Samacheer Kalvi Class 10 Maths Solutions Chapter 6 Trigonometry Exercise 6.4

Question 1. From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree, \( (\sqrt { 3 } = 1.732) \)
Answer:Let the height of the second tree be “h” m. Let the horizontal distance between the trees be "x" m. The first tree (AC) is 13 m tall. The second tree (BD) has height 'h'. A horizontal line CE is drawn from the top of the first tree (C) to the second tree. So, BE = AC = 13 m. The remaining height of the second tree is DE = h - 13 m.In the right-angled triangle ABC: \( \tan 30^\circ = \frac { BC }{ AB } \) \( \frac { 1 }{ \sqrt { 3 } } = \frac { 13 }{ x } \)
\( \implies x = 13 \sqrt { 3 } \) .....(1) In the right-angled triangle CDE: \( \tan 45^\circ = \frac { DE }{ EC } \) \( 1 = \frac { h-13 }{ x } \)
\( \implies x = h - 13 \) .....(2) From equations (1) and (2), we can set them equal to each other: \( h - 13 = 13 \sqrt { 3 } \)
\( \implies h = 13 \sqrt { 3 } + 13 \) Now, substitute the given value for \( \sqrt{3} \): \( h = 13 \times 1.732 + 13 \) \( h = 22.516 + 13 \) \( h = 35.516 \) Rounding to two decimal places: \( h = 35.52 \) m. Thus, the height of the second tree is 35.52 meters. This problem shows how trigonometry helps measure heights of tall objects indirectly.In simple words: We used angles and the height of the first tree to find the distance between the trees. Then, we used that distance with another angle to calculate the full height of the second tree.

🎯 Exam Tip: Always draw a clear diagram and label all known values and angles. This helps in correctly identifying the triangles and applying the correct trigonometric ratios.

Question 2. A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30° . Calculate the distance of the hill from the ship and the height of the hill. \( (\sqrt { 3 } = 1.732) \)
Answer:Let the height of the hill BE be “h” m and the distance of the hill from the ship be "x" m. The man is at point A, 40 m above the water level (AB = 40 m). A horizontal line AD is drawn from the observation point A to the hill, where D is on the vertical line BE. So, AD = BC = x (distance to the hill). CD = AB = 40 m. The total height of the hill is BE = BD + DE = 40 + DE.In the right-angled triangle ADC (for depression of the base of the hill): \( \tan 30^\circ = \frac { CD }{ AD } \) \( \frac { 1 }{ \sqrt { 3 } } = \frac { 40 }{ x } \)
\( \implies x = 40 \sqrt { 3 } \) .....(1) In the right-angled triangle ADE (for elevation of the top of the hill): \( \tan 60^\circ = \frac { DE }{ AD } \) \( \sqrt { 3 } = \frac { BE - BD }{ x } = \frac { h - 40 }{ x } \)
\( \implies x = \frac { h - 40 }{ \sqrt { 3 } } \) .....(2) From equations (1) and (2), we equate the expressions for x: \( \frac { h - 40 }{ \sqrt { 3 } } = 40 \sqrt { 3 } \) Multiply both sides by \( \sqrt { 3 } \): \( h - 40 = 40 \times 3 \) \( h - 40 = 120 \) \( h = 120 + 40 \) \( h = 160 \) m. So, the height of the hill is 160 m. Now, calculate the distance of the hill from the ship using equation (1): \( x = 40 \sqrt { 3 } \) Substitute the value \( \sqrt { 3 } = 1.732 \): \( x = 40 \times 1.732 \) \( x = 69.28 \) m. Therefore, the height of the hill is 160 m, and its distance from the ship is 69.28 m.In simple words: We found the distance to the hill using the angle of depression to its base. Then, we used that distance and the angle of elevation to find the height of the hill.

🎯 Exam Tip: When dealing with elevation and depression from a point, always draw a horizontal line through the observation point to form the right-angled triangles.

