Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.10

Question 1. Determine whether the following measurements produce one triangle, two triangles or no triangle. \( \angle B = 88^\circ \), \( a = 23 \), \( b = 2 \). Solve if solution exists.
Answer: We use the sine formula to check for a solution. The sine formula states that for a triangle ABC, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).
Given values are \( \angle B = 88^\circ \), side \( a = 23 \), and side \( b = 2 \).
Using the sine formula, we have:
\( \frac{a}{\sin A} = \frac{b}{\sin B} \)
\( \frac{23}{\sin A} = \frac{2}{\sin 88^\circ} \)
Now, we find \( \sin A \):
\( \sin A = \frac{23 \times \sin 88^\circ}{2} \)
We know that \( \sin 88^\circ \) is approximately \( 0.999 \).
\( \sin A = \frac{23 \times 0.999}{2} \)
\( \sin A = \frac{22.977}{2} \)
\( \sin A = 11.4885 \)
Since the value of \( \sin A \) must be between \( -1 \) and \( 1 \), a value of \( 11.4885 \) is not possible. This means that no triangle can be formed with the given measurements. This highlights how side lengths and angles must be consistent for a triangle to exist.
In simple words: We used a special rule for triangles called the sine formula. After putting in the given numbers, we found that the sine of angle A was too big (more than 1), which is impossible for any angle. So, no triangle can be made with these measurements.

🎯 Exam Tip: Always remember that the sine of any angle must be between -1 and 1. If your calculation results in a sine value outside this range, it means no triangle exists with those dimensions.

Question 2. If the sides of a \( \triangle ABC \) are \( a = 4, b = 6 \) and \( c = 8 \), then show that \( 4 \cos B + 3 \cos C = 2 \).
Answer: We are given the side lengths of triangle ABC as \( a = 4, b = 6, c = 8 \). We need to show that \( 4 \cos B + 3 \cos C = 2 \). To do this, we will first find the values of \( \cos B \) and \( \cos C \) using the cosine formula. The cosine formula for an angle in a triangle relates the sides to the angle.
For \( \cos B \):
\( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \)
\( \cos B = \frac{4^2 + 8^2 - 6^2}{2(4)(8)} \)
\( \cos B = \frac{16 + 64 - 36}{64} \)
\( \cos B = \frac{80 - 36}{64} \)
\( \cos B = \frac{44}{64} \)
\( \cos B = \frac{11}{16} \)
For \( \cos C \):
\( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)
\( \cos C = \frac{4^2 + 6^2 - 8^2}{2(4)(6)} \)
\( \cos C = \frac{16 + 36 - 64}{48} \)
\( \cos C = \frac{52 - 64}{48} \)
\( \cos C = \frac{-12}{48} \)
\( \cos C = -\frac{1}{4} \)
Now, substitute the values of \( \cos B \) and \( \cos C \) into the expression \( 4 \cos B + 3 \cos C \):
\( 4 \cos B + 3 \cos C = 4 \left( \frac{11}{16} \right) + 3 \left( -\frac{1}{4} \right) \)
\( \implies \) \( = \frac{44}{16} - \frac{3}{4} \)
\( \implies \) \( = \frac{11}{4} - \frac{3}{4} \)
\( \implies \) \( = \frac{11 - 3}{4} \)
\( \implies \) \( = \frac{8}{4} \)
\( \implies \) \( = 2 \)
Thus, we have shown that \( 4 \cos B + 3 \cos C = 2 \). Using the cosine rule helps find unknown angles or sides in any triangle when three sides or two sides and an included angle are known.
In simple words: First, we used a special math rule called the cosine formula to find the value of "cos B" and "cos C" for our triangle. Then, we put those values into the given equation \( 4 \cos B + 3 \cos C \). After doing the calculations, the answer came out to be 2, which is what we needed to show.

🎯 Exam Tip: When proving an identity with triangle sides and angles, always clearly state which formula (sine rule or cosine rule) you are using for each step. Double-check your arithmetic, especially with fractions and negative numbers.

