Class 10 Mathematics: Applications of Trigonometry important questions

Applications of Trigonometry is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 15 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. Draw a rough diagram: a person observes the top of a tree 10 m from its foot at an angle of elevation of 45°.

   Draw a vertical line AB for the tree (A at the top). Mark the person at C on the ground with BC = 10 m. Join CA; the angle ACB = 45° and the angle at B is 90°. Since tan 45° = 1, the tree is 10 m tall.

2. Very Important · 8 marks

   2. A straight highway leads to the foot of a tower. Ramaiah, at the top of the tower, sees a car at an angle of depression of 30°. The car approaches the tower at a uniform speed. Six seconds later the angle of depression is 60°. Find the time taken by the car to reach the foot of the tower from this point.

   Let the tower height be h. At 30°: distance from the foot = h/tan 30° = h√3. At 60°: distance = h/tan 60° = h/√3. Distance covered in 6 s = h√3 − h/√3 = 2h/√3, so speed = (2h/√3)/6 = h/(3√3) per second. Time for the remaining h/√3 = (h/√3) ÷ (h/(3√3)) = 3 seconds.

3. Practice · 1 marks

   3. A person standing 20 m from the base of a building sees the top at an angle of elevation of 45°. Find the height of the building.

   tan 45° = height/20, and tan 45° = 1, so the height is 20 m.

4. Important · 2 marks

   4. Draw a diagram: a boy observes the top of an electric pole at an angle of elevation of 60° from a point 8 m from the foot of the pole.

   Draw a vertical line AB for the pole (A at the top, B at the foot). Mark the boy at C on the ground, with BC = 8 m. Join CA. The angle ACB = 60° and the angle at B is 90°. (The height would be AB = 8 tan 60° = 8√3 m.)

5. Practice · 1 marks

   5. True or false: in heights and distances problems, angles are always taken with respect to the horizontal line.

   True. Angles of elevation and depression are both measured from the horizontal line through the observer's eye.

6. Important · 2 marks

   6. Define the angle of elevation, with a rough diagram.

   The angle of elevation is the angle between the horizontal line through the observer's eye and the line of sight to an object above that horizontal level. Diagram: an observer at O looks up at the top P of a tower. Draw the horizontal line OA and the line of sight OP; the angle AOP is the angle of elevation.

7. Very Important · 8 marks

   7. From a point on a bridge across a river, the angles of depression of the banks on opposite sides are 30° and 45°. The bridge is 3 m above the banks. Find the width of the river.

   Let P be the point on the bridge, 3 m above the ground, and A, B the banks on either side. For the 45° bank: tan 45° = 3/x₁, so x₁ = 3 m. For the 30° bank: tan 30° = 3/x₂, so x₂ = 3√3 m. Width = x₁ + x₂ = 3 + 3√3 = 3(1 + √3) m ≈ 8.20 m.

8. Practice · 1 marks

   8. Which situation involves an angle of depression? A) A bird looking up at the sky. B) A man looking at the top of a tower in front of him. C) A girl looking down at a car from the balcony. D) A person looking at the top of a tree in front of him.

   C) The girl looks down from a higher level, so the angle between the horizontal and her line of sight is an angle of depression.

9. Important · 2 marks

   9. Name any two real-life situations where trigonometry is used.

   1) Finding the height of a tall building, tower or tree without climbing it. 2) Finding the distance between two ships, or of a ship from a lighthouse. (Others: surveying land, navigation, astronomy.)

10. Practice · 1 marks

    10. A group is finding the height of a tower using trigonometric ratios. Arrange these steps in the correct order: 1) Calculate the height using the trigonometric ratio 2) Draw and label the right triangle 3) Identify the given measurements 4) Choose the appropriate trigonometric ratio A) 3, 2, 4, 1 B) 2, 3, 1, 4 C) 4, 3, 2, 1 D) 3, 4, 2, 1

    A) 3, 2, 4, 1 First identify the given measurements, then draw and label the right triangle, then choose the ratio that links the known and unknown sides, and finally calculate the height.

11. Important · 2 marks

    11. A person on top of a tree of height h, standing at one side of a river, observes the two banks of the river at angles of depression θ₁ and θ₂ (θ₁ < θ₂). The width of the river is d. Draw a rough diagram for this situation.

    Draw a vertical line AB for the tree (A at the top, height h). Draw the ground line through B. Mark C on the near bank and D on the far bank, so CD = d. The angle of depression to the near bank C is the larger angle θ₂, and to the far bank D is the smaller angle θ₁. These equal the angles of elevation ∠ACB = θ₂ and ∠ADB = θ₁ at the banks.

12. Very Important · 8 marks

    12. The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

    Let the multi-storeyed building have height h m and let the distance between the buildings be x m. From the bottom of the small building (angle 45°): tan 45° = h/x, so x = h. From the top of the small building (angle 30°): tan 30° = (h − 8)/x, so 1/√3 = (h − 8)/h. h = √3(h − 8), so h(√3 − 1) = 8√3 and h = 8√3/(√3 − 1) = 8√3(√3 + 1)/2 = 4(3 + √3). Height = 4(3 + √3) m ≈ 18.93 m, and the distance between the buildings is also 4(3 + √3) m ≈ 18.93 m.

13. Practice · 1 marks

    13. Two students measure the height of a tree. One gets 20 m and the other gets 22 m. What should they do before deciding which answer is correct? A) Accept the first answer only B) Accept the second answer only C) Recheck the measurements and calculations systematically D) Ignore both answers

    C) Recheck the measurements and calculations systematically.

14. Important · 2 marks

    14. Draw a figure for this situation: a student observes the top of a tower at an angle of elