `sin B + sin C` and the angle sum property of a triangle to simplify. This gives us `a \cos((B-C)/2)`. Now, for the left side, we expand `sin(A/2 + B)` and use `A+B = \pi - C` to transform the argument. We show that `sin(A/2 + B)` simplifies to `cos((B-C)/2)`, proving that both sides are equal.

🎯 Exam Tip: When dealing with \( A/2 + B \) or similar angle combinations, always consider using \( A+B+C = \pi \) to simplify the expression into a more manageable form involving half angles.

Question 11. In any \( \triangle ABC \), Prove that \( c^2 = (a - b)^2 \cos^2 \frac { C }{ 2 } + (a + b)^2 \sin^2 \frac { C }{ 2 } \).
Answer:
We start with the right-hand side (RHS) of the equation: \( (a - b)^2 \cos^2 \frac{C}{2} + (a + b)^2 \sin^2 \frac{C}{2} \).
Expand the squared terms:
RHS \( = (a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \frac{C}{2} \)
Distribute \( \cos^2 \frac{C}{2} \) and \( \sin^2 \frac{C}{2} \):
RHS \( = a^2 \cos^2 \frac{C}{2} - 2ab \cos^2 \frac{C}{2} + b^2 \cos^2 \frac{C}{2} + a^2 \sin^2 \frac{C}{2} + 2ab \sin^2 \frac{C}{2} + b^2 \sin^2 \frac{C}{2} \)
Group terms with \( a^2 \), \( b^2 \), and \( 2ab \):
RHS \( = a^2 (\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) + b^2 (\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab (\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}) \)
Using the fundamental trigonometric identity \( \cos^2 x + \sin^2 x = 1 \):
RHS \( = a^2 (1) + b^2 (1) - 2ab (\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}) \)
Now, recall the double angle identity for cosine: \( \cos C = \cos^2 \frac{C}{2} - \sin^2 \frac{C}{2} \).
Substitute this into the expression:
RHS \( = a^2 + b^2 - 2ab \cos C \)
By the cosine rule, we know that \( c^2 = a^2 + b^2 - 2ab \cos C \).
So, RHS \( = c^2 \).
This matches the left-hand side (LHS) of the given equation. So, LHS = RHS, and the identity is proven. This identity is derived directly from the cosine rule, demonstrating the versatility of half-angle formulas in expressing fundamental geometric properties.
In simple words: We start by expanding the right side of the equation. We group terms involving `(a^2+b^2)` and `2ab`. We then use the identity `cos^2(X) + sin^2(X) = 1` and the half-angle identity for `cos C = cos^2(C/2) - sin^2(C/2)`. After these substitutions, the expression simplifies to `a^2 + b^2 - 2ab cos C`, which is exactly the cosine rule for `c^2`.

🎯 Exam Tip: \( \cos C = \cos^2(C/2) - \sin^2(C/2) \) (or \( 2\cos^2(C/2) - 1 \) or \( 1 - 2\sin^2(C/2) \)) is a key identity for problems involving half-angles and the cosine rule.

Question 12. In any \( \triangle ABC \), Prove that \( a \sin (B - C) + b \sin (C - A) + c \sin (A - B) = 0 \).
Answer:
We start with the left-hand side (LHS) of the equation: \( a \sin (B - C) + b \sin (C - A) + c \sin (A - B) \).
Using the sine formula, we know that \( a = k \sin A \), \( b = k \sin B \), and \( c = k \sin C \) for some constant \( k \).
Substitute these expressions for \( a, b, c \) into the LHS:
LHS \( = k \sin A \sin (B - C) + k \sin B \sin (C - A) + k \sin C \sin (A - B) \)
Factor out the common term \( k \):
LHS \( = k [\sin A \sin (B - C) + \sin B \sin (C - A) + \sin C \sin (A - B)] \)
For a triangle \( ABC \), the sum of angles is \( A+B+C = \pi \). This implies:
\( A = \pi - (B+C) \implies \sin A = \sin (\pi - (B+C)) = \sin (B+C) \)
\( B = \pi - (A+C) \implies \sin B = \sin (\pi - (A+C)) = \sin (A+C) \)
\( C = \pi - (A+B) \implies \sin C = \sin (\pi - (A+B)) = \sin (A+B) \)
Substitute these into the expression for LHS:
LHS \( = k [\sin (B+C)\sin (B-C) + \sin (A+C)\sin (C-A) + \sin (A+B)\sin (A-B)] \)
Now, use the trigonometric identity: \( \sin(X+Y)\sin(X-Y) = \sin^2 X - \sin^2 Y \).
Apply this identity to each term:
\( \sin (B+C)\sin (B-C) = \sin^2 B - \sin^2 C \)
\( \sin (A+C)\sin (C-A) = \sin^2 C - \sin^2 A \) (since \( \sin(C-A) = -\sin(A-C) \) and \( \sin(A+C) = \sin(C+A) \))
\( \sin (A+B)\sin (A-B) = \sin^2 A - \sin^2 B \)
Substitute these back into the LHS:
LHS \( = k [(\sin^2 B - \sin^2 C) + (\sin^2 C - \sin^2 A) + (\sin^2 A - \sin^2 B)] \)
Observe that all terms inside the square brackets cancel each other out:
LHS \( = k [0] \)
LHS \( = 0 \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. This identity beautifully demonstrates the cyclical symmetry often found in relations involving the angles and sides of a triangle.
In simple words: We use the sine rule to replace the sides 'a', 'b', and 'c' with `k \sin A`, `k \sin B`, and `k \sin C`. Then, we use the property that `\sin A = \sin(B+C)` (because `A+B+C = 180^\circ`). We apply the trigonometric identity `\sin(X+Y)\sin(X-Y) = \sin^2 X - \sin^2 Y`. After doing this for each term, we find that all terms cancel each other out, proving the sum is zero.

🎯 Exam Tip: Identities of the form \( \sin(X+Y)\sin(X-Y) = \sin^2 X - \sin^2 Y \) are very effective for simplifying products of sines, especially when dealing with sums and differences of angles.

Question 13. In any \( \triangle ABC \), Prove that \( \frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2} = \frac{1}{a^2}-\frac{1}{b^2} \).
Answer:
We start with the left-hand side (LHS) of the equation: \( \frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2} \).
Using the double angle identity for cosine, \( \cos 2x = 1 - 2 \sin^2 x \).
Apply this to \( \cos 2A \) and \( \cos 2B \):
LHS \( = \frac{1 - 2 \sin^2 A}{a^2} - \frac{1 - 2 \sin^2 B}{b^2} \)
Now, use the sine formula, which states that \( \frac{\sin A}{a} = \frac{\sin B}{b} = k \) (a constant).
From this, we have \( \sin A = ak \) and \( \sin B = bk \).
Substitute these expressions for \( \sin A \) and \( \sin B \) into the LHS:
LHS \( = \frac{1 - 2 (ak)^2}{a^2} - \frac{1 - 2 (bk)^2}{b^2} \)
LHS \( = \frac{1 - 2 a^2 k^2}{a^2} - \frac{1 - 2 b^2 k^2}{b^2} \)
Separate the terms in each fraction:
LHS \( = \left(\frac{1}{a^2} - \frac{2 a^2 k^2}{a^2}\right) - \left(\frac{1}{b^2} - \frac{2 b^2 k^2}{b^2}\right) \)
Cancel out the \( a^2 \) and \( b^2 \) terms:
LHS \( = \frac{1}{a^2} - 2k^2 - \frac{1}{b^2} + 2k^2 \)
The \( -2k^2 \) and \( +2k^2 \) terms cancel each other out:
LHS \( = \frac{1}{a^2} - \frac{1}{b^2} \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. This identity shows how double angle cosine functions and the sine rule combine to yield a simple algebraic difference.
In simple words: We start by replacing `cos 2A` and `cos 2B` with their equivalent forms `(1 - 2 sin^2 A)` and `(1 - 2 sin^2 B)`. Then, using the sine rule, we substitute `sin A` with `ak` and `sin B` with `bk`. After simplifying the fractions, the `k^2` terms cancel out, leaving us with `1/a^2 - 1/b^2`, which is the right side of the equation.