Question 3. In a \( \triangle ABC \), if \( a = \sqrt{3} - 1 \), \( b = \sqrt{3} + 1 \) and \( C = 60^\circ \) find the other side and other two angles.
Answer: We are given two sides, \( a = \sqrt{3} - 1 \) and \( b = \sqrt{3} + 1 \), and the included angle \( C = 60^\circ \). We need to find the third side \( c \) and the other two angles, \( A \) and \( B \).
First, let's find side \( c \) using the cosine formula:
\( c^2 = a^2 + b^2 - 2ab \cos C \)
Substitute the given values:
\( c^2 = (\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2 - 2(\sqrt{3} - 1)(\sqrt{3} + 1) \cos 60^\circ \)
We know \( (\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2\sqrt{3} + 1 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \).
And \( (\sqrt{3} + 1)^2 = (\sqrt{3})^2 + 2\sqrt{3} + 1 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \).
Also, \( (\sqrt{3} - 1)(\sqrt{3} + 1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2 \).
And \( \cos 60^\circ = \frac{1}{2} \).
Substitute these simplified terms back into the \( c^2 \) equation:
\( c^2 = (4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) - 2(2) \left( \frac{1}{2} \right) \)
\( \implies \) \( c^2 = 4 - 2\sqrt{3} + 4 + 2\sqrt{3} - 2 \)
\( \implies \) \( c^2 = 8 - 2 \)
\( \implies \) \( c^2 = 6 \)
\( \implies \) \( c = \sqrt{6} \)
Now, let's find the angles \( A \) and \( B \) using the sine formula:
\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)
Using \( \frac{a}{\sin A} = \frac{c}{\sin C} \):
\( \frac{\sqrt{3} - 1}{\sin A} = \frac{\sqrt{6}}{\sin 60^\circ} \)
We know \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
\( \frac{\sqrt{3} - 1}{\sin A} = \frac{\sqrt{6}}{\frac{\sqrt{3}}{2}} \)
\( \implies \) \( \frac{\sqrt{3} - 1}{\sin A} = \frac{2\sqrt{6}}{\sqrt{3}} \)
\( \implies \) \( \frac{\sqrt{3} - 1}{\sin A} = 2\sqrt{2} \)
\( \implies \) \( \sin A = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
We know that \( \sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \).
\( \implies \) \( \sin 15^\circ = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) \)
\( \implies \) \( \sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( \implies \) \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
Since \( \sin A = \frac{\sqrt{3} - 1}{2\sqrt{2}} \) and \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \), we can say that \( A = 15^\circ \).
Finally, we find angle \( B \) using the angle sum property of a triangle:
\( A + B + C = 180^\circ \)
\( 15^\circ + B + 60^\circ = 180^\circ \)
\( \implies \) \( B + 75^\circ = 180^\circ \)
\( \implies \) \( B = 180^\circ - 75^\circ \)
\( \implies \) \( B = 105^\circ \)
The other side is \( c = \sqrt{6} \), and the other two angles are \( A = 15^\circ \) and \( B = 105^\circ \). This problem shows how different trigonometric formulas work together to solve a triangle.
In simple words: We first found the missing side 'c' using the cosine formula because we knew two sides and the angle between them. Then, we used the sine formula to find angle 'A'. We matched the value we got for \( \sin A \) with a known angle, \( 15^\circ \). Finally, we used the rule that all angles in a triangle add up to \( 180^\circ \) to find the last angle, 'B'.

🎯 Exam Tip: When dealing with values like \( \sqrt{3}-1 \) and \( \sqrt{3}+1 \), remember the algebraic identities \( (x-y)^2 = x^2 - 2xy + y^2 \) and \( (x-y)(x+y) = x^2 - y^2 \) to simplify calculations quickly.

Question 4. In any \( \triangle ABC \), prove that the area \( = \frac{b^2+c^2 - a^2}{4 \cot A} \).
Answer: We need to prove the given formula for the area of a triangle. Let \( \Delta \) represent the area of \( \triangle ABC \).
We know the standard formula for the area of a triangle is:
\( \Delta = \frac{1}{2} bc \sin A \)
We also know the cosine formula for angle \( A \):
\( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \)
From this, we can express \( bc \):
\( bc = \frac{b^2 + c^2 - a^2}{2 \cos A} \)
Now, substitute this expression for \( bc \) into the area formula \( \Delta = \frac{1}{2} bc \sin A \):
\( \Delta = \frac{1}{2} \left( \frac{b^2 + c^2 - a^2}{2 \cos A} \right) \sin A \)
\( \implies \) \( \Delta = \frac{1}{4} (b^2 + c^2 - a^2) \frac{\sin A}{\cos A} \)
We know that \( \frac{\sin A}{\cos A} = \tan A \).
So, \( \Delta = \frac{1}{4} (b^2 + c^2 - a^2) \tan A \)
Since \( \tan A = \frac{1}{\cot A} \), we can write:
\( \Delta = \frac{1}{4} (b^2 + c^2 - a^2) \frac{1}{\cot A} \)
\( \implies \) \( \Delta = \frac{b^2 + c^2 - a^2}{4 \cot A} \)
Thus, we have proved that the area of a triangle \( \Delta = \frac{b^2+c^2 - a^2}{4 \cot A} \). This shows how different trigonometric identities can be used to derive new formulas in geometry.
In simple words: We started with the basic formula for a triangle's area, which uses two sides and the sine of the angle between them. Then, we used another rule called the cosine formula to change the part with sides 'b' and 'c'. We replaced this part in the area formula. After simplifying and using that \( \tan A \) is the same as \( 1/\cot A \), we got the formula we wanted to prove.

🎯 Exam Tip: When proving trigonometric identities in triangles, it's often helpful to start from a known formula (like the basic area formula or cosine rule) and work towards the desired expression, using fundamental identities like \( \tan A = \sin A / \cos A \).

Question 5. In a \( \triangle ABC \), if \( a = 12 \text{ cm}, b = 8 \text{ cm} \) and \( C = 30^\circ \), then show that its area is \( 24 \text{ sq.cm} \).
Answer: We are given the sides \( a = 12 \text{ cm} \), \( b = 8 \text{ cm} \), and the included angle \( C = 30^\circ \). We need to show that the area of the triangle is \( 24 \text{ sq.cm} \).
The formula for the area of a triangle when two sides and the included angle are known is:
Area \( \Delta = \frac{1}{2} ab \sin C \)
Substitute the given values into the formula:
Area \( \Delta = \frac{1}{2} \times 12 \times 8 \times \sin 30^\circ \)
We know that \( \sin 30^\circ = \frac{1}{2} \).
Area \( \Delta = \frac{1}{2} \times 12 \times 8 \times \frac{1}{2} \)
\( \implies \) Area \( \Delta = 6 \times 8 \times \frac{1}{2} \)
\( \implies \) Area \( \Delta = 48 \times \frac{1}{2} \)
\( \implies \) Area \( \Delta = 24 \)
So, the area of the triangle is \( 24 \text{ sq.cm} \). This problem is a direct application of the area formula, illustrating how trigonometry helps calculate areas of non-right-angled triangles.
In simple words: We have two sides of a triangle and the angle between them. To find the area, we use a simple formula: half of one side times the other side, times the sine of the angle in between. When we put in the numbers for our triangle and calculate, the area comes out to be exactly 24 square centimeters.