Question 3. If the angle of elevation of a cloud from a point 'h' metres above a lake is \( \theta_1 \) and the angle of depression of its reflection in the lake is \( \theta_2 \). Prove that the height that the cloud is located from the ground is \( \frac{h\left(\tan \theta_{1}+\tan \theta_{2}\right)}{\tan \theta_{2}-\tan \theta_{1}} \)
Answer:Let P be the cloud and Q be its reflection in the lake. Let A be the point of observation such that AB = h, where B is on the lake surface. Let the height of the cloud from the lake surface be x. So, PS = x. Since Q is the reflection of P, its depth below the lake surface is also x. So, SQ = x. Draw a horizontal line AR from the observation point A to the vertical line PSQ. So, RS = AB = h. Then, PR = PS - RS = x - h. And, RQ = RS + SQ = h + x. Let AR = y.In the right-angled triangle PAR (for the cloud P): \( \tan \theta_1 = \frac { PR }{ AR } = \frac { x - h }{ y } \) .....(1) In the right-angled triangle AQR (for the reflection Q): \( \tan \theta_2 = \frac { RQ }{ AR } = \frac { x + h }{ y } \) .....(2) From equation (1), \( y = \frac { x - h }{ \tan \theta_1 } \) From equation (2), \( y = \frac { x + h }{ \tan \theta_2 } \) Equating the expressions for y: \( \frac { x - h }{ \tan \theta_1 } = \frac { x + h }{ \tan \theta_2 } \) Cross-multiply: \( (x - h) \tan \theta_2 = (x + h) \tan \theta_1 \) Expand both sides: \( x \tan \theta_2 - h \tan \theta_2 = x \tan \theta_1 + h \tan \theta_1 \) Gather terms with x on one side and terms with h on the other: \( x \tan \theta_2 - x \tan \theta_1 = h \tan \theta_1 + h \tan \theta_2 \) Factor out x and h: \( x (\tan \theta_2 - \tan \theta_1) = h (\tan \theta_1 + \tan \theta_2) \) Solve for x (the height of the cloud from the lake): \( x = \frac { h (\tan \theta_1 + \tan \theta_2) }{ \tan \theta_2 - \tan \theta_1 } \) This is the required proof, showing how the cloud's height can be determined using these angles. This method is useful for finding heights of objects that are hard to reach.In simple words: We used trigonometry with the angles of the cloud and its reflection to create two equations. By solving these equations together, we found a formula for the cloud's height based on the observation point's height and the two angles.

🎯 Exam Tip: Remember that the reflection of an object is as far below the surface as the object is above it. This principle is key for setting up the distances correctly in problems involving reflections.

Question 4. The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30° . If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.
Answer:Let the height of the cell phone tower be “h” m (BC = h). The height of the apartment AD is 50 m. Let the horizontal distance between the apartment and the tower be "x" m (AB = x). Draw a horizontal line DE from the top of the apartment (D) to the tower (BC). So, DE = AB = x. BE = AD = 50 m. Then, EC = BC - BE = h - 50 m.In the right-angled triangle ABC (angle of elevation from apartment foot to tower top): \( \tan 60^\circ = \frac { BC }{ AB } \) \( \sqrt { 3 } = \frac { h }{ x } \)
\( \implies x = \frac { h }{ \sqrt { 3 } } \) .....(1) In the right-angled triangle DEB (angle of depression from apartment top to tower foot): The angle of depression of B from D is \( \angle EDB = 30^\circ \). Due to alternate interior angles, \( \angle ABD = 30^\circ \). \( \tan 30^\circ = \frac { DE }{ BE } \) -- (Correction: the diagram shows angle ADB as 30 deg, which is consistent with the wording 'depression of the foot of the tower from the top of the apartment'. In triangle ABD, tan 30 = AD/AB) Let's follow the provided solution's triangles which align with the diagram. In the right-angled triangle ABD (using the 30° depression angle): \( \tan 30^\circ = \frac { AD }{ AB } \) \( \frac { 1 }{ \sqrt { 3 } } = \frac { 50 }{ x } \)
\( \implies x = 50 \sqrt { 3 } \) .....(2) From equations (1) and (2), we equate the expressions for x: \( \frac { h }{ \sqrt { 3 } } = 50 \sqrt { 3 } \) Multiply both sides by \( \sqrt { 3 } \): \( h = 50 \times (\sqrt { 3 } \times \sqrt { 3 }) \) \( h = 50 \times 3 \) \( h = 150 \) m. The height of the cell phone tower is 150 m. Now, we check if this height meets the radiation norms: Minimum height required = 120 m. Calculated height of the tower = 150 m. Since 150 m > 120 m, the cell phone tower meets the radiation norms. Cell phone towers need to be tall to ensure good signal coverage and reduce radiation exposure on the ground.In simple words: We used the angles and the apartment's height to find the distance to the tower. Then, we used that distance and the angle from the ground to find the tower's full height. We checked if the tower height meets the safety rules.

🎯 Exam Tip: Clearly distinguish between the two right-angled triangles formed by the lines of sight. Ensure you use the correct height and base for each tangent calculation, and remember to check the final answer against any given conditions.

