OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Exercise 19 (B)

Question 1. Given that tan \( \theta = \frac { 5 }{ 12 } \) and angle \( \theta \) is an acute angle, find sin \( \theta \) and cos \( \theta \). (ICSE)
Answer: First, we know that \( \tan \theta = \frac { 5 }{ 12 } \). This means in a right-angled triangle, the perpendicular side is 5 units and the base side is 12 units. Let's imagine a right-angled triangle ABC, where \( \angle B = 90^\circ \) and \( \theta \) is at angle A.
So, Perpendicular (BC) = 5 and Base (AB) = 12.
Next, we use the Pythagoras Theorem to find the hypotenuse (AC):
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (12)^2 + (5)^2 \)
\( AC^2 = 144 + 25 \)
\( AC^2 = 169 \)
\( AC = \sqrt{169} \)
\( AC = 13 \)
Now we have all three sides of the triangle.
To find \( \sin \theta \):
\( \sin \theta = \frac{\text { Perpendicular }}{\text { Hypotenuse }} = \frac{\text { BC }}{\text { AC }} = \frac { 5 }{ 13 } \)
To find \( \cos \theta \):
\( \cos \theta = \frac{\text { Base }}{\text { Hypotenuse }} = \frac{\text { AB }}{\text { AC }} = \frac { 12 }{ 13 } \)
In simple words: We used the given tan value to set the sides of a right triangle. Then, we found the third side using Pythagoras' theorem. Finally, we used the definitions of sine and cosine (opposite/hypotenuse and adjacent/hypotenuse) to find their values.

🎯 Exam Tip: Always draw a right-angled triangle to visualize the problem. Label the sides (perpendicular, base, hypotenuse) correctly with respect to the given angle \( \theta \).

Question 2. If sin \( \theta = \frac { 3 }{ 5 } \) and \( \theta \) is an acute angle, find
(i) cos \( \theta \),
(ii) tan \( \theta \).
Answer: We are given that \( \sin \theta = \frac { 3 }{ 5 } \). In a right-angled triangle, sine is defined as the ratio of the perpendicular side to the hypotenuse. Let's consider a right-angled triangle ABC, where \( \angle B = 90^\circ \) and \( \theta \) is at angle A.
So, Perpendicular (BC) = 3 and Hypotenuse (AC) = 5.
Now, we find the base (AB) using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (5)^2 = AB^2 + (3)^2 \)
\( 25 = AB^2 + 9 \)
\( AB^2 = 25 - 9 \)
\( AB^2 = 16 \)
\( AB = \sqrt{16} \)
\( AB = 4 \)
Now that we have all sides:
(i) To find \( \cos \theta \):
\( \cos \theta = \frac{\text { Base }}{\text { Hypotenuse }} = \frac{\text { AB }}{\text { AC }} = \frac { 4 }{ 5 } \)
(ii) To find \( \tan \theta \):
\( \tan \theta = \frac{\text { Perpendicular }}{\text { Base }} = \frac{\text { BC }}{\text { AB }} = \frac { 3 }{ 4 } \)
In simple words: Start with the given sine value to set up a right triangle. Use the Pythagorean theorem to find the missing side. Then use the definitions of cosine and tangent to get their values from the triangle sides.

🎯 Exam Tip: Remember that "acute angle" means the angle is less than 90 degrees, so all trigonometric ratios will be positive. This ensures your side lengths are positive.

Question 3. \( \triangle ABC \) is right-angled at \( \angle A \). Find tan, sine and cos of angles B and C in each of the following cases.
(i) AB = 7 cm, AC = 24 cm
(ii) AB = 12 cm, AC = 9 cm
Answer:
(i) In \( \triangle ABC \), \( \angle A = 90^\circ \), AB = 7 cm, AC = 24 cm. We first need to find the hypotenuse BC using the Pythagoras Theorem.
\( BC^2 = AB^2 + AC^2 \)
\( BC^2 = (7)^2 + (24)^2 \)
\( BC^2 = 49 + 576 \)
\( BC^2 = 625 \)
\( BC = \sqrt{625} \)
\( BC = 25 \text{ cm} \)
Now we can find the trigonometric ratios for angles B and C:
For angle B:
\( \tan B = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { AC }}{\text { AB }} = \frac { 24 }{ 7 } \)
\( \sin B = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { AC }}{\text { BC }} = \frac { 24 }{ 25 } \)
\( \cos B = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { AB }}{\text { BC }} = \frac { 7 }{ 25 } \)
For angle C:
\( \tan C = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { AB }}{\text { AC }} = \frac { 7 }{ 24 } \)
\( \sin C = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { AB }}{\text { BC }} = \frac { 7 }{ 25 } \)
\( \cos C = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { AC }}{\text { BC }} = \frac { 24 }{ 25 } \)