🎯 Exam Tip: Remember the area formula \( \frac{1}{2}ab \sin C \). This is crucial for finding the area of any triangle when two sides and their included angle are given. Make sure to use the correct angle (the one *between* the two given sides).

Question 6. In a \( \triangle ABC \), if \( a = 18 \text{ cm}, b = 24 \text{ cm} \) and \( c = 30 \text{ cm} \), then show that its area is \( 216 \text{ sq.cm} \).
Answer: We are given all three side lengths of the triangle: \( a = 18 \text{ cm}, b = 24 \text{ cm}, c = 30 \text{ cm} \). We need to show that its area is \( 216 \text{ sq.cm} \). Since all three sides are given, we will use Heron's formula to find the area. Heron's formula requires calculating the semi-perimeter first.
First, calculate the semi-perimeter \( s \):
\( s = \frac{a + b + c}{2} \)
\( s = \frac{18 + 24 + 30}{2} \)
\( s = \frac{72}{2} \)
\( s = 36 \)
Now, apply Heron's formula for the area \( \Delta \):
Area \( \Delta = \sqrt{s(s - a)(s - b)(s - c)} \)
Substitute the values of \( s, a, b, c \):
Area \( \Delta = \sqrt{36(36 - 18)(36 - 24)(36 - 30)} \)
\( \implies \) Area \( \Delta = \sqrt{36 \times 18 \times 12 \times 6} \)
To simplify the square root, we can factorize the numbers:
\( 36 = 6 \times 6 \)
\( 18 = 3 \times 6 \)
\( 12 = 2 \times 6 \)
\( 6 = 6 \)
Area \( \Delta = \sqrt{(6 \times 6) \times (3 \times 6) \times (2 \times 6) \times 6} \)
\( \implies \) Area \( \Delta = \sqrt{6^2 \times 3 \times 6 \times 2 \times 6 \times 6} \)
\( \implies \) Area \( \Delta = \sqrt{6^2 \times 6^2 \times 6^2} \)
\( \implies \) Area \( \Delta = \sqrt{(6 \times 6 \times 6)^2} \)
\( \implies \) Area \( \Delta = 6 \times 6 \times 6 \)
\( \implies \) Area \( \Delta = 216 \)
So, the area of the triangle is \( 216 \text{ sq.cm} \). Heron's formula is very useful for finding the area of a triangle when only its side lengths are known, without needing any angles.
In simple words: Since we know all three sides of the triangle, we used a special rule called Heron's formula. First, we found half the perimeter of the triangle. Then, we put that number and the sides into Heron's formula. After doing the math, we found that the area of the triangle is 216 square centimeters.

🎯 Exam Tip: Heron's formula is perfect for finding the area when all three sides are known. When simplifying the square root, try to find pairs of factors or perfect squares within the numbers to make the calculation easier.

Question 7. Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are \( 30^\circ \) and \( 45^\circ \) respectively. If A and B stand \( 5 \text{ km} \) apart, find the distance of the intruder from B.
Answer: Let's set up the problem:
Let \( C \) be the position of the intruder at the top of the hill.
Let \( A \) and \( B \) be the positions of the two soldiers on the ground.
The distance between A and B is \( AB = 5 \text{ km} \).
The angle of elevation of the intruder from A is \( \angle BAC = 30^\circ \).
The angle of elevation of the intruder from B is \( \angle PBC = 45^\circ \). If P is a point on the ground such that \( \angle PBC \) is the angle of elevation, we can consider the triangle formed by A, B, and C. The angle \( \angle ABC \) in the triangle would be \( 180^\circ - 45^\circ = 135^\circ \) (if A, B, P are collinear and B is between A and P, and P is on the same side as C's projection). Alternatively, using the given diagram, \( \angle B \) in \( \triangle ABC \) is \( 180^\circ - \angle PBC \) as \( \angle PBC \) refers to an external angle with respect to the horizontal. Assuming the diagram shows internal angles or that \( \angle ABC \) is \( 45^\circ \) and \( \angle BAC \) is \( 30^\circ \), this means the points A, B are on the same side relative to the vertical line from C.
Looking at the provided diagram, let C be the top of the hill. Let D be the point on the road directly below C. So \( \triangle ADC \) and \( \triangle BDC \) are right-angled triangles.
The angle of elevation from A is \( \angle CAD = 30^\circ \).
The angle of elevation from B is \( \angle CBD = 45^\circ \).
In \( \triangle ABC \), the angles are:
\( \angle CAB = 30^\circ \)
\( \angle CBA = 180^\circ - 45^\circ = 135^\circ \) (since the angle \( 45^\circ \) is an angle of elevation from B, and \( \angle CBA \) is interior to \( \triangle ABC \), with C above the horizontal AB).
Now, find the third angle \( \angle BCA \):
\( \angle BCA = 180^\circ - (\angle CAB + \angle CBA) \)
\( \angle BCA = 180^\circ - (30^\circ + 135^\circ) \)
\( \angle BCA = 180^\circ - 165^\circ \)
\( \angle BCA = 15^\circ \)
Now, we use the sine formula in \( \triangle ABC \) to find the distance of the intruder from B, which is side \( BC \).
\( \frac{BC}{\sin A} = \frac{AB}{\sin (\angle BCA)} \)
\( \frac{BC}{\sin 30^\circ} = \frac{5}{\sin 15^\circ} \)
\( BC = \frac{5 \times \sin 30^\circ}{\sin 15^\circ} \)
We know \( \sin 30^\circ = \frac{1}{2} \).
And \( \sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \).
\( \implies \) \( \sin 15^\circ = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) \)
\( \implies \) \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
Now, substitute these values back into the equation for \( BC \):
\( BC = \frac{5 \times \frac{1}{2}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \)
\( \implies \) \( BC = \frac{5}{2} \times \frac{2\sqrt{2}}{\sqrt{3} - 1} \)
\( \implies \) \( BC = \frac{5\sqrt{2}}{\sqrt{3} - 1} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\( BC = \frac{5\sqrt{2}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( \implies \) \( BC = \frac{5\sqrt{2}(\sqrt{3} + 1)}{3 - 1} \)
\( \implies \) \( BC = \frac{5\sqrt{2}(\sqrt{3} + 1)}{2} \text{ km} \)
The distance of the intruder from B is \( \frac{5\sqrt{2}(\sqrt{3} + 1)}{2} \text{ km} \). This problem showcases how trigonometry is used to solve real-world problems involving distances and angles, such as in surveying or navigation.
In simple words: We drew a triangle with the two soldiers and the intruder. We knew the distance between the soldiers and their angles of looking up at the intruder. We used these angles to find the third angle of the triangle. Then, using a rule called the sine formula, we calculated the distance from soldier B to the intruder. We made the final answer simpler by removing the square root from the bottom part of the fraction.