🎯 Exam Tip: When working with \( \cos 2X \), remember the three forms \( (\cos^2 X - \sin^2 X, 2 \cos^2 X - 1, 1 - 2 \sin^2 X) \); choose the one that best suits the problem (here \( 1 - 2 \sin^2 X \) is best).

Question 14. In any \( \triangle ABC \), Prove that \( \frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^2+b^2}{a^2+c^2} \).
Answer:
We start with the left-hand side (LHS) of the equation: \( \frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B} \).
For a triangle \( ABC \), the sum of angles is \( A+B+C = \pi \). This implies:
\( C = \pi - (A+B) \implies \cos C = \cos (\pi - (A+B)) = -\cos (A+B) \)
\( B = \pi - (A+C) \implies \cos B = \cos (\pi - (A+C)) = -\cos (A+C) \)
Substitute these into the LHS expression:
LHS \( = \frac{1+\cos (A-B) (-\cos(A+B))}{1+\cos (A-C) (-\cos(A+C))} \)
LHS \( = \frac{1-\cos (A-B) \cos(A+B)}{1-\cos (A-C) \cos(A+C)} \)
Now, use the trigonometric identity: \( 1 - \cos X \cos Y = \sin^2 X + \sin^2 Y - 1 \) if \( X=Y \). Or, more directly, using `\cos(X-Y)\cos(X+Y) = \cos^2 X - \sin^2 Y`.
So, \( 1 - (\cos^2 A - \sin^2 B) = 1 - \cos^2 A + \sin^2 B = (1 - \cos^2 A) + \sin^2 B = \sin^2 A + \sin^2 B \).
Apply this identity to the numerator:
Numerator \( = 1 - \cos (A-B) \cos(A+B) = 1 - (\cos^2 A - \sin^2 B) = \sin^2 A + \sin^2 B \).
Apply the same identity to the denominator:
Denominator \( = 1 - \cos (A-C) \cos(A+C) = 1 - (\cos^2 A - \sin^2 C) = \sin^2 A + \sin^2 C \).
So, the LHS becomes:
LHS \( = \frac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C} \)
Finally, use the sine formula: \( \sin A = ak \), \( \sin B = bk \), \( \sin C = ck \) for some constant \( k \).
LHS \( = \frac{(ak)^2 + (bk)^2}{(ak)^2 + (ck)^2} \)
LHS \( = \frac{k^2 a^2 + k^2 b^2}{k^2 a^2 + k^2 c^2} \)
Factor out \( k^2 \) from the numerator and denominator and cancel it:
LHS \( = \frac{k^2 (a^2 + b^2)}{k^2 (a^2 + c^2)} \)
LHS \( = \frac{a^2+b^2}{a^2+c^2} \)
This matches the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. This identity beautifully connects the product of cosines with the sum of squares of sines, and ultimately to the squares of the triangle's sides.
In simple words: We use the property that angles in a triangle add up to 180 degrees to rewrite `cos C` and `cos B` in terms of `cos(A+B)` and `cos(A+C)`. Then, we apply the trigonometric identity `1 - \cos(X-Y)\cos(X+Y) = \sin^2 X + \sin^2 Y`. This simplifies the numerator and denominator to sums of squares of sines. Finally, using the sine rule (`\sin X = kX`), we replace `sin^2` terms with `k^2` times side squares, canceling `k^2` to get the right side of the equation.

🎯 Exam Tip: The identity \( 1 - \cos(X-Y)\cos(X+Y) = \sin^2 X + \sin^2 Y \) is a very efficient shortcut for problems involving products of cosines of sums and differences of angles.

Question 15. In any \( \triangle ABC \), Prove that \( (b^2 - c^2) \cot A + (c^2 - a^2) \cot B + (a^2 - b^2) \cot C = 0 \).
Answer:
We start with the left-hand side (LHS) of the equation: \( (b^2 - c^2) \cot A + (c^2 - a^2) \cot B + (a^2 - b^2) \cot C \).
Using the sine formula, we know that \( a = k \sin A \), \( b = k \sin B \), \( c = k \sin C \) for some constant \( k \).
We also know that \( \cot A = \frac{\cos A}{\sin A} \).
Let's evaluate the first term \( (b^2 - c^2) \cot A \):
\( (b^2 - c^2) \cot A = (k^2 \sin^2 B - k^2 \sin^2 C) \frac{\cos A}{\sin A} \)
\( = k^2 (\sin^2 B - \sin^2 C) \frac{\cos A}{\sin A} \)
Using the identity \( \sin^2 X - \sin^2 Y = \sin(X-Y)\sin(X+Y) \):
\( = k^2 \sin(B-C)\sin(B+C) \frac{\cos A}{\sin A} \)
For a triangle \( ABC \), \( A+B+C = \pi \), so \( B+C = \pi - A \). Therefore, \( \sin(B+C) = \sin(\pi - A) = \sin A \).
Substitute \( \sin(B+C) = \sin A \) into the expression:
\( = k^2 \sin(B-C) \sin A \frac{\cos A}{\sin A} \)
Cancel \( \sin A \):
\( = k^2 \sin(B-C) \cos A \)
So, the first term is \( k^2 \sin(B-C) \cos A \). Similarly, we can find the other two terms:
\( (c^2 - a^2) \cot B = k^2 \sin(C-A) \cos B \)
\( (a^2 - b^2) \cot C = k^2 \sin(A-B) \cos C \)
Now, sum these three terms to get the LHS:
LHS \( = k^2 [\sin(B-C)\cos A + \sin(C-A)\cos B + \sin(A-B)\cos C] \)
Expand each product using the identity \( \sin(X-Y) = \sin X \cos Y - \cos X \sin Y \):
\( \sin(B-C)\cos A = (\sin B \cos C - \cos B \sin C) \cos A = \sin B \cos C \cos A - \cos B \sin C \cos A \)
\( \sin(C-A)\cos B = (\sin C \cos A - \cos C \sin A) \cos B = \sin C \cos A \cos B - \cos C \sin A \cos B \)
\( \sin(A-B)\cos C = (\sin A \cos B - \cos A \sin B) \cos C = \sin A \cos B \cos C - \cos A \sin B \cos C \)
Add these expanded terms together:
LHS \( = k^2 [\sin B \cos C \cos A - \cos B \sin C \cos A + \sin C \cos A \cos B - \cos C \sin A \cos B + \sin A \ cos B \cos C - \cos A \sin B \cos C] \)
Observe that all terms cancel each other out (e.g., \( \sin B \cos C \cos A \) cancels with \( - \cos A \sin B \cos C \)).
LHS \( = k^2 [0] \)
LHS \( = 0 \)
This is the right-hand side (RHS) of the equation. So, LHS = RHS, and the identity is proven. This identity, often called Euler's Theorem for cotangents, elegantly demonstrates a fundamental cyclical relationship within triangle trigonometry.
In simple words: We first rewrite `cot A` as `cos A / sin A` and use the sine rule to express `b^2 - c^2` in terms of `sin^2 B - sin^2 C`. We apply the identity `sin^2 X - sin^2 Y = sin(X-Y)sin(X+Y)` and then use `sin(B+C) = sin A`. This simplifies each term to a form like `k^2 \sin(B-C) \cos A`. When all three such terms are added together, and we expand them, all parts cancel out, proving the sum is zero.