(ii) In \( \triangle ABC \), \( \angle A = 90^\circ \), AB = 12 cm, AC = 9 cm. Again, we find the hypotenuse BC first.
\( BC^2 = AB^2 + AC^2 \)
\( BC^2 = (12)^2 + (9)^2 \)
\( BC^2 = 144 + 81 \)
\( BC^2 = 225 \)
\( BC = \sqrt{225} \)
\( BC = 15 \text{ cm} \)
Now we find the trigonometric ratios for angles B and C:
For angle B:
\( \tan B = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { AC }}{\text { AB }} = \frac { 9 }{ 12 } \text{ or } \frac { 3 }{ 4 } \)
\( \sin B = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { AC }}{\text { BC }} = \frac { 9 }{ 15 } \text{ or } \frac { 3 }{ 5 } \)
\( \cos B = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { AB }}{\text { BC }} = \frac { 12 }{ 15 } \text{ or } \frac { 4 }{ 5 } \)
For angle C:
\( \tan C = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { AB }}{\text { AC }} = \frac { 12 }{ 9 } \text{ or } \frac { 4 }{ 3 } \)
\( \sin C = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { AB }}{\text { BC }} = \frac { 12 }{ 15 } \text{ or } \frac { 4 }{ 5 } \)
\( \cos C = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { AC }}{\text { BC }} = \frac { 9 }{ 15 } \text{ or } \frac { 3 }{ 5 } \)
In simple words: For each case, first use Pythagoras to get all three side lengths. Then, for each angle (B and C), identify which side is opposite, adjacent, and the hypotenuse relative to that angle to find its tan, sin, and cos values. Remember, the hypotenuse is always the longest side.

🎯 Exam Tip: When finding ratios for different angles in the same triangle, remember that the "opposite" and "adjacent" sides change depending on which angle you're focusing on. The hypotenuse remains the same.

Question 4. \( \triangle PQR \) is right-angled at R. Find tan, sine and cos of \( \angle P \) and \( \angle Q \) in each of the following cases.
(i) PQ = 10 cm, QR = 8 cm
(ii) PQ = 29 mm, PR = 21 mm
(iii) PQ = 3.7 cm, PR = 3.5 cm
Answer:
(i) In \( \triangle PQR \), \( \angle R = 90^\circ \), PQ = 10 cm, QR = 8 cm. First, find the missing side PR using the Pythagoras Theorem.
\( PQ^2 = QR^2 + PR^2 \)
\( (10)^2 = (8)^2 + PR^2 \)
\( 100 = 64 + PR^2 \)
\( PR^2 = 100 - 64 \)
\( PR^2 = 36 \)
\( PR = \sqrt{36} \)
\( PR = 6 \text{ cm} \)
Now, find the trigonometric ratios for \( \angle P \) and \( \angle Q \):
For \( \angle P \):
\( \tan P = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { QR }}{\text { PR }} = \frac { 8 }{ 6 } \text{ or } \frac { 4 }{ 3 } \)
\( \sin P = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { QR }}{\text { PQ }} = \frac { 8 }{ 10 } \text{ or } \frac { 4 }{ 5 } \)
\( \cos P = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { PR }}{\text { PQ }} = \frac { 6 }{ 10 } \text{ or } \frac { 3 }{ 5 } \)
For \( \angle Q \):
\( \tan Q = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { PR }}{\text { QR }} = \frac { 6 }{ 8 } \text{ or } \frac { 3 }{ 4 } \)
\( \sin Q = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { PR }}{\text { PQ }} = \frac { 6 }{ 10 } \text{ or } \frac { 3 }{ 5 } \)
\( \cos Q = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { QR }}{\text { PQ }} = \frac { 8 }{ 10 } \text{ or } \frac { 4 }{ 5 } \)

(ii) In \( \triangle PQR \), \( \angle R = 90^\circ \), PQ = 29 mm, PR = 21 mm. Find the missing side QR.
\( PQ^2 = PR^2 + QR^2 \)
\( (29)^2 = (21)^2 + QR^2 \)
\( 841 = 441 + QR^2 \)
\( QR^2 = 841 - 441 \)
\( QR^2 = 400 \)
\( QR = \sqrt{400} \)
\( QR = 20 \text{ mm} \)
Now, find the trigonometric ratios for \( \angle P \) and \( \angle Q \):
For \( \angle P \):
\( \tan P = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { QR }}{\text { PR }} = \frac { 20 }{ 21 } \)
\( \sin P = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { QR }}{\text { PQ }} = \frac { 20 }{ 29 } \)
\( \cos P = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { PR }}{\text { PQ }} = \frac { 21 }{ 29 } \)
For \( \angle Q \):
\( \tan Q = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { PR }}{\text { QR }} = \frac { 21 }{ 20 } \)
\( \sin Q = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { PR }}{\text { PQ }} = \frac { 21 }{ 29 } \)
\( \cos Q = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { QR }}{\text { PQ }} = \frac { 20 }{ 29 } \)