🎯 Exam Tip: Carefully identify the interior angles of the triangle formed. Remember that an angle of elevation might be part of a larger \( 180^\circ \) angle when considering the internal angles of a triangle on a flat plane. Rationalize denominators for final answers involving square roots.

Question 8. A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point P, he finds the distance to the eastern-most point of the pond to be \( 8 \text{ km} \), while the distance to the westernmost point from P to be \( 6 \text{ km} \). If the angle between the two lines of sight is \( 60^\circ \) find the width of the pond.
Answer: Let the eastern-most point of the pond be A and the western-most point be B. Let P be the point of observation. We are given the following information:
Distance from P to A (eastern-most point) = \( PA = 8 \text{ km} \).
Distance from P to B (western-most point) = \( PB = 6 \text{ km} \).
The angle between the two lines of sight from P to A and P to B is \( \angle APB = 60^\circ \).
We need to find the width of the pond, which is the distance \( AB \). We can form a triangle \( \triangle PAB \) with sides \( PA, PB \) and the included angle \( \angle APB \). We can use the cosine formula to find the side \( AB \).
The cosine formula for side \( AB \) (or \( p \)) is:
\( AB^2 = PA^2 + PB^2 - 2(PA)(PB) \cos(\angle APB) \)
Substitute the given values:
\( AB^2 = 8^2 + 6^2 - 2(8)(6) \cos 60^\circ \)
We know \( 8^2 = 64 \), \( 6^2 = 36 \), and \( \cos 60^\circ = \frac{1}{2} \).
\( AB^2 = 64 + 36 - 2(8)(6) \left( \frac{1}{2} \right) \)
\( \implies \) \( AB^2 = 100 - (96) \left( \frac{1}{2} \right) \)
\( \implies \) \( AB^2 = 100 - 48 \)
\( \implies \) \( AB^2 = 52 \)
Now, take the square root to find \( AB \):
\( AB = \sqrt{52} \)
To simplify \( \sqrt{52} \), we can write \( 52 = 4 \times 13 \):
\( AB = \sqrt{4 \times 13} \)
\( AB = 2\sqrt{13} \text{ km} \)
The width of the pond is \( 2\sqrt{13} \text{ km} \). This problem demonstrates a practical application of the cosine rule to measure inaccessible distances.
In simple words: We imagined a triangle where point P is where the researcher stood, and A and B are the edges of the pond. We knew the distances from P to A, from P to B, and the angle at P. Using the cosine formula, we calculated the length of the side connecting A and B, which is the width of the pond. The width is \( 2\sqrt{13} \) kilometers.

🎯 Exam Tip: The cosine rule is especially useful when you know two sides of a triangle and the angle between them (SAS), and you want to find the third side. Draw a clear diagram to visualize the problem and label all given information correctly.