🎯 Exam Tip: For identities involving \( \cot \) and \( b^2 - c^2 \), it's often effective to convert \( b^2 - c^2 \) using the sine rule and \( \cot \) using \( \cos/\sin \), then simplify with sum-product and angle sum identities.

Question 16. \( a^3 \sin (B – C) \text{ cosec}^2 A + b^3 \sin (C – A) \text{ cosec}^2 B + c^3 \sin (A – B) \text{ cosec}^2 C = 0 \)
Answer: To prove this, we first rewrite the cosec² terms as \( 1/\sin^2 \).
So, L.H.S = \( \frac{a^3}{\sin^2 A} \sin(B-C) + \frac{b^3}{\sin^2 B} \sin(C-A) + \frac{c^3}{\sin^2 C} \sin(A-B) \).
Now, we use the sine rule which states that \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \) (where k is a constant).
This means \( a = k \sin A, b = k \sin B, c = k \sin C \). We substitute these into the expression:
L.H.S = \( \frac{(k \sin A)^3}{\sin^2 A} \sin(B-C) + \frac{(k \sin B)^3}{\sin^2 B} \sin(C-A) + \frac{(k \sin C)^3}{\sin^2 C} \sin(A-B) \)
L.H.S = \( k^3 \sin A \sin(B-C) + k^3 \sin B \sin(C-A) + k^3 \sin C \sin(A-B) \)
We know that for any triangle, \( A+B+C = \pi \) (or 180 degrees).
So, \( \sin A = \sin(\pi - (B+C)) = \sin(B+C) \). Similarly, \( \sin B = \sin(A+C) \) and \( \sin C = \sin(A+B) \).
Substitute these back:
L.H.S = \( k^3 [\sin(B+C)\sin(B-C) + \sin(A+C)\sin(C-A) + \sin(A+B)\sin(A-B)] \)
Now, we use the trigonometric identity: \( \sin(X+Y)\sin(X-Y) = \sin^2 X - \sin^2 Y \).
L.H.S = \( k^3 [(\sin^2 B - \sin^2 C) + (\sin^2 C - \sin^2 A) + (\sin^2 A - \sin^2 B)] \)
When we open the brackets, all terms cancel each other out:
L.H.S = \( k^3 [0] = 0 \).
Since L.H.S = 0, and the R.H.S is also 0, the equation is proven.
In simple words: We changed the terms using sine rule and angle properties. Then, we used a special math rule that helps combine sine functions. All the parts of the equation canceled each other out, showing that the total is zero.

🎯 Exam Tip: Remember to use the sine rule \( a = k \sin A \) and the identity \( \sin(X+Y)\sin(X-Y) = \sin^2 X - \sin^2 Y \) when simplifying such expressions. This method often leads to terms cancelling out.

Question 17. \( a^3 \cos (B – C) + b^3 \cos (C – A) + c^3 \cos (A – B) = 3abc \)
Answer: We begin by expressing \( a, b, c \) using the sine rule: \( a = k \sin A, b = k \sin B, c = k \sin C \).
Substitute these into the L.H.S:
L.H.S. = \( (k \sin A)^3 \cos(B-C) + (k \sin B)^3 \cos(C-A) + (k \sin C)^3 \cos(A-B) \)
L.H.S. = \( k^3 [\sin^3 A \cos(B-C) + \sin^3 B \cos(C-A) + \sin^3 C \cos(A-B)] \)
This can be rewritten as:
L.H.S. = \( k a^2 (\sin A \cos(B-C)) + k b^2 (\sin B \cos(C-A)) + k c^2 (\sin C \cos(A-B)) \)
We use the identity \( \sin(X+Y)\cos(X-Y) = \frac{1}{2}(\sin 2X + \sin 2Y) \). Here, \( \sin A = \sin(B+C) \).
So, \( \sin A \cos(B-C) = \sin(B+C)\cos(B-C) = \frac{1}{2}(\sin 2B + \sin 2C) \).
Applying this to each term:
L.H.S. = \( k a^2 \cdot \frac{1}{2}(\sin 2B + \sin 2C) + k b^2 \cdot \frac{1}{2}(\sin 2C + \sin 2A) + k c^2 \cdot \frac{1}{2}(\sin 2A + \sin 2B) \)
Now, use the double angle formula \( \sin 2X = 2 \sin X \cos X \):
L.H.S. = \( \frac{k a^2}{2}(2 \sin B \cos B + 2 \sin C \cos C) + \frac{k b^2}{2}(2 \sin C \cos C + 2 \sin A \cos A) + \frac{k c^2}{2}(2 \sin A \cos A + 2 \sin B \cos B) \)
L.H.S. = \( k a^2 (\sin B \cos B + \sin C \cos C) + k b^2 (\sin C \cos C + \sin A \cos A) + k c^2 (\sin A \cos A + \sin B \cos B) \)
Substitute \( \sin A = a/k, \sin B = b/k, \sin C = c/k \) back:
L.H.S. = \( k a^2 (\frac{b}{k} \cos B + \frac{c}{k} \cos C) + k b^2 (\frac{c}{k} \cos C + \frac{a}{k} \cos A) + k c^2 (\frac{a}{k} \cos A + \frac{b}{k} \cos B) \)
L.H.S. = \( a^2 (b \cos B + c \cos C) + b^2 (c \cos C + a \cos A) + c^2 (a \cos A + b \cos B) \)
This expression can be rearranged to make use of the projection formulae:
Projection formulae: \( a = b \cos C + c \cos B \), \( b = c \cos A + a \cos C \), \( c = a \cos B + b \cos A \).
Rearrange terms in the L.H.S to match these:
L.H.S. = \( ab(a \cos B + b \cos A) + bc(b \cos C + c \cos B) + ac(a \cos C + c \cos A) \)
Now, apply the projection formulae:
L.H.S. = \( ab(c) + bc(a) + ac(b) \)
L.H.S. = \( abc + abc + abc = 3abc \).
Thus, L.H.S. = R.H.S., and the identity is proven. This shows a powerful connection between the sides and angles of a triangle.
In simple words: We used the sine rule to replace the sides with angle sines. Then, we used angle sum properties and double angle rules to simplify. Finally, we rearranged the terms to match the special "projection formulas" which directly give us the sides again, leading to the answer 3abc.

🎯 Exam Tip: Questions involving \( a^n \) or \( \sin^n A \) often require the sine rule to introduce \( k \) and reduce the power, followed by strategic use of angle sum identities and projection formulae for simplification.