(iii) In \( \triangle PQR \), \( \angle R = 90^\circ \), PQ = 3.7 cm, PR = 3.5 cm. Find the missing side QR.
\( PQ^2 = PR^2 + QR^2 \)
\( (3.7)^2 = (3.5)^2 + QR^2 \)
\( 13.69 = 12.25 + QR^2 \)
\( QR^2 = 13.69 - 12.25 \)
\( QR^2 = 1.44 \)
\( QR = \sqrt{1.44} \)
\( QR = 1.2 \text{ cm} \)
Now, find the trigonometric ratios for \( \angle P \) and \( \angle Q \):
For \( \angle P \):
\( \tan P = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { QR }}{\text { PR }} = \frac { 1.2 }{ 3.5 } \text{ or } \frac { 12 }{ 35 } \)
\( \sin P = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { QR }}{\text { PQ }} = \frac { 1.2 }{ 3.7 } \text{ or } \frac { 12 }{ 37 } \)
\( \cos P = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { PR }}{\text { PQ }} = \frac { 3.5 }{ 3.7 } \text{ or } \frac { 35 }{ 37 } \)
For \( \angle Q \):
\( \tan Q = \frac{\text { Opposite }}{\text { Adjacent }} = \frac{\text { PR }}{\text { QR }} = \frac { 3.5 }{ 1.2 } \text{ or } \frac { 35 }{ 12 } \)
\( \sin Q = \frac{\text { Opposite }}{\text { Hypotenuse }} = \frac{\text { PR }}{\text { PQ }} = \frac { 3.5 }{ 3.7 } \text{ or } \frac { 35 }{ 37 } \)
\( \cos Q = \frac{\text { Adjacent }}{\text { Hypotenuse }} = \frac{\text { QR }}{\text { PQ }} = \frac { 1.2 }{ 3.7 } \text{ or } \frac { 12 }{ 37 } \)
In simple words: For each part, you need to find the missing side of the right triangle using the Pythagorean theorem first. Then, for each angle P and Q, decide which side is opposite and which is adjacent, and use those with the hypotenuse to calculate tan, sin, and cos.

🎯 Exam Tip: When using decimal lengths, be careful with calculations. It's often easier to work with fractions where possible to avoid rounding errors until the final step.

Question 5. For any angle \( \theta \), state the value of \( \sin^2 \theta + \cos^2 \theta \).
Answer: Let's consider a right-angled triangle ABC, with the right angle at B, and let \( \angle A = \theta \).
According to the definitions of sine and cosine:
\( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
Now, let's find \( \sin^2 \theta + \cos^2 \theta \):
\( \sin^2 \theta + \cos^2 \theta = \left(\frac{BC}{AC}\right)^2 + \left(\frac{AB}{AC}\right)^2 \)
\( = \frac{BC^2}{AC^2} + \frac{AB^2}{AC^2} \)
\( = \frac{BC^2 + AB^2}{AC^2} \)
From the Pythagoras Theorem in \( \triangle ABC \), we know that \( BC^2 + AB^2 = AC^2 \).
Substitute this into the equation:
\( = \frac{AC^2}{AC^2} \)
\( = 1 \)
Therefore, for any angle \( \theta \), \( \sin^2 \theta + \cos^2 \theta = 1 \). This is a fundamental trigonometric identity.
In simple words: We used a right triangle and the definitions of sine and cosine. By applying the Pythagorean theorem, we found that sine squared plus cosine squared always equals one, no matter what the angle is.

🎯 Exam Tip: This identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), is crucial and should be memorized. It is used frequently in solving trigonometric problems.

Question 6. The diagonals AC, BD of a rhombus ABCD meet at O. If AC = 6, BD = 8, find sin OCD.
Answer: In a rhombus, the diagonals bisect each other at right angles. This means they cut each other in half, and where they cross (at point O), they form 90-degree angles.
Given: AC = 6 units and BD = 8 units.
Since the diagonals bisect each other:
\( AO = OC = \frac{AC}{2} = \frac{6}{2} = 3 \) units
\( BO = OD = \frac{BD}{2} = \frac{8}{2} = 4 \) units
Consider the right-angled triangle \( \triangle DOC \). The right angle is at O.
We have: DO = 4 units and OC = 3 units.
Using the Pythagoras Theorem to find the hypotenuse DC:
\( DC^2 = DO^2 + OC^2 \)
\( DC^2 = (4)^2 + (3)^2 \)
\( DC^2 = 16 + 9 \)
\( DC^2 = 25 \)
\( DC = \sqrt{25} \)
\( DC = 5 \) units
Now we need to find \( \sin \angle OCD \). In \( \triangle DOC \), for angle OCD:
The side opposite to \( \angle OCD \) is DO.
The hypotenuse is DC.
So, \( \sin \angle OCD = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{DO}{DC} = \frac{4}{5} \)
In simple words: A rhombus's diagonals cut each other in half at a 90-degree angle. We used this fact to find the lengths of half-diagonals, then used the Pythagorean theorem in a small right triangle to find a side of the rhombus. Finally, we calculated sine using the opposite side and hypotenuse.