Question 9. Two Navy helicopters A and B are flying over the Bay of Bengal at same altitude from sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart \( 10 \text{ km} \) from each other. If the distance of the boat from A is \( 6 \text{ km} \) and if the line segment AB subtends \( 60^\circ \) at the boat, find the distance of the boat from B.
Answer: Let A and B be the positions of the two helicopters. Let C be the position of the missing boat. We are given the following information:
Distance between helicopters \( AB = 10 \text{ km} \).
Distance of the boat from helicopter A is \( AC = 6 \text{ km} \).
The line segment AB subtends an angle of \( 60^\circ \) at the boat, which means \( \angle ACB = 60^\circ \).
We need to find the distance of the boat from helicopter B, which is side \( BC \). Let \( BC = a \). We can use the cosine formula in \( \triangle ABC \) to find side \( a \).
The cosine formula for side \( AB \) (or \( c \)) is:
\( AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos(\angle ACB) \)
Substitute the given values:
\( 10^2 = a^2 + 6^2 - 2(a)(6) \cos 60^\circ \)
We know \( 10^2 = 100 \), \( 6^2 = 36 \), and \( \cos 60^\circ = \frac{1}{2} \).
\( 100 = a^2 + 36 - 2(a)(6) \left( \frac{1}{2} \right) \)
\( \implies \) \( 100 = a^2 + 36 - 6a \)
Rearrange the terms to form a quadratic equation:
\( a^2 - 6a + 36 - 100 = 0 \)
\( \implies \) \( a^2 - 6a - 64 = 0 \)
We can solve this quadratic equation using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a_{quad}} \), where \( a_{quad}=1, b=-6, c=-64 \).
\( a = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-64)}}{2(1)} \)
\( \implies \) \( a = \frac{6 \pm \sqrt{36 + 256}}{2} \)
\( \implies \) \( a = \frac{6 \pm \sqrt{292}}{2} \)
We can simplify \( \sqrt{292} \): \( 292 = 4 \times 73 \), so \( \sqrt{292} = 2\sqrt{73} \).
\( \implies \) \( a = \frac{6 \pm 2\sqrt{73}}{2} \)
\( \implies \) \( a = 3 \pm \sqrt{73} \)
Since distance must be a positive value, we consider both possibilities:
Case 1: \( a = 3 + \sqrt{73} \). This is a positive value, approximately \( 3 + 8.54 = 11.54 \text{ km} \).
Case 2: \( a = 3 - \sqrt{73} \). Since \( \sqrt{73} \) is approximately \( 8.54 \), \( 3 - 8.54 \) would be a negative value, which is not possible for a distance.
Therefore, the distance of the boat from B is \( 3 + \sqrt{73} \text{ km} \). This problem is a good example of applying trigonometry to solve navigation problems.
In simple words: We had a triangle with the two helicopters and the boat. We knew the distance between the helicopters, the distance from one helicopter to the boat, and the angle the helicopters made at the boat. We used the cosine formula to set up an equation to find the distance from the second helicopter to the boat. Solving this equation gave us two answers, but only one made sense for a distance (it had to be a positive number). So, the distance from helicopter B to the boat is \( 3 + \sqrt{73} \) kilometers.

🎯 Exam Tip: When using the cosine rule to find a side length and you end up with a quadratic equation, always check both solutions. A distance must always be positive, so discard any negative results.

Question 10. A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and B of the tunnel to be built from a point P in front of the mountain. If \( AP = 3 \text{ km}, BP = 5 \text{ km} \), and \( \angle APB = 120^\circ \), then find the length of the tunnel to be built.
Answer: Let P be the observation point and A and B be the two ends of the tunnel. We are given the following information:
Distance from P to A = \( AP = 3 \text{ km} \).
Distance from P to B = \( BP = 5 \text{ km} \).
The angle between the lines of sight from P to A and P to B is \( \angle APB = 120^\circ \).
We need to find the length of the tunnel, which is the distance \( AB \). We can form a triangle \( \triangle PAB \) with sides \( AP, BP \) and the included angle \( \angle APB \). We can use the cosine formula to find the side \( AB \).
Let the length of the tunnel be \( c \). The cosine formula for side \( c \) is:
\( c^2 = AP^2 + BP^2 - 2(AP)(BP) \cos(\angle APB) \)
Substitute the given values:
\( c^2 = 3^2 + 5^2 - 2(3)(5) \cos 120^\circ \)
We know \( 3^2 = 9 \), \( 5^2 = 25 \), and \( \cos 120^\circ = -\frac{1}{2} \).
\( c^2 = 9 + 25 - 2(3)(5) \left( -\frac{1}{2} \right) \)
\( \implies \) \( c^2 = 34 - (30) \left( -\frac{1}{2} \right) \)
\( \implies \) \( c^2 = 34 + 15 \)
\( \implies \) \( c^2 = 49 \)
Now, take the square root to find \( c \):
\( c = \sqrt{49} \)
\( c = 7 \text{ km} \)
The length of the tunnel to be built is \( 7 \text{ km} \). This calculation shows how the cosine rule can be used to determine distances in inaccessible terrains, such as mountains.
In simple words: The surveyor made a triangle with the observation point P and the two ends of the tunnel, A and B. He knew the distances from P to A and P to B, and the angle at P. Using the cosine formula, he calculated the length of the tunnel (the side AB). The tunnel will be 7 kilometers long.

🎯 Exam Tip: Remember that \( \cos 120^\circ \) is a negative value (\( -\frac{1}{2} \)). A common mistake is to forget the negative sign, which will lead to an incorrect result. Always visualize the angle in the unit circle or recall its quadrant to determine the sign of the cosine value.

Question 11. A farmer wants to purchase a triangular-shaped land with sides \( 120 \text{ feet} \) and \( 60 \text{ feet} \) and the angle included between these two sides is \( 60^\circ \). If the land costs \( \text{Rs. } 500 \) per square feet, find the amount he needed to purchase the land. Also, find the perimeter of the land.
Answer: Let the triangular-shaped land be \( \triangle ABC \). We are given two sides and the included angle:
Side \( AB = 120 \text{ feet} \)
Side \( AC = 60 \text{ feet} \)
Included angle \( \angle BAC = 60^\circ \)
First, let's find the third side \( BC \) (let's call it \( a \)) using the cosine formula:
\( a^2 = AB^2 + AC^2 - 2(AB)(AC) \cos(\angle BAC) \)
\( a^2 = 120^2 + 60^2 - 2(120)(60) \cos 60^\circ \)
We know \( 120^2 = 14400 \), \( 60^2 = 3600 \), and \( \cos 60^\circ = \frac{1}{2} \).
\( a^2 = 14400 + 3600 - 2(120)(60) \left( \frac{1}{2} \right) \)
\( \implies \) \( a^2 = 18000 - 7200 \)
\( \implies \) \( a^2 = 10800 \)
To find \( a \), take the square root of \( 10800 \):
\( a = \sqrt{10800} \)
To simplify, factorize \( 10800 \): \( 10800 = 100 \times 108 = 100 \times 36 \times 3 = 10^2 \times 6^2 \times 3 \).
\( a = \sqrt{10^2 \times 6^2 \times 3} \)
\( a = 10 \times 6 \times \sqrt{3} \)
\( a = 60\sqrt{3} \text{ feet} \)