Question 18.
(i) In a \( \triangle \text{ABC} \), if \( \frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{b c} + \frac{b}{c a} \), prove that \( \angle A = 90^\circ \).
(ii) In a \( \triangle \text{ABC} \), AD is the altitude from A. Given \( b > c \), \( \angle C = 23^\circ \) and AD = \( \frac{a b c}{b^2-c^2} \), find \( \angle B \).
Answer:
(i) Given: \( \frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{b c} + \frac{b}{c a} \).
To simplify, we multiply the entire equation by \( abc \):
\( 2bc \cos A + ac \cos B + 2ab \cos C = a^2 + b^2 \).
Now, we use the cosine rule for each angle: \( \cos A = \frac{b^2+c^2-a^2}{2bc} \), \( \cos B = \frac{a^2+c^2-b^2}{2ac} \), \( \cos C = \frac{a^2+b^2-c^2}{2ab} \).
Substitute these into the equation:
\( 2bc \left( \frac{b^2+c^2-a^2}{2bc} \right) + ac \left( \frac{a^2+c^2-b^2}{2ac} \right) + 2ab \left( \frac{a^2+b^2-c^2}{2ab} \right) = a^2+b^2 \)
This simplifies to:
\( (b^2+c^2-a^2) + \frac{1}{2}(a^2+c^2-b^2) + (a^2+b^2-c^2) = a^2+b^2 \)
Combine similar terms on the L.H.S:
\( b^2+c^2-a^2 + \frac{a^2}{2} + \frac{c^2}{2} - \frac{b^2}{2} + a^2+b^2-c^2 = a^2+b^2 \)
\( (1+\frac{1}{2}-1)a^2 + (1-\frac{1}{2}+1)b^2 + (1+\frac{1}{2}-1)c^2 = a^2+b^2 \)
\( \frac{1}{2}a^2 + \frac{3}{2}b^2 + \frac{1}{2}c^2 = a^2+b^2 \)
Multiply by 2:
\( a^2 + 3b^2 + c^2 = 2a^2 + 2b^2 \)
Rearrange the terms to one side:
\( 0 = (2a^2-a^2) + (2b^2-3b^2) - c^2 \)
\( 0 = a^2 - b^2 - c^2 \)
This means \( a^2 = b^2 + c^2 \).
By the converse of the Pythagorean theorem, if the square of one side is equal to the sum of the squares of the other two sides, the angle opposite the first side is \( 90^\circ \). In this case, angle A is opposite side a. So, \( \angle A = 90^\circ \).
(ii) Let AD be the altitude from A to BC. In the right-angled triangle \( \triangle \text{ADC} \), we have \( \sin C = \frac{\text{AD}}{\text{AC}} \).
Since AC = b, we have \( \text{AD} = b \sin C \).
We are given that \( \text{AD} = \frac{a b c}{b^2-c^2} \).
Equating the two expressions for AD:
\( b \sin C = \frac{a b c}{b^2-c^2} \)
Divide both sides by b (since \( b \neq 0 \)):
\( \sin C = \frac{a c}{b^2-c^2} \)
Now, we use the sine rule, which states \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \). From this, \( a = k \sin A \) and \( c = k \sin C \).
Substitute these into the equation for \( \sin C \):
\( \sin C = \frac{(k \sin A)(k \sin C)}{(k \sin B)^2 - (k \sin C)^2} \)
\( \sin C = \frac{k^2 \sin A \sin C}{k^2 (\sin^2 B - \sin^2 C)} \)
\( \sin C = \frac{\sin A \sin C}{\sin^2 B - \sin^2 C} \)
Since \( C \) is an angle in a triangle, \( \sin C \neq 0 \), so we can divide both sides by \( \sin C \):
\( 1 = \frac{\sin A}{\sin^2 B - \sin^2 C} \)
This means \( \sin^2 B - \sin^2 C = \sin A \).
We use the trigonometric identity: \( \sin^2 X - \sin^2 Y = \sin(X+Y)\sin(X-Y) \).
So, \( \sin(B+C)\sin(B-C) = \sin A \).
For any triangle, the sum of angles \( A+B+C = 180^\circ \), which means \( B+C = 180^\circ - A \).
Therefore, \( \sin(B+C) = \sin(180^\circ - A) = \sin A \).
Substitute this back into our equation:
\( \sin A \sin(B-C) = \sin A \).
Since \( A \) is an angle in a triangle, \( \sin A \neq 0 \), so we can divide by \( \sin A \):
\( \sin(B-C) = 1 \).
The only angle whose sine is 1 is \( 90^\circ \). So, \( B-C = 90^\circ \).
We are given \( \angle C = 23^\circ \).
So, \( B - 23^\circ = 90^\circ \).
\( B = 90^\circ + 23^\circ = 113^\circ \).
In simple words: (i) We started with the given equation and replaced the cosine terms using the cosine rule. After doing some algebra, we found that \( a^2 = b^2 + c^2 \). This is the Pythagoras theorem, which tells us that the angle opposite side 'a' (which is angle A) must be 90 degrees. (ii) First, we expressed the altitude AD in two ways: using the sine of angle C, and using the given formula. By setting these equal and using the sine rule, we simplified the expression. We then used a special identity for \( \sin^2 B - \sin^2 C \) and the fact that \( A+B+C = 180^\circ \). This led us to find that \( \sin(B-C) = 1 \), which means \( B-C = 90^\circ \). Since C is 23 degrees, B must be 113 degrees.

🎯 Exam Tip: For part (i), simplifying the fractions and using the standard cosine rule identities is key. For part (ii), relating the altitude to sine C is the crucial first step, and recognizing the \( \sin^2 X - \sin^2 Y \) identity will help quickly solve it.

Question 19. In a \( \triangle \text{ABC} \), if \( \frac{a^2+b^2}{a^2-b^2} = \frac{\sin (A+B)}{\sin (A-B)} \), prove the triangle is isosceles or right triangle.
Answer: Given the equation: \( \frac{a^2+b^2}{a^2-b^2} = \frac{\sin (A+B)}{\sin (A-B)} \).
We can apply the componendo and dividendo rule to both sides. This rule states if \( \frac{x}{y} = \frac{u}{v} \), then \( \frac{x+y}{x-y} = \frac{u+v}{u-v} \).
Applying it here:
\( \frac{(a^2+b^2) + (a^2-b^2)}{(a^2+b^2) - (a^2-b^2)} = \frac{\sin (A+B) + \sin (A-B)}{\sin (A+B) - \sin (A-B)} \)
Simplify the L.H.S:
\( \frac{2a^2}{2b^2} = \frac{a^2}{b^2} \).
Simplify the R.H.S using sum-to-product identities:
\( \sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2} \)
\( \sin X - \sin Y = 2 \cos \frac{X+Y}{2} \sin \frac{X-Y}{2} \)
So, \( \sin (A+B) + \sin (A-B) = 2 \sin A \cos B \).
And, \( \sin (A+B) - \sin (A-B) = 2 \cos A \sin B \).
Substitute these into the R.H.S:
\( \frac{2 \sin A \cos B}{2 \cos A \sin B} = \frac{\sin A \cos B}{\cos A \sin B} \).
So now we have: \( \frac{a^2}{b^2} = \frac{\sin A \cos B}{\cos A \sin B} \).
From the sine rule, \( \frac{a}{\sin A} = \frac{b}{\sin B} = k \), which means \( a = k \sin A \) and \( b = k \sin B \).
Substitute these into the L.H.S:
\( \frac{(k \sin A)^2}{(k \sin B)^2} = \frac{\sin^2 A}{\sin^2 B} \).
So, the equation becomes:
\( \frac{\sin^2 A}{\sin^2 B} = \frac{\sin A \cos B}{\cos A \sin B} \).
Cross-multiply and rearrange (assuming \( \sin A \neq 0 \) and \( \sin B \neq 0 \) for a triangle):
\( \sin^2 A \cos A \sin B = \sin^2 B \sin A \cos B \)
Divide by \( \sin A \sin B \):
\( \sin A \cos A = \sin B \cos B \).
Now, use the identity \( \sin 2X = 2 \sin X \cos X \), so \( \sin X \cos X = \frac{1}{2} \sin 2X \).
\( \frac{1}{2} \sin 2A = \frac{1}{2} \sin 2B \).
\( \sin 2A = \sin 2B \).
This implies two possibilities for angles in a triangle (where \( 2A, 2B \) are within \( 0^\circ \) to \( 360^\circ \) range):
1. \( 2A = 2B \implies A = B \). If two angles of a triangle are equal, the triangle is isosceles.
2. \( 2A = 180^\circ - 2B \). (Because \( \sin x = \sin(180^\circ-x) \)).
This means \( 2A + 2B = 180^\circ \).
Dividing by 2 gives \( A+B = 90^\circ \).
Since \( A+B+C = 180^\circ \) for a triangle, if \( A+B = 90^\circ \), then \( C = 180^\circ - 90^\circ = 90^\circ \).
If one angle is \( 90^\circ \), the triangle is a right-angled triangle.
Therefore, the triangle must be either isosceles or a right triangle.
In simple words: We started by using a math trick called componendo and dividendo. Then, we replaced sine terms using a rule that connects sides and angles. This led us to the equation \( \sin 2A = \sin 2B \). This equation can be true in two ways: either angle A equals angle B (making it an isosceles triangle), or angles A and B add up to 90 degrees (making the third angle C 90 degrees, so it's a right triangle).