🎯 Exam Tip: Remember the properties of a rhombus: diagonals bisect each other at right angles. This property creates four congruent right-angled triangles within the rhombus, simplifying calculations.

Question 7. If cos \( \theta = 0.6 \), find the value of \( 5 \sin \theta – 3 \tan \theta \).
Answer: We are given \( \cos \theta = 0.6 \). We can write this as a fraction:
\( \cos \theta = \frac{6}{10} = \frac{3}{5} \)
In a right-angled triangle ABC, let \( \angle A = \theta \) and \( \angle B = 90^\circ \). Cosine is the ratio of the adjacent side to the hypotenuse.
So, Adjacent (AB) = 3 and Hypotenuse (AC) = 5.
Now, we find the perpendicular side (BC) using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (5)^2 = (3)^2 + BC^2 \)
\( 25 = 9 + BC^2 \)
\( BC^2 = 25 - 9 \)
\( BC^2 = 16 \)
\( BC = \sqrt{16} \)
\( BC = 4 \)
Now that we have all three sides (AB=3, BC=4, AC=5), we can find \( \sin \theta \) and \( \tan \theta \):
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4}{5} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{4}{3} \)
Finally, we substitute these values into the expression \( 5 \sin \theta - 3 \tan \theta \):
\( 5 \sin \theta - 3 \tan \theta = 5 \times \frac{4}{5} - 3 \times \frac{4}{3} \)
\( = 4 - 4 \)
\( = 0 \)
In simple words: First, change the given cosine decimal to a fraction to find two sides of a right triangle. Then use Pythagoras to find the third side. Once all sides are known, calculate sine and tangent. Finally, put these values into the given expression to solve it.

🎯 Exam Tip: When given a decimal value for a trigonometric ratio, convert it to a simple fraction first. This makes it easier to identify the side ratios of the right triangle.

Question 8. In a right-angled triangle, it is given that A is an acute angle and that tan A = \( \frac { 3 }{ 4 } \). Without using tables, find the value of cos A.
Answer: We are given \( \tan A = \frac { 3 }{ 4 } \). In a right-angled triangle ABC, with \( \angle B = 90^\circ \), tangent A is the ratio of the perpendicular side to the base side.
So, for angle A, Perpendicular (BC) = 3 and Base (AB) = 4.
Now, we find the hypotenuse (AC) using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (4)^2 + (3)^2 \)
\( AC^2 = 16 + 9 \)
\( AC^2 = 25 \)
\( AC = \sqrt{25} \)
\( AC = 5 \)
Now that we have all three sides (AB=4, BC=3, AC=5), we can find \( \cos A \):
\( \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4}{5} \)
This can also be written as a decimal: \( \cos A = 0.8 \).
In simple words: We used the given tangent value to find two sides of the right triangle. Then, we used Pythagoras to find the third side. With all sides known, we could find the value of cosine A by dividing the adjacent side by the hypotenuse.

🎯 Exam Tip: "Without using tables" indicates that you need to rely on the fundamental definitions of trigonometric ratios and the Pythagorean theorem, which is a common type of question.

Question 9. If \( \sin \theta = \frac{6}{10} \), find, without using table, the value of \( (\cos \theta + \tan \theta) \).
Answer: We are given \( \sin \theta = \frac{6}{10} \). This fraction can be simplified to \( \frac{3}{5} \).
In a right-angled triangle ABC, let \( \angle A = \theta \) and \( \angle B = 90^\circ \). Sine is the ratio of the perpendicular side to the hypotenuse.
So, Perpendicular (BC) = 6 and Hypotenuse (AC) = 10.
Now, we find the base (AB) using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (10)^2 = AB^2 + (6)^2 \)
\( 100 = AB^2 + 36 \)
\( AB^2 = 100 - 36 \)
\( AB^2 = 64 \)
\( AB = \sqrt{64} \)
\( AB = 8 \)
Now that we have all three sides (AB=8, BC=6, AC=10), we can find \( \cos \theta \) and \( \tan \theta \):
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{8}{10} = \frac{4}{5} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{6}{8} = \frac{3}{4} \)
Finally, we add these values:
\( \cos \theta + \tan \theta = \frac{4}{5} + \frac{3}{4} \)
To add these fractions, we find a common denominator, which is 20.
\( = \frac{4 \times 4}{5 \times 4} + \frac{3 \times 5}{4 \times 5} \)
\( = \frac{16}{20} + \frac{15}{20} \)
\( = \frac{16 + 15}{20} \)
\( = \frac{31}{20} \)
This can also be written as a mixed number: \( 1\frac{11}{20} \).
In simple words: First, use the given sine value to determine the perpendicular and hypotenuse of a right triangle. Then, use the Pythagorean theorem to find the base. Once all sides are known, calculate the cosine and tangent values. Finally, add these two fractions together.