Next, let's find the perimeter of the land:
Perimeter \( = AB + AC + BC \)
Perimeter \( = 120 + 60 + 60\sqrt{3} \)
Perimeter \( = 180 + 60\sqrt{3} \)
Perimeter \( = 60(3 + \sqrt{3}) \text{ feet} \)

Finally, let's find the area of the land. We can use the formula Area \( \Delta = \frac{1}{2}ab \sin C \). In our case, this means Area \( \Delta = \frac{1}{2} (AB)(AC) \sin(\angle BAC) \).
Area \( \Delta = \frac{1}{2} \times 120 \times 60 \times \sin 60^\circ \)
We know \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
Area \( \Delta = \frac{1}{2} \times 120 \times 60 \times \frac{\sqrt{3}}{2} \)
\( \implies \) Area \( \Delta = 60 \times 60 \times \frac{\sqrt{3}}{2} \)
\( \implies \) Area \( \Delta = 3600 \times \frac{\sqrt{3}}{2} \)
\( \implies \) Area \( \Delta = 1800\sqrt{3} \text{ sq. feet} \)

Now, calculate the total cost of the land. The cost is \( \text{Rs. } 500 \) per square foot.
Total cost \( = \text{Area} \times \text{Cost per sq. foot} \)
Total cost \( = 1800\sqrt{3} \times 500 \)
Total cost \( = 900000\sqrt{3} \text{ Rs.} \)
The total amount needed to purchase the land is \( \text{Rs. } 900000\sqrt{3} \), and the perimeter of the land is \( 60(3 + \sqrt{3}) \text{ feet} \). This problem combines different geometric and trigonometric concepts to solve a practical scenario involving land measurement and cost.
In simple words: The farmer's land is a triangle. We knew two sides and the angle between them. First, we used a rule called the cosine formula to find the length of the third side. With all three sides, we could calculate the perimeter (the total distance around the land). Next, we found the area of the land using a special area formula for triangles. Finally, we multiplied the area by the cost per square foot to find the total money the farmer needs.

🎯 Exam Tip: Break down complex problems into smaller, manageable steps. First find the missing side, then perimeter and area, and finally the cost. Remember that \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).

Question 12. A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be \( 30^\circ \). If after \( 100 \text{ km} \), the target has an angle of depression of \( 45^\circ \), how far is the target from the fighter jet at that instant?
Answer: Let T be the position of the target. Let A be the initial position of the fighter jet, and B be its position after flying \( 100 \text{ km} \) horizontally. Let D be the point on the ground directly below A, and E be the point on the ground directly below B. The target T is on the ground. The horizontal distance covered by the jet is \( AB = 100 \text{ km} \). The initial angle of depression from A to T is \( 30^\circ \). This means \( \angle HAT = 30^\circ \), where H is a point on the horizontal line through A. Since \( HT \) is parallel to the ground (where T lies), \( \angle ATD = 30^\circ \). In \( \triangle ADT \), \( \angle TAD = 30^\circ \). The second angle of depression from B to T is \( 45^\circ \). This means \( \angle H'BT = 45^\circ \), where H' is on the horizontal line through B. So \( \angle BTD = 45^\circ \). In \( \triangle BET \), \( \angle TBE = 45^\circ \). Consider the triangle formed by the jet's two positions (A, B) and the target (T). In \( \triangle ABT \): The distance between A and B is \( AB = 100 \text{ km} \). The angle at A, \( \angle TAB \), is the angle of depression \( 30^\circ \) with respect to the horizontal line of flight. If we consider the horizontal path of the jet, then the angle \( \angle AT \) makes with \( AB \) is \( 30^\circ \). The angle at B, \( \angle TBA \), is \( 180^\circ - 45^\circ = 135^\circ \) (since the angle of depression from B to T is \( 45^\circ \), the angle formed by the horizontal flight path and line BT inside the triangle is \( 180^\circ - 45^\circ \)). Now, let's find the third angle, \( \angle ATB \): \( \angle ATB = 180^\circ - (\angle TAB + \angle TBA) \) \( \angle ATB = 180^\circ - (30^\circ + 135^\circ) \) \( \angle ATB = 180^\circ - 165^\circ \) \( \angle ATB = 15^\circ \) We need to find the distance of the target from the fighter jet at that instant, which is \( BT \). We can use the sine formula in \( \triangle ABT \): \( \frac{BT}{\sin(\angle TAB)} = \frac{AB}{\sin(\angle ATB)} \) \( \frac{BT}{\sin 30^\circ} = \frac{100}{\sin 15^\circ} \)
\( BT = \frac{100 \times \sin 30^\circ}{\sin 15^\circ} \)
We know \( \sin 30^\circ = \frac{1}{2} \).
And \( \sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \).
\( \implies \) \( \sin 15^\circ = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2} \right) \)
\( \implies \) \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
Substitute these values back into the equation for \( BT \):
\( BT = \frac{100 \times \frac{1}{2}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \)
\( \implies \) \( BT = \frac{50}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \)
\( \implies \) \( BT = 50 \times \frac{2\sqrt{2}}{\sqrt{3} - 1} \)
\( \implies \) \( BT = \frac{100\sqrt{2}}{\sqrt{3} - 1} \)
To rationalize the denominator, multiply numerator and denominator by \( \sqrt{3} + 1 \):
\( BT = \frac{100\sqrt{2}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( \implies \) \( BT = \frac{100\sqrt{2}(\sqrt{3} + 1)}{3 - 1} \)
\( \implies \) \( BT = \frac{100\sqrt{2}(\sqrt{3} + 1)}{2} \)
\( \implies \) \( BT = 50\sqrt{2}(\sqrt{3} + 1) \text{ km} \)
The distance of the target from the fighter jet at that instant (position B) is \( 50\sqrt{2}(\sqrt{3} + 1) \text{ km} \). This problem is a classic application of the sine rule in real-world scenarios like air navigation.
In simple words: The jet flew a certain distance, and at two points (A and B), the pilot looked down at a target. We knew the distance the jet flew and the two "angles of depression" (how much the pilot had to look down). We used these angles to find the angles inside the triangle formed by the two jet positions and the target. Then, using the sine formula, we calculated the distance from the jet's second position (B) to the target.