Question 5. The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find
(i) the height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. \( (\sqrt { 3 } = 1.732) \)
Answer:Let the height of the lamp post AE be “h” m. The height of the apartment is 66 m (AD = 66 m). Let the horizontal distance between the apartment and the lamp post be "x" m (AB = x). Draw a horizontal line DC from the top of the apartment (D) to the lamp post (AE). So, DC = AB = x. BC = AD = 66 m. Then, CE = AE - BC = h - 66 m.In the right-angled triangle DBC (for depression of the lamp post bottom from apartment top): \( \tan 30^\circ = \frac { BC }{ DC } \) \( \frac { 1 }{ \sqrt { 3 } } = \frac { 66 }{ x } \)
\( \implies x = 66 \sqrt { 3 } \) .....(1) In the right-angled triangle DEC (for elevation of the lamp post top from apartment top): \( \tan 60^\circ = \frac { CE }{ DC } \) \( \sqrt { 3 } = \frac { h - 66 }{ x } \)
\( \implies x = \frac { h - 66 }{ \sqrt { 3 } } \) .....(2) From equations (1) and (2), we equate the expressions for x: \( \frac { h - 66 }{ \sqrt { 3 } } = 66 \sqrt { 3 } \) Multiply both sides by \( \sqrt { 3 } \): \( h - 66 = 66 \times (\sqrt { 3 } \times \sqrt { 3 }) \) \( h - 66 = 66 \times 3 \) \( h - 66 = 198 \) \( h = 198 + 66 \) \( h = 264 \) m. Now, we can answer the specific parts of the question: (i) The height of the lamp post: The height of the lamp post is \( h = 264 \) m. (ii) The difference between the height of the lamp post and the apartment: Difference \( = \) Height of lamp post \( - \) Height of apartment Difference \( = 264 - 66 = 198 \) m. (iii) The distance between the lamp post and the apartment: From equation (1): \( x = 66 \sqrt { 3 } \) Substitute \( \sqrt { 3 } = 1.732 \): \( x = 66 \times 1.732 \) \( x = 114.312 \) m. Rounding to two decimal places, the distance is 114.31 m. This demonstrates how relative heights and distances can be calculated using angles from an observation point.In simple words: First, we used the angle of depression to find the distance between the apartment and the lamp post. Then, using this distance and the angle of elevation, we calculated the lamp post's total height. We also found the height difference and the exact distance between them.

🎯 Exam Tip: Pay close attention to which height (total, or partial) corresponds to which triangle and angle. Double-check your calculations, especially when using approximate values for square roots.

Question 6. Three villagers A, B and C can see each other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate:
(i) the vertical height between A and B.
(ii) the vertical height between B and C. \( (\tan 20° = 0 .3640, \sqrt { 3 } = 1.732) \)
Answer:Let AD be the vertical height between A and B. Let CE be the vertical height between B and C. The horizontal distance between A and B is 8 km. The horizontal distance between B and C is 12 km.(i) The vertical height between A and B: In the right-angled triangle ABD (formed by A, B, and the horizontal line from A): The horizontal distance between A and B is 8 km (BD in this triangle, AB in the text). The angle of depression of B from A is \( \angle DAB = 20^\circ \). \( \tan 20^\circ = \frac { AD }{ \text{Horizontal distance AB} } \) \( \tan 20^\circ = \frac { AD }{ 8 } \) We are given \( \tan 20^\circ = 0.3640 \). \( 0.3640 = \frac { AD }{ 8 } \)
\( \implies AD = 0.3640 \times 8 \) \( AD = 2.912 \) km. So, the vertical height between A and B is 2.91 km (rounded to two decimal places). (ii) The vertical height between B and C: In the right-angled triangle BCE (formed by B, C, and the horizontal line from B): The horizontal distance between B and C is 12 km (BE in this triangle). The angle of elevation of C from B is \( \angle CBE = 30^\circ \). \( \tan 30^\circ = \frac { CE }{ BE } \) \( \tan 30^\circ = \frac { CE }{ 12 } \) We know \( \tan 30^\circ = \frac { 1 }{ \sqrt { 3 } } \). \( \frac { 1 }{ \sqrt { 3 } } = \frac { CE }{ 12 } \)
\( \implies CE = \frac { 12 }{ \sqrt { 3 } } \) To remove \( \sqrt{3} \) from the denominator, multiply by \( \frac { \sqrt { 3 } }{ \sqrt { 3 } } \): \( CE = \frac { 12 \times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } = \frac { 12 \sqrt { 3 } }{ 3 } \) \( CE = 4 \sqrt { 3 } \) Substitute the given value \( \sqrt { 3 } = 1.732 \): \( CE = 4 \times 1.732 \) \( CE = 6.928 \) km. Rounding to two decimal places, the vertical height between B and C is 6.93 km. These calculations help understand the terrain's features in a valley.In simple words: We used the horizontal distance and the angle of depression from A to B to find the vertical height difference between them. Then, we used the horizontal distance and the angle of elevation from B to C to find the vertical height difference between B and C.

🎯 Exam Tip: For problems involving multiple points and angles, break them down into separate right-angled triangles. Ensure you correctly identify the opposite and adjacent sides for each angle when applying tangent ratios.