🎯 Exam Tip: Always simplify fractions (like \( \frac{6}{10} \) to \( \frac{3}{5} \)) before proceeding with calculations. This often makes the numbers smaller and easier to work with.

Question 10. Answer true or false.
(a) In the diagram
(i) tan \( \theta \) = \( \frac{\text { BC }}{\text { AB }} \);
(ii) sec \( \theta \) = \( \frac{\text { BC }}{\text { AC }} \);
(b) AB : BC = sin A : cos A
Answer:
(a) Let's analyze the given diagram. It shows a right-angled triangle ABC, with the right angle at B, and \( \theta \) at angle A.
(i) For \( \tan \theta \) (where \( \theta \) is at A):
\( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} \)
The statement says \( \tan \theta = \frac{BC}{AB} \). This matches the definition.
Therefore, (i) is True.
(ii) For \( \sec \theta \) (where \( \theta \) is at A):
\( \sec \theta = \frac{1}{\cos \theta} \)
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
So, \( \sec \theta = \frac{AC}{AB} \).
The statement says \( \sec \theta = \frac{BC}{AC} \). This does not match our finding.
Therefore, (ii) is False.

(b) Let's examine the ratio AB : BC and compare it to \( \sin A : \cos A \).
We know that \( \frac{\sin A}{\cos A} = \tan A \).
In \( \triangle ABC \) (right-angled at B, with angle A), \( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} \).
So, \( \sin A : \cos A \) is equivalent to \( \frac{BC}{AB} \).
The statement says AB : BC = \( \sin A : \cos A \).
This means AB : BC = BC : AB.
For this to be true, \( AB^2 = BC^2 \), which implies AB = BC. This is not generally true for all right-angled triangles unless angle A is 45 degrees.
Therefore, (b) is False.
In simple words: For part (a), we looked at the definitions of tangent and secant based on the triangle shown. Tangent was correct, but secant was not. For part (b), we compared the ratio of sides AB:BC to the ratio of sine A:cosine A. Since sine A:cosine A equals tangent A (which is BC:AB), the statement AB:BC = BC:AB is generally false.

🎯 Exam Tip: Always recall the basic definitions: SOH CAH TOA for sine, cosine, tangent, and their reciprocals (cosecant, secant, cotangent). This helps avoid mistakes in true/false questions.

Question 11. Using the measurements given in figure, (a) Find the value of
(i) sin \( \Phi \),
(ii) tan \( \theta \)
(b) Write an expression for AD in terms of \( \theta \).
Answer: We are given a complex figure, which can be understood by breaking it down. Let's refer to the geometry implied by the solution steps.
(a) (i) The solution states \( \sin \Phi = \frac{CD}{BC} = \frac{5}{13} \). This implies a right-angled triangle where CD is the perpendicular and BC is the hypotenuse with respect to angle \( \Phi \).
(ii) For \( \tan \theta \), let's consider the construction where DE is drawn parallel to BC in the triangle ABC, with B as the right angle. E is on AB, and D is on AC. We have AB=14 and BC=12. From the solution's construction: EB = DC = 5, and ED = BC = 12. So, AE = AB - EB = 14 - 5 = 9. In the right-angled triangle AED (with \( \angle E = 90^\circ \)), \( \theta \) is at angle A.
\( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{DE}{AE} = \frac{12}{9} = \frac{4}{3} \)

(b) To write an expression for AD in terms of \( \theta \):
In the right-angled triangle AED, with \( \angle E = 90^\circ \), we found AE = 9 and DE = 12.
Using Pythagoras Theorem, \( AD^2 = AE^2 + DE^2 = 9^2 + 12^2 = 81 + 144 = 225 \), so \( AD = 15 \).
Now we relate AD to \( \sin \theta \) and \( \cos \theta \):
\( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{DE}{AD} \)
\( \implies AD = \frac{DE}{\sin \theta} = \frac{12}{\sin \theta} \)
Also,
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AE}{AD} \)
\( \implies AD = \frac{AE}{\cos \theta} = \frac{9}{\cos \theta} \)
So, the expression for AD in terms of \( \theta \) can be \( \frac{12}{\sin \theta} \) or \( \frac{9}{\cos \theta} \).
In simple words: For part (a), we used the given ratio to find sine for angle phi, and then, by analyzing the constructed triangle AED, we found the tangent for angle theta. For part (b), using the same triangle AED, we expressed the hypotenuse AD using both sine and cosine definitions.

🎯 Exam Tip: For complex geometry problems, sometimes you need to break down the main figure into smaller right-angled triangles using constructions (like drawing a parallel line). This helps simplify the problem into manageable parts.