🎯 Exam Tip: When dealing with angles of depression or elevation, always draw a clear diagram and remember that the angle of depression from a horizontal line to an object is equal to the angle of elevation from the object to that horizontal line (alternate interior angles). This helps correctly identify angles within your triangle.

Question 13. A plane is \( 1 \text{ km} \) from one landmark and \( 2 \text{ km} \) from another. From the plane's point of view, the land between them subtends an angle of \( 45^\circ \). How far apart are the landmarks?
Answer: Let C be the position of the plane. Let A and B be the two landmarks. We are given the following information:
Distance from the plane to landmark A = \( CA = 1 \text{ km} \).
Distance from the plane to landmark B = \( CB = 2 \text{ km} \).
The angle subtended by the land between them at the plane is \( \angle ACB = 45^\circ \). This means the angle at the plane between the lines of sight to the two landmarks is \( 45^\circ \).
We need to find the distance between the landmarks, which is \( AB \). We can form a triangle \( \triangle ABC \) with sides \( CA, CB \) and the included angle \( \angle ACB \). We can use the cosine formula to find the side \( AB \).
The cosine formula for side \( AB \) (let's call it \( c \)) is:
\( AB^2 = CA^2 + CB^2 - 2(CA)(CB) \cos(\angle ACB) \)
Substitute the given values:
\( AB^2 = 1^2 + 2^2 - 2(1)(2) \cos 45^\circ \)
We know \( 1^2 = 1 \), \( 2^2 = 4 \), and \( \cos 45^\circ = \frac{1}{\sqrt{2}} \).
\( AB^2 = 1 + 4 - 2(1)(2) \left( \frac{1}{\sqrt{2}} \right) \)
\( \implies \) \( AB^2 = 5 - 4 \left( \frac{1}{\sqrt{2}} \right) \)
\( \implies \) \( AB^2 = 5 - \frac{4}{\sqrt{2}} \)
To rationalize \( \frac{4}{\sqrt{2}} \), multiply numerator and denominator by \( \sqrt{2} \): \( \frac{4\sqrt{2}}{2} = 2\sqrt{2} \).
\( \implies \) \( AB^2 = 5 - 2\sqrt{2} \)
Now, take the square root to find \( AB \):
\( AB = \sqrt{5 - 2\sqrt{2}} \text{ km} \)
The landmarks are \( \sqrt{5 - 2\sqrt{2}} \text{ km} \) apart. This is another example of using the cosine rule to find an unknown distance when direct measurement is not possible.
In simple words: The plane, along with the two landmarks, forms a triangle. We know the distance from the plane to each landmark, and the angle made at the plane by looking at both landmarks. We used the cosine formula to find the distance between the two landmarks. The distance between them is \( \sqrt{5 - 2\sqrt{2}} \) kilometers.

🎯 Exam Tip: Pay attention to the terms "subtends an angle". It means the angle is formed at the point specified (in this case, the plane) by the lines connecting to the endpoints of the segment (the landmarks). Simplification of square roots and rationalization of denominators are important final steps.

Question 14. A man starts his morning walk at a point A, reaches two points B and C, and finally returns to A such that \( \angle A = 60^\circ \) and \( \angle B = 45^\circ \). Given that AC = 4 km in the \( \Delta ABC \), find the total distance he covered during his morning walk.
Answer: In \( \Delta ABC \), we are given \( \angle A = 60^\circ \), \( \angle B = 45^\circ \), and side \( AC = 4 \) km.
First, find \( \angle C \):
\( \angle C = 180^\circ - (\angle A + \angle B) \)
\( \angle C = 180^\circ - (60^\circ + 45^\circ) \)
\( \angle C = 180^\circ - 105^\circ \)
\( \angle C = 75^\circ \)
Now, use the sine formula to find the lengths of sides AB and BC.
\( \frac{BC}{\sin A} = \frac{AC}{\sin B} = \frac{AB}{\sin C} \)
To find BC:
\( \frac{BC}{\sin 60^\circ} = \frac{AC}{\sin 45^\circ} \)
\( \frac{BC}{\sin 60^\circ} = \frac{4}{\sin 45^\circ} \)
\( BC = \frac{4 \sin 60^\circ}{\sin 45^\circ} \)
\( BC = \frac{4 \times \frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} \)
\( BC = 4 \times \frac{\sqrt{3}}{2} \times \sqrt{2} \)
\( BC = 2\sqrt{6} \) km
To find AB:
\( \frac{AB}{\sin C} = \frac{AC}{\sin B} \)
\( \frac{AB}{\sin 75^\circ} = \frac{4}{\sin 45^\circ} \)
\( AB = \frac{4 \sin 75^\circ}{\sin 45^\circ} \)
We know \( \sin 75^\circ = \sin (45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
\( = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3}+1}{2\sqrt{2}} \)
\( AB = \frac{4 \times \frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}}} \)
\( AB = 4 \times \frac{\sqrt{3}+1}{2\sqrt{2}} \times \sqrt{2} \)
\( AB = 2(\sqrt{3}+1) \) km
The total distance covered is the perimeter of the triangle: \( AB + BC + AC \)
Total distance \( = 2(\sqrt{3}+1) + 2\sqrt{6} + 4 \)
Total distance \( = 2\sqrt{3} + 2 + 2\sqrt{6} + 4 \)
Total distance \( = 6 + 2\sqrt{3} + 2\sqrt{6} \) km. The man walks around the triangle, so the total distance is the sum of all its sides.
In simple words: First, find the third angle of the triangle. Then, use the sine rule to calculate the lengths of the two unknown sides, AB and BC. Finally, add all three side lengths together to get the total distance the man walked.