🎯 Exam Tip: The componendo and dividendo rule is a powerful tool for simplifying such fractional equations. Also, remember the two possibilities when \( \sin X = \sin Y \), which are \( X = Y \) and \( X = 180^\circ - Y \).

Question 20. If \( \frac{\sin A}{\sin C} = \frac{\sin (A-B)}{\sin (B-C)} \), prove that \( a^2, b^2, c^2 \) are in A.P.
Answer: Given the equation: \( \frac{\sin A}{\sin C} = \frac{\sin (A-B)}{\sin (B-C)} \).
First, we cross-multiply the terms:
\( \sin A \sin(B-C) = \sin C \sin(A-B) \).
Now, we expand the terms \( \sin(B-C) \) and \( \sin(A-B) \) using the identity \( \sin(X-Y) = \sin X \cos Y - \cos X \sin Y \):
\( \sin A (\sin B \cos C - \cos B \sin C) = \sin C (\sin A \cos B - \cos A \sin B) \).
Distribute \( \sin A \) and \( \sin C \):
\( \sin A \sin B \cos C - \sin A \cos B \sin C = \sin C \sin A \cos B - \sin C \cos A \sin B \).
Divide the entire equation by \( \sin A \sin B \sin C \) (since A, B, C are angles of a triangle, their sines are non-zero):
\( \frac{\sin A \sin B \cos C}{\sin A \sin B \sin C} - \frac{\sin A \cos B \sin C}{\sin A \sin B \sin C} = \frac{\sin C \sin A \cos B}{\sin A \sin B \sin C} - \frac{\sin C \cos A \sin B}{\sin A \sin B \sin C} \).
This simplifies to:
\( \frac{\cos C}{\sin C} - \frac{\cos B}{\sin B} = \frac{\cos B}{\sin B} - \frac{\cos A}{\sin A} \).
Using \( \frac{\cos X}{\sin X} = \cot X \), the equation becomes:
\( \cot C - \cot B = \cot B - \cot A \).
Rearrange the terms:
\( \cot A + \cot C = 2 \cot B \).
This shows that \( \cot A, \cot B, \cot C \) are in an Arithmetic Progression (A.P.).
Now, we need to prove that if \( \cot A, \cot B, \cot C \) are in A.P., then \( a^2, b^2, c^2 \) are also in A.P.
We use the cosine rule for \( \cos A = \frac{b^2+c^2-a^2}{2bc} \) and the sine rule \( \sin A = \frac{a}{2R} \) (where R is the circumradius).
So, \( \cot A = \frac{\cos A}{\sin A} = \frac{(b^2+c^2-a^2)/(2bc)}{a/(2R)} = \frac{R(b^2+c^2-a^2)}{abc} \).
Substitute these cotangent values into \( \cot A + \cot C = 2 \cot B \):
\( \frac{R(b^2+c^2-a^2)}{abc} + \frac{R(a^2+b^2-c^2)}{abc} = 2 \frac{R(a^2+c^2-b^2)}{abc} \).
Since \( R \neq 0 \) and \( abc \neq 0 \), we can cancel \( \frac{R}{abc} \) from all terms:
\( (b^2+c^2-a^2) + (a^2+b^2-c^2) = 2(a^2+c^2-b^2) \).
Simplify the L.H.S:
\( b^2+c^2-a^2+a^2+b^2-c^2 = 2b^2 \).
So, \( 2b^2 = 2(a^2+c^2-b^2) \).
Divide by 2:
\( b^2 = a^2+c^2-b^2 \).
Rearrange:
\( 2b^2 = a^2+c^2 \).
This equation shows that \( b^2 \) is the arithmetic mean of \( a^2 \) and \( c^2 \). Therefore, \( a^2, b^2, c^2 \) are in A.P.
In simple words: We started by cross-multiplying the given equation and then expanded the sine terms using a trigonometric identity. After simplifying, we found that \( \cot A, \cot B, \cot C \) are in an arithmetic progression (A.P.). Then, we used the cosine and sine rules to show that if cotangents are in A.P., then \( 2b^2 = a^2+c^2 \), which means \( a^2, b^2, c^2 \) are also in an A.P.

🎯 Exam Tip: When proving relations between sides and angles, always consider converting trigonometric ratios to expressions involving sides (using sine and cosine rules) or vice versa. The identity \( \cot A = \frac{R(b^2+c^2-a^2)}{abc} \) is very useful here.

Question 21. If \( \sin 2A + \sin 2B = \sin 2C \), prove that \( A = 90^\circ \) or \( B = 90^\circ \).
Answer: Given the equation: \( \sin 2A + \sin 2B = \sin 2C \).
We use the sum-to-product identity: \( \sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2} \).
Applying this to the L.H.S:
\( 2 \sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} = \sin 2C \)
\( 2 \sin(A+B) \cos(A-B) = \sin 2C \).
For any triangle, the sum of angles \( A+B+C = 180^\circ \). This means \( A+B = 180^\circ - C \).
Therefore, \( \sin(A+B) = \sin(180^\circ - C) = \sin C \).
Also, use the double angle formula for \( \sin 2C = 2 \sin C \cos C \).
Substitute these into the equation:
\( 2 \sin C \cos(A-B) = 2 \sin C \cos C \).
Since \( C \) is an angle in a triangle, \( \sin C \) cannot be zero. So, we can divide both sides by \( 2 \sin C \):
\( \cos(A-B) = \cos C \).
For angles in a triangle, this implies two possibilities:
1. \( A-B = C \).
We know \( A+B+C = 180^\circ \). Substitute \( C = A-B \):
\( A+B+(A-B) = 180^\circ \)
\( 2A = 180^\circ \)
\( A = 90^\circ \).
2. \( A-B = -C \).
This means \( A+C = B \).
Substitute this into \( A+B+C = 180^\circ \):
\( (A+C)+B = 180^\circ \)
\( B+B = 180^\circ \)
\( 2B = 180^\circ \)
\( B = 90^\circ \).
Therefore, if \( \sin 2A + \sin 2B = \sin 2C \), then either \( A = 90^\circ \) or \( B = 90^\circ \).
In simple words: We used a math rule to change the sum of two sine terms into a product. Then, we used the fact that all angles in a triangle add up to 180 degrees. This helped us simplify the equation until we got \( \cos(A-B) = \cos C \). From this, we found two possibilities: either angle A is 90 degrees, or angle B is 90 degrees.