Question 12. ABC is a right-angled triangle, right-angled at B. Given that \( \angle ACB = \theta \), side AB = 2 units and side BC = 1 unit, find the value of \( \sin^2 \theta + \tan^2 \theta \).
Answer: We have a right-angled triangle ABC, with \( \angle B = 90^\circ \). We are given \( \angle ACB = \theta \), AB = 2 units, and BC = 1 unit.
First, find the hypotenuse AC using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (2)^2 + (1)^2 \)
\( AC^2 = 4 + 1 \)
\( AC^2 = 5 \)
\( AC = \sqrt{5} \)
Now, for angle \( \theta \) at C:
\( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{2}{\sqrt{5}} \)
\( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{2}{1} = 2 \)
Next, we calculate \( \sin^2 \theta \) and \( \tan^2 \theta \):
\( \sin^2 \theta = \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{2^2}{(\sqrt{5})^2} = \frac{4}{5} \)
\( \tan^2 \theta = (2)^2 = 4 \)
Finally, find the value of \( \sin^2 \theta + \tan^2 \theta \):
\( \sin^2 \theta + \tan^2 \theta = \frac{4}{5} + 4 \)
To add these, we can write 4 as \( \frac{20}{5} \).
\( = \frac{4}{5} + \frac{20}{5} \)
\( = \frac{4+20}{5} \)
\( = \frac{24}{5} \)
As a mixed number, this is \( 4\frac{4}{5} \).
In simple words: We started by finding the hypotenuse of the right triangle using the given sides. Then, we calculated the sine and tangent of angle theta. After squaring these values, we added them together to get the final answer.

🎯 Exam Tip: When dealing with square roots in denominators, it's usually best to keep them as is until the very end, especially when squaring the ratio, as the root will often cancel out nicely.

Question 13. (i) If \( \sin \theta = \frac{12}{13} \) and \( \theta \) is less than \( 90^\circ \), find the value of \( (\cos \theta + \tan \theta) \).
(ii) If \( \cos \theta = \frac{12}{13} \), find the value of \( \sin \theta \) and \( \tan \theta \). Also, find the value of \( 2 \sin \theta - 4 \tan \theta \), where \( \theta \) is acute.
Answer:
(i) Given \( \sin \theta = \frac{12}{13} \). In a right-angled triangle ABC, let \( \angle A = \theta \) and \( \angle B = 90^\circ \). Perpendicular (BC) = 12 and Hypotenuse (AC) = 13.
Find the base (AB) using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (13)^2 = AB^2 + (12)^2 \)
\( 169 = AB^2 + 144 \)
\( AB^2 = 169 - 144 \)
\( AB^2 = 25 \)
\( AB = \sqrt{25} \)
\( AB = 5 \)
Now, find \( \cos \theta \) and \( \tan \theta \):
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{5}{13} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{12}{5} \)
Then, calculate \( \cos \theta + \tan \theta \):
\( \cos \theta + \tan \theta = \frac{5}{13} + \frac{12}{5} \)
Find a common denominator, which is 65.
\( = \frac{5 \times 5}{13 \times 5} + \frac{12 \times 13}{5 \times 13} \)
\( = \frac{25}{65} + \frac{156}{65} \)
\( = \frac{25 + 156}{65} \)
\( = \frac{181}{65} \)
This can be written as a mixed number: \( 2\frac{51}{65} \).

(ii) Given \( \cos \theta = \frac{12}{13} \). In a right-angled triangle ABC, let \( \angle A = \theta \) and \( \angle B = 90^\circ \). Base (AB) = 12 and Hypotenuse (AC) = 13.
Find the perpendicular (BC) using the Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
\( (13)^2 = (12)^2 + BC^2 \)
\( 169 = 144 + BC^2 \)
\( BC^2 = 169 - 144 \)
\( BC^2 = 25 \)
\( BC = \sqrt{25} \)
\( BC = 5 \)
Now, find \( \sin \theta \) and \( \tan \theta \):
\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5}{13} \)
\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{5}{12} \)
Finally, find the value of \( 2 \sin \theta - 4 \tan \theta \):
\( 2 \sin \theta - 4 \tan \theta = 2 \times \frac{5}{13} - 4 \times \frac{5}{12} \)
\( = \frac{10}{13} - \frac{20}{12} \)
\( = \frac{10}{13} - \frac{5}{3} \)
Find a common denominator, which is 39.
\( = \frac{10 \times 3}{13 \times 3} - \frac{5 \times 13}{3 \times 13} \)
\( = \frac{30}{39} - \frac{65}{39} \)
\( = \frac{30 - 65}{39} \)
\( = \frac{-35}{39} \)
In simple words: For both parts, first draw a right triangle and use the given sine or cosine value to find two sides. Use Pythagoras to calculate the third side. Then, find the other required trigonometric ratios and substitute them into the given expressions to solve. Remember to simplify fractions when possible.

🎯 Exam Tip: Pay close attention to which angle the sine/cosine/tangent is referring to. Always double-check the "opposite" and "adjacent" sides relative to the specified angle.