🎯 Exam Tip: Remember that \( \sin 75^\circ \) and \( \cos 75^\circ \) values are often needed in trigonometry questions and can be derived from sum/difference formulas if not remembered.

Question 15. Two vehicles leave the same place P at the same time, moving along two different roads. One vehicle moves at an average speed of 60 km/hr and the other vehicle moves at an average speed of 80 km/hr. After half an hour, the vehicles reach destinations A and B. If the line segment AB subtends 60° at the initial point P, then find AB.
Answer: Let P be the initial point. The vehicles travel for half an hour.
Distance covered by the first vehicle (PA):
Speed = 60 km/hr, Time = 0.5 hr
\( PA = \text{Speed} \times \text{Time} = 60 \times 0.5 = 30 \) km
Distance covered by the second vehicle (PB):
Speed = 80 km/hr, Time = 0.5 hr
\( PB = \text{Speed} \times \text{Time} = 80 \times 0.5 = 40 \) km
The angle subtended by AB at P is \( \angle APB = 60^\circ \). We need to find the length of AB.
We can use the cosine formula in \( \Delta PAB \):
\( AB^2 = PA^2 + PB^2 - 2(PA)(PB) \cos(\angle APB) \)
\( AB^2 = 30^2 + 40^2 - 2(30)(40) \cos(60^\circ) \)
\( AB^2 = 900 + 1600 - 2(1200)\left(\frac{1}{2}\right) \)
\( AB^2 = 2500 - 1200 \)
\( AB^2 = 1300 \)
\( AB = \sqrt{1300} \)
\( AB = \sqrt{100 \times 13} \)
\( AB = 10\sqrt{13} \) km. This represents the direct distance between the two vehicles after half an hour.
In simple words: First, calculate how far each vehicle traveled using their speed and time. Then, use the cosine rule with these distances and the given angle to find the direct distance between the two vehicles.

🎯 Exam Tip: When using the cosine rule, ensure the angle used is the one opposite the side you are trying to find.

Question 16. Suppose that a satellite in space, an earth station, and the center of Earth all lie in the same plane. Let r be the radius of Earth and R be the distance from the center of Earth to the satellite. Let d be the distance from the earth station to the satellite. Let 30° be the angle of elevation from the earth station to the satellite. If the line segment connecting the earth station and satellite subtends angle \( \alpha \) at the center of Earth, then prove that \( d=R\sqrt{1+\left(\frac{r}{R}\right)^{2}-2\left(\frac{r}{R}\right) \cos \alpha} \).
Answer: Let O be the center of Earth, A be the Earth station, and S be the satellite. The points O, A, S lie in the same plane, forming a triangle \( \Delta OAS \).
From the problem description:
\( OA = r \) (radius of Earth)
\( OS = R \) (distance from center of Earth to satellite)
\( AS = d \) (distance from Earth station to satellite)
The angle subtended by the line segment AS at the center of Earth is \( \angle AOS = \alpha \).
We can use the cosine formula for \( \Delta OAS \):
\( AS^2 = OA^2 + OS^2 - 2(OA)(OS) \cos(\angle AOS) \)
Substitute the given values into the formula:
\( d^2 = r^2 + R^2 - 2(r)(R) \cos \alpha \)
To match the required form, we can factor out \( R^2 \) from the right side of the equation:
\( d^2 = R^2 \left( \frac{r^2}{R^2} + \frac{R^2}{R^2} - \frac{2rR}{R^2} \cos \alpha \right) \)
\( d^2 = R^2 \left( \left(\frac{r}{R}\right)^2 + 1 - 2\left(\frac{r}{R}\right) \cos \alpha \right) \)
Rearranging the terms inside the parenthesis:
\( d^2 = R^2 \left( 1 + \left(\frac{r}{R}\right)^2 - 2\left(\frac{r}{R}\right) \cos \alpha \right) \)
Now, take the square root on both sides to find d:
\( d = \sqrt{R^2 \left( 1 + \left(\frac{r}{R}\right)^2 - 2\left(\frac{r}{R}\right) \cos \alpha \right)} \)
\( d = R\sqrt{1 + \left(\frac{r}{R}\right)^2 - 2\left(\frac{r}{R}\right) \cos \alpha} \)
This proves the given relationship. The cosine rule is a fundamental tool for finding unknown sides or angles in a triangle when other information is known.
In simple words: Imagine a triangle formed by the Earth's center, the satellite, and the Earth station. Use the cosine rule on this triangle, replacing the sides with r, R, and d, and the angle at the Earth's center with \( \alpha \). Then, rearrange the equation to get d by itself, which will match the formula we need to prove.

🎯 Exam Tip: When proving an identity, start with a known formula (like the cosine rule) and systematically manipulate it algebraically until it matches the target identity. Factoring out common terms like \( R^2 \) is often key in such proofs.