🎯 Exam Tip: Always remember the two possible solutions when solving \( \cos X = \cos Y \) or \( \sin X = \sin Y \) in the context of triangle angles. The sum-to-product and double-angle formulas are also fundamental for these types of proofs.

Question 23.
(i) In a \( \triangle \text{ABC} \) the angles A, B, C are in A.P. show that \( 2 \cos \frac{A-C}{2} = \frac{a+c}{\sqrt{a^2-ac+c^2}} \)
(ii) If the angles A, B, C of \( \triangle \text{ABC} \) are in A.P. and \( b : c = \sqrt{3} : \sqrt{2} \), show that \( A = 75^\circ \).
(iii) If \( A = 45^\circ \) and \( B = 75^\circ \), show that \( a + c\sqrt{2} = 2b \).
Answer:
(i) If the angles A, B, C are in A.P., then the middle angle B is the average of A and C, meaning \( 2B = A+C \).
Also, the sum of angles in a triangle is \( A+B+C = 180^\circ \).
Substitute \( A+C = 2B \) into the sum:
\( 2B+B = 180^\circ \implies 3B = 180^\circ \implies B = 60^\circ \).
Now consider the R.H.S of the equation we need to prove: \( \frac{a+c}{\sqrt{a^2-ac+c^2}} \).
From the cosine rule, \( b^2 = a^2+c^2-2ac \cos B \).
Since \( B = 60^\circ \), \( \cos B = \cos 60^\circ = \frac{1}{2} \).
So, \( b^2 = a^2+c^2-2ac(\frac{1}{2}) = a^2+c^2-ac \).
Therefore, the denominator \( \sqrt{a^2-ac+c^2} \) is simply \( \sqrt{b^2} = b \).
The R.H.S simplifies to \( \frac{a+c}{b} \).
Now, let's look at this expression using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \).
So, \( a = k \sin A, b = k \sin B, c = k \sin C \).
R.H.S = \( \frac{k \sin A + k \sin C}{k \sin B} = \frac{\sin A + \sin C}{\sin B} \).
Use the sum-to-product identity: \( \sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2} \).
R.H.S = \( \frac{2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}}{\sin B} \).
Since \( A+C = 2B \), then \( \frac{A+C}{2} = B \).
R.H.S = \( \frac{2 \sin B \cos \frac{A-C}{2}}{\sin B} \).
Since \( \sin B \neq 0 \) (as B is an angle of a triangle), we can cancel \( \sin B \):
R.H.S = \( 2 \cos \frac{A-C}{2} \).
This is equal to the L.H.S, so the equation is proven.
(ii) We are given that A, B, C are in A.P., so as shown in part (i), \( B = 60^\circ \).
We are also given the ratio of sides \( b:c = \sqrt{3}:\sqrt{2} \), which means \( \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}} \).
Using the sine rule, \( \frac{b}{\sin B} = \frac{c}{\sin C} \), which implies \( \frac{b}{c} = \frac{\sin B}{\sin C} \).
So, \( \frac{\sin B}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}} \).
Substitute \( B = 60^\circ \):
\( \frac{\sin 60^\circ}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}} \)
\( \frac{\sqrt{3}/2}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}} \)
Divide by \( \sqrt{3} \):
\( \frac{1/2}{\sin C} = \frac{1}{\sqrt{2}} \)
\( \sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \).
This implies \( C = 45^\circ \) (since C must be an acute angle in this context).
Now, we find angle A using \( A+B+C = 180^\circ \):
\( A = 180^\circ - B - C = 180^\circ - 60^\circ - 45^\circ = 75^\circ \).
Thus, \( A = 75^\circ \) is proven.
(iii) Given \( A = 45^\circ \) and \( B = 75^\circ \).
First, find angle C: \( C = 180^\circ - A - B = 180^\circ - 45^\circ - 75^\circ = 180^\circ - 120^\circ = 60^\circ \).
We need to prove \( a + c\sqrt{2} = 2b \).
Using the sine rule: \( a = k \sin A, b = k \sin B, c = k \sin C \).
Let's evaluate the L.H.S:
L.H.S. = \( k \sin A + k \sin C \sqrt{2} \)
L.H.S. = \( k (\sin 45^\circ + \sin 60^\circ \sqrt{2}) \)
L.H.S. = \( k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \sqrt{2} \right) \)
L.H.S. = \( k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{6}}{2} \right) \)
L.H.S. = \( k \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} \right) = k \frac{\sqrt{2}+\sqrt{6}}{2} \).
Now, let's evaluate the R.H.S:
R.H.S. = \( 2b = 2k \sin B = 2k \sin 75^\circ \).
We know that \( \sin 75^\circ = \sin(45^\circ+30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \).
\( \sin 75^\circ = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}} \).
To rationalize, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \): \( \frac{(\sqrt{3}+1)\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4} \).
So, R.H.S. = \( 2k \left( \frac{\sqrt{6}+\sqrt{2}}{4} \right) = k \frac{\sqrt{6}+\sqrt{2}}{2} \).
Since L.H.S. = R.H.S., the identity \( a + c\sqrt{2} = 2b \) is proven.
In simple words: (i) If angles A, B, C are in A.P., it means B is 60 degrees. Using this and the cosine rule, the bottom part of the fraction becomes 'b'. Then, using the sine rule and angle sum rules, we showed that both sides are equal to \( 2 \cos \frac{A-C}{2} \). (ii) Knowing B is 60 degrees from A.P. and the given side ratio, we used the sine rule to find angle C, which came out to be 45 degrees. From there, we found A by subtracting B and C from 180 degrees. (iii) We first found the third angle C. Then, we used the sine rule to write sides a, b, c in terms of sines of their angles. By putting in the values for \( \sin 45^\circ \), \( \sin 60^\circ \), and \( \sin 75^\circ \), we showed that both sides of the equation are equal.

🎯 Exam Tip: When angles are in A.P., immediately deduce the value of the middle angle (B = 60°). For side relations, always convert sides to sines using the sine rule. Remember to evaluate special angle values like \( \sin 75^\circ \) using angle sum formulas.