Question 14. If \( \tan \theta = \frac{5}{12} \), find the value of \( \frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta} \).
Answer: We are given \( \tan \theta = \frac{5}{12} \). To evaluate the expression \( \frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta} \), we can use a clever trick. Divide every term in the numerator and the denominator by \( \cos \theta \).
\( \frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta} = \frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}} \)
We know that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \). So, the expression becomes:
\( = \frac{1 - \tan \theta}{1 + \tan \theta} \)
Now, substitute the given value of \( \tan \theta = \frac{5}{12} \):
\( = \frac{1 - \frac{5}{12}}{1 + \frac{5}{12}} \)
For the numerator: \( 1 - \frac{5}{12} = \frac{12}{12} - \frac{5}{12} = \frac{12 - 5}{12} = \frac{7}{12} \)
For the denominator: \( 1 + \frac{5}{12} = \frac{12}{12} + \frac{5}{12} = \frac{12 + 5}{12} = \frac{17}{12} \)
So, the expression is:
\( = \frac{\frac{7}{12}}{\frac{17}{12}} \)
When dividing by a fraction, we multiply by its reciprocal:
\( = \frac{7}{12} \times \frac{12}{17} \)
The 12s cancel out:
\( = \frac{7}{17} \)
In simple words: We used a smart trick: divide the top and bottom of the fraction by cosine theta. This turned sine/cosine into tangent. Then, we just put in the given tangent value and simplified the fractions.

🎯 Exam Tip: When an expression involves sine, cosine, and tangent, dividing by the highest power of \( \cos \theta \) (or \( \sin \theta \)) can simplify it into terms of \( \tan \theta \) (or \( \cot \theta \)), making calculations easier.

Question 15. If \( 5 \sin \theta = 4 \), find the value of \( \frac{1+\sin \theta}{1-\sin \theta} \).
Answer: First, we find the value of \( \sin \theta \):
\( 5 \sin \theta = 4 \)
\( \implies \sin \theta = \frac{4}{5} \)
Now, substitute this value into the given expression:
\( \frac{1+\sin \theta}{1-\sin \theta} = \frac{1+\frac{4}{5}}{1-\frac{4}{5}} \)
To solve this, combine the fractions in the numerator and denominator:
\( = \frac{\frac{5+4}{5}}{\frac{5-4}{5}} = \frac{\frac{9}{5}}{\frac{1}{5}} \)
When dividing by a fraction, we multiply by its reciprocal:
\( = \frac{9}{5} \times \frac{5}{1} = 9 \)
In simple words: First, work out what \( \sin \theta \) is. Then, put that fraction into the big sum. Solve the top and bottom parts separately, and then divide them.

🎯 Exam Tip: Always simplify fractions before performing multiplication or division to make calculations easier and reduce errors.

Question 16. If \( b \tan \theta = a \), find the value of \( \frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta} \).
Answer: From the given information, we can find \( \tan \theta \):
\( b \tan \theta = a \)
\( \implies \tan \theta = \frac{a}{b} \)
Next, we rewrite the expression by dividing every term in the numerator and denominator by \( \cos \theta \). This helps convert the expression into terms of \( \tan \theta \):
\( \frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta} = \frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}} \)
\( \implies = \frac{1+\tan \theta}{1-\tan \theta} \)
Now, substitute the value of \( \tan \theta \):
\( = \frac{1+\frac{a}{b}}{1-\frac{a}{b}} \)
Combine the fractions in the numerator and denominator:
\( = \frac{\frac{b+a}{b}}{\frac{b-a}{b}} \)
Then, multiply by the reciprocal of the denominator to simplify:
\( = \frac{b+a}{b} \times \frac{b}{b-a} \)
\( = \frac{b+a}{b-a} \)
This method is very useful for simplifying expressions with sines and cosines that can be turned into tangents.
In simple words: First, find \( \tan \theta \). Then, change the big sum by dividing everything by \( \cos \theta \), which turns it into a sum with \( \tan \theta \). Put your \( \tan \theta \) value in and solve.

🎯 Exam Tip: When an expression involves both \( \sin \theta \) and \( \cos \theta \), dividing all terms by \( \cos \theta \) is a common and effective strategy to convert the expression into terms of \( \tan \theta \).

Question 17. If \( 5 \tan \theta = 4 \), find the value of \( \frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta} \).
Answer: First, find the value of \( \tan \theta \):
\( 5 \tan \theta = 4 \)
\( \implies \tan \theta = \frac{4}{5} \)
To simplify the given expression, divide every term in the numerator and the denominator by \( \cos \theta \). This changes the expression into terms of \( \tan \theta \):
\( \frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta} = \frac{5 \frac{\sin \theta}{\cos \theta}-3 \frac{\cos \theta}{\cos \theta}}{5 \frac{\sin \theta}{\cos \theta}+2 \frac{\cos \theta}{\cos \theta}} \)
\( \implies = \frac{5 \tan \theta-3}{5 \tan \theta+2} \)
Now, substitute the value of \( \tan \theta = \frac{4}{5} \) into this new expression:
\( = \frac{5 \times \frac{4}{5}-3}{5 \times \frac{4}{5}+2} \)
Perform the multiplication and subtraction/addition:
\( = \frac{4-3}{4+2} \)
\( = \frac{1}{6} \)
This method allows us to solve the problem without finding \( \sin \theta \) and \( \cos \theta \) separately.
In simple words: Figure out \( \tan \theta \) first. Then, make the main sum easier by dividing every part by \( \cos \theta \), changing it into a sum with \( \tan \theta \). Put the \( \tan \theta \) value in and solve.