Question 24. The angles A, B, C of a triangle are in the ratio 3:5:4, prove that \( a + c\sqrt{2} = 2b \).
Answer: Let the angles be \( A=3x, B=5x, C=4x \).
The sum of angles in a triangle is \( 180^\circ \).
So, \( A+B+C = 180^\circ \).
\( 3x+5x+4x = 180^\circ \)
\( 12x = 180^\circ \)
\( x = \frac{180^\circ}{12} = 15^\circ \).
Now we can find the individual angle measures:
\( A = 3x = 3 \times 15^\circ = 45^\circ \).
\( B = 5x = 5 \times 15^\circ = 75^\circ \).
\( C = 4x = 4 \times 15^\circ = 60^\circ \).
We need to prove \( a + c\sqrt{2} = 2b \).
Using the sine rule: \( a = k \sin A, b = k \sin B, c = k \sin C \).
Let's evaluate the L.H.S:
L.H.S. = \( k \sin A + k \sin C \sqrt{2} \)
L.H.S. = \( k (\sin 45^\circ + \sin 60^\circ \sqrt{2}) \)
L.H.S. = \( k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \sqrt{2} \right) \)
L.H.S. = \( k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{6}}{2} \right) \)
L.H.S. = \( k \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} \right) = k \frac{\sqrt{2}+\sqrt{6}}{2} \).
Now, let's evaluate the R.H.S:
R.H.S. = \( 2b = 2k \sin B = 2k \sin 75^\circ \).
We know that \( \sin 75^\circ = \sin(45^\circ+30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \).
\( \sin 75^\circ = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{3}+1}{2\sqrt{2}} \).
To rationalize, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \): \( \frac{(\sqrt{3}+1)\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4} \).
So, R.H.S. = \( 2k \left( \frac{\sqrt{6}+\sqrt{2}}{4} \right) = k \frac{\sqrt{6}+\sqrt{2}}{2} \).
Since L.H.S. = R.H.S., the identity \( a + c\sqrt{2} = 2b \) is proven. This shows how understanding angle ratios can directly lead to side relationships in a triangle.
In simple words: First, we used the ratio of the angles and the fact that angles in a triangle add up to 180 degrees to find the exact measure of each angle (A=45°, B=75°, C=60°). Then, we used the sine rule to express sides 'a', 'b', and 'c' using these angles. By putting in the values of \( \sin 45^\circ \), \( \sin 60^\circ \), and \( \sin 75^\circ \), we found that both sides of the equation were equal.

🎯 Exam Tip: When given angle ratios, always calculate the actual angle measures first. For proving side relationships, the sine rule is usually the most direct approach. Ensure you know the values of sine and cosine for common angles like 15°, 75°, 45°, 60°.

Question 25. In a \( \triangle \text{ABC} \), if \( \cos A = \frac{17}{22}, \cos C = \frac{1}{14} \), prove that the ratio of the sides is 7:9:11.
Answer: We are given \( \cos A = \frac{17}{22} \) and \( \cos C = \frac{1}{14} \).
To find the ratio of sides, we need the sines of the angles, since by the sine rule, \( a:b:c = \sin A : \sin B : \sin C \).
First, calculate \( \sin A \) and \( \sin C \) using the identity \( \sin^2 x + \cos^2 x = 1 \):
\( \sin A = \sqrt{1-\cos^2 A} = \sqrt{1 - \left(\frac{17}{22}\right)^2} = \sqrt{1 - \frac{289}{484}} = \sqrt{\frac{484-289}{484}} = \sqrt{\frac{195}{484}} = \frac{\sqrt{195}}{22} \).
\( \sin C = \sqrt{1-\cos^2 C} = \sqrt{1 - \left(\frac{1}{14}\right)^2} = \sqrt{1 - \frac{1}{196}} = \sqrt{\frac{195}{196}} = \frac{\sqrt{195}}{14} \).
Now, we need to find \( \sin B \). We know that \( A+B+C = 180^\circ \), so \( B = 180^\circ - (A+C) \).
Therefore, \( \sin B = \sin(180^\circ - (A+C)) = \sin(A+C) \).
Using the sum formula for sine: \( \sin(A+C) = \sin A \cos C + \cos A \sin C \).
Substitute the values we have:
\( \sin B = \left(\frac{\sqrt{195}}{22}\right) \left(\frac{1}{14}\right) + \left(\frac{17}{22}\right) \left(\frac{\sqrt{195}}{14}\right) \)
\( \sin B = \frac{\sqrt{195}}{22 \times 14} + \frac{17\sqrt{195}}{22 \times 14} = \frac{\sqrt{195} + 17\sqrt{195}}{308} = \frac{18\sqrt{195}}{308} \).
Simplify the fraction: \( \frac{18}{308} = \frac{9}{154} \).
So, \( \sin B = \frac{9\sqrt{195}}{154} \).
Now we can write the ratio of the sides using the sine rule:
\( a:b:c = \sin A : \sin B : \sin C \)
\( a:b:c = \frac{\sqrt{195}}{22} : \frac{9\sqrt{195}}{154} : \frac{\sqrt{195}}{14} \).
We can divide all parts of the ratio by \( \sqrt{195} \):
\( a:b:c = \frac{1}{22} : \frac{9}{154} : \frac{1}{14} \).
To get whole numbers, find the Least Common Multiple (LCM) of the denominators 22, 154, and 14. The LCM is 154.
Multiply each part of the ratio by 154:
\( a:b:c = \left(\frac{1}{22} \times 154\right) : \left(\frac{9}{154} \times 154\right) : \left(\frac{1}{14} \times 154\right) \)
\( a:b:c = 7 : 9 : 11 \).
Thus, the ratio of the sides is 7:9:11, which is proven. This demonstrates how side ratios are determined by the sines of the opposite angles.
In simple words: We used the given cosine values to find the sine values for angles A and C. Then, knowing that all angles in a triangle add up to 180 degrees, we found the sine of angle B using a trigonometric addition formula. Finally, using the sine rule that says sides are proportional to the sines of their opposite angles, we put all the sine values into a ratio and simplified it to get 7:9:11.

🎯 Exam Tip: When given cosines, always find the sines first using \( \sin^2 x + \cos^2 x = 1 \). For the third angle's sine, use \( \sin(A+C) = \sin A \cos C + \cos A \sin C \). Remember to simplify ratios by dividing out common factors and finding a common denominator.

Question 26. The angle of a triangle are in the ratio 1:2:7, prove that the ratio of the greatest side to the least side is \( (\sqrt{5}+1):(\sqrt{5}-1) \).
Answer: Let the angles of the triangle be \( A=x, B=2x, C=7x \).
The sum of angles in a triangle is \( 180^\circ \).
So, \( x+2x+7x = 180^\circ \).
\( 10x = 180^\circ \)
\( x = 18^\circ \).
Therefore, the angles are:
\( A = 18^\circ \)
\( B = 2 \times 18^\circ = 36^\circ \)
\( C = 7 \times 18^\circ = 126^\circ \).
In a triangle, the greatest side is opposite the greatest angle, and the least side is opposite the least angle.
Here, the greatest angle is \( C = 126^\circ \), so the greatest side is \( c \).
The least angle is \( A = 18^\circ \), so the least side is \( a \).
We need to find the ratio \( \frac{\text{greatest side}}{\text{least side}} = \frac{c}{a} \).
Using the sine rule, \( \frac{a}{\sin A} = \frac{c}{\sin C} \), which implies \( \frac{c}{a} = \frac{\sin C}{\sin A} \).
Substitute the angles:
\( \frac{c}{a} = \frac{\sin 126^\circ}{\sin 18^\circ} \).
We know that \( \sin(180^\circ - \theta) = \sin \theta \). So, \( \sin 126^\circ = \sin(180^\circ - 54^\circ) = \sin 54^\circ \).
Thus, \( \frac{c}{a} = \frac{\sin 54^\circ}{\sin 18^\circ} \).
We also know that \( \sin \theta = \cos(90^\circ - \theta) \). So, \( \sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ \).
Therefore, \( \frac{c}{a} = \frac{\cos 36^\circ}{\sin 18^\circ} \).
Now, we use the standard trigonometric values for \( \sin 18^\circ \) and \( \cos 36^\circ \):
\( \sin 18^\circ = \frac{\sqrt{5}-1}{4} \)
\( \cos 36^\circ = \frac{\sqrt{5}+1}{4} \)
Substitute these values:
\( \frac{c}{a} = \frac{(\sqrt{5}+1)/4}{(\sqrt{5}-1)/4} \).
The \( 4 \) in the denominator cancels out:
\( \frac{c}{a} = \frac{\sqrt{5}+1}{\sqrt{5}-1} \).
So, the ratio of the greatest side to the least side is \( (\sqrt{5}+1):(\sqrt{5}-1) \), which is proven. This highlights how specific angle values lead to distinct side ratios.
In simple words: First, we used the given angle ratio and the fact that angles in a triangle add to 180 degrees to find the exact measures of the angles (18°, 36°, 126°). Then, we used the sine rule to find the ratio of the longest side (opposite 126°) to the shortest side (opposite 18°). By using known values for \( \sin 18^\circ \) and \( \cos 36^\circ \), we proved the given ratio.

🎯 Exam Tip: Always relate the greatest/least side to the greatest/least angle. For calculations involving \( 18^\circ, 36^\circ, 54^\circ, 72^\circ \), remember their specific trigonometric values or how they can be derived from other angles. Simplify by using identities like \( \sin(180^\circ-\theta)=\sin\theta \) and \( \sin\theta=\cos(90^\circ-\theta) \).

Question 27. If the sides of \( \triangle \text{ABC} \) are in the ratio 4:5:6, prove that one angle it twice that of the other.
Answer: Let the sides of the triangle be \( a = 4k, b = 5k, c = 6k \), where k is a constant.
We will use the cosine rule to find the cosine of each angle:
\( \cos A = \frac{b^2+c^2-a^2}{2bc} \)
\( \cos A = \frac{(5k)^2+(6k)^2-(4k)^2}{2(5k)(6k)} = \frac{25k^2+36k^2-16k^2}{60k^2} = \frac{45k^2}{60k^2} = \frac{3}{4} \).
\( \cos B = \frac{a^2+c^2-b^2}{2ac} \)
\( \cos B = \frac{(4k)^2+(6k)^2-(5k)^2}{2(4k)(6k)} = \frac{16k^2+36k^2-25k^2}{48k^2} = \frac{27k^2}{48k^2} = \frac{9}{16} \).
\( \cos C = \frac{a^2+b^2-c^2}{2ab} \)
\( \cos C = \frac{(4k)^2+(5k)^2-(6k)^2}{2(4k)(5k)} = \frac{16k^2+25k^2-36k^2}{40k^2} = \frac{5k^2}{40k^2} = \frac{1}{8} \).
Now we need to check if one angle is twice another. We can use the double angle formula for cosine: \( \cos 2X = 2 \cos^2 X - 1 \).
Let's check if \( \cos C = \cos 2A \):
\( \cos 2A = 2 \cos^2 A - 1 \)
Substitute the value of \( \cos A = \frac{3}{4} \):
\( \cos 2A = 2 \left(\frac{3}{4}\right)^2 - 1 = 2 \left(\frac{9}{16}\right) - 1 = \frac{9}{8} - 1 = \frac{1}{8} \).
We found that \( \cos C = \frac{1}{8} \) and \( \cos 2A = \frac{1}{8} \).
Therefore, \( \cos C = \cos 2A \).
Since \( A \) and \( C \) are angles in a triangle, and \( 2A \) and \( C \) are within the range of possible angles (0° to 180° for A and C, 0° to 360° for 2A), this implies that \( C = 2A \).
This proves that one angle (C) is twice another angle (A) in the triangle. This relationship helps us understand the geometry of the triangle.
In simple words: We used the given side ratios and the cosine rule to calculate the cosine of each angle (A, B, C). Then, we checked if any angle was double another. We found that \( \cos 2A \) was equal to \( \cos C \). This means angle C is exactly twice angle A.

🎯 Exam Tip: When given side ratios, always calculate the cosines of the angles using the cosine rule. To prove an angle is double another, use the double angle cosine formula (\( \cos 2X = 2 \cos^2 X - 1 \)) and compare with the cosine of another angle.

Question 28. Two sides and included angles of a triangle are respectively \( 3+\sqrt{3}, 3-\sqrt{3} \) and \( 60^\circ \). Show that the remaining elements of the triangle are \( 105^\circ, 15^\circ, 3\sqrt{2} \).
Answer: Let the given two sides be \( c = 3+\sqrt{3} \) and \( b = 3-\sqrt{3} \), and the included angle be \( A = 60^\circ \).
We need to find the third side \( a \) and the other two angles \( B \) and \( C \).
First, find side \( a \) using the cosine rule: \( a^2 = b^2+c^2-2bc \cos A \).
Calculate \( b^2 \), \( c^2 \), and \( bc \):
\( b^2 = (3-\sqrt{3})^2 = 3^2 - 2(3)(\sqrt{3}) + (\sqrt{3})^2 = 9 - 6\sqrt{3} + 3 = 12 - 6\sqrt{3} \).
\( c^2 = (3+\sqrt{3})^2 = 3^2 + 2(3)(\sqrt{3}) + (\sqrt{3})^2 = 9 + 6\sqrt{3} + 3 = 12 + 6\sqrt{3} \).
\( bc = (3-\sqrt{3})(3+\sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6 \).
We know \( \cos A = \cos 60^\circ = \frac{1}{2} \).
Substitute these values into the cosine rule:
\( a^2 = (12-6\sqrt{3}) + (12+6\sqrt{3}) - 2(6)\left(\frac{1}{2}\right) \)
\( a^2 = 12-6\sqrt{3} + 12+6\sqrt{3} - 6 \)
\( a^2 = 24 - 6 = 18 \).
So, \( a = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \). (This is one of the remaining elements).
Next, we find angles \( B \) and \( C \) using the sine rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).
Let's find \( \sin B \): \( \frac{a}{\sin A} = \frac{b}{\sin B} \implies \sin B = \frac{b \sin A}{a} \).
\( \sin B = \frac{(3-\sqrt{3}) \sin 60^\circ}{3\sqrt{2}} = \frac{(3-\sqrt{3}) (\sqrt{3}/2)}{3\sqrt{2}} \)
\( \sin B = \frac{3\sqrt{3} - 3}{6\sqrt{2}} = \frac{3(\sqrt{3}-1)}{6\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}} \).
To recognize this value, multiply numerator and denominator by \( \sqrt{2} \):
\( \sin B = \frac{(\sqrt{3}-1)\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4} \).
This value is known to be \( \sin 15^\circ \). So, \( B = 15^\circ \). (This is another remaining element).
Finally, find angle \( C \) using the sum of angles in a triangle:
\( C = 180^\circ - A - B = 180^\circ - 60^\circ - 15^\circ = 180^\circ - 75^\circ = 105^\circ \). (This is the last remaining element).
Thus, the remaining elements of the triangle are \( 105^\circ, 15^\circ, 3\sqrt{2} \), which is proven. This process helps complete the entire geometry of a triangle from partial information.
In simple words: We were given two sides and the angle between them. First, we used the cosine rule to calculate the length of the third side, which came out to be \( 3\sqrt{2} \). Then, using the sine rule, we calculated the sine of one of the unknown angles, which was \( \frac{\sqrt{6}-\sqrt{2}}{4} \), equal to \( \sin 15^\circ \). So, that angle is 15 degrees. Finally, we found the last angle by subtracting the known angles from 180 degrees, which gave us 105 degrees.

🎯 Exam Tip: Always start by finding the unknown side using the cosine rule when two sides and the included angle are given. Then, use the sine rule to find the angles. Knowing special values like \( \sin 15^\circ \) or \( \cos 105^\circ \) helps in quickly identifying the angles.