🎯 Exam Tip: When given a relationship for \( \tan \theta \) and asked to evaluate an expression involving \( \sin \theta \) and \( \cos \theta \), converting the expression to \( \tan \theta \) is often the quickest path to the solution.

Question 18. If \( 13 \sin A = 5 \) and \( \angle A \) is acute, find the value of \( \frac{5 \sin A-2 \cos A}{\tan A} \).
Answer: First, use the given information to find \( \sin A \):
\( 13 \sin A = 5 \)
\( \implies \sin A = \frac{5}{13} \)
In a right-angled triangle, \( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} \). So, let the side opposite to angle A (BC) be 5 units and the hypotenuse (AC) be 13 units. Now, use the Pythagorean theorem to find the adjacent side (AB):
\( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \)
\( (13)^2 = \text{AB}^2 + (5)^2 \)
\( 169 = \text{AB}^2 + 25 \)
\( \implies \text{AB}^2 = 169 - 25 = 144 \)
\( \implies \text{AB} = \sqrt{144} = 12 \)
Now we have all sides. We can find \( \cos A \) and \( \tan A \):
\( \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\text{AB}}{\text{AC}} = \frac{12}{13} \)
\( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{BC}}{\text{AB}} = \frac{5}{12} \)
Finally, substitute these values into the given expression:
\( \frac{5 \sin A-2 \cos A}{\tan A} = \frac{5 \times \frac{5}{13}-2 \times \frac{12}{13}}{\frac{5}{12}} \)
Simplify the numerator:
\( = \frac{\frac{25}{13}-\frac{24}{13}}{\frac{5}{12}} = \frac{\frac{1}{13}}{\frac{5}{12}} \)
To divide by a fraction, multiply by its reciprocal:
\( = \frac{1}{13} \times \frac{12}{5} = \frac{12}{65} \)
In simple words: First, use \( \sin A \) to find the lengths of two sides of the triangle. Then, use Pythagoras to find the third side. Once you have all three sides, calculate \( \cos A \) and \( \tan A \). Put these numbers into the given sum and do the math.

🎯 Exam Tip: For problems where only one trigonometric ratio is given, always draw a right-angled triangle and use the Pythagorean theorem to find the lengths of all three sides. This allows you to easily find any other required trigonometric ratio.

Question 19. Given \( 5 \cos A - 12 \sin A = 0 \), find the value of \( \frac{\sin A+\cos A}{2 \cos A-\sin A} \).
Answer: First, use the given equation to find \( \tan A \):
\( 5 \cos A - 12 \sin A = 0 \)
\( \implies 5 \cos A = 12 \sin A \)
Now, divide both sides by \( \cos A \) and by 12:
\( \frac{\sin A}{\cos A} = \frac{5}{12} \)
\( \implies \tan A = \frac{5}{12} \)
Next, to simplify the expression, divide every term in the numerator and the denominator by \( \cos A \):
\( \frac{\sin A+\cos A}{2 \cos A-\sin A} = \frac{\frac{\sin A}{\cos A}+\frac{\cos A}{\cos A}}{\frac{2 \cos A}{\cos A}-\frac{\sin A}{\cos A}} \)
\( \implies = \frac{\tan A+1}{2-\tan A} \)
Now, substitute the value of \( \tan A = \frac{5}{12} \) into this simplified expression:
\( = \frac{\frac{5}{12}+1}{2-\frac{5}{12}} \)
Combine the fractions in the numerator and denominator:
\( = \frac{\frac{5+12}{12}}{\frac{24-5}{12}} = \frac{\frac{17}{12}}{\frac{19}{12}} \)
To divide by a fraction, multiply by its reciprocal:
\( = \frac{17}{12} \times \frac{12}{19} = \frac{17}{19} \)
This algebraic manipulation is a powerful tool to solve such trigonometric expressions efficiently.
In simple words: First, use the given equation to find \( \tan A \). Then, make the main sum simpler by dividing all its parts by \( \cos A \), which changes it into a sum with \( \tan A \). Put your \( \tan A \) value in and solve.

🎯 Exam Tip: When an equation relates \( \sin A \) and \( \cos A \), it's often useful to convert it into a value for \( \tan A \) by dividing by \( \cos A \). This helps simplify further calculations when \( \tan A \) is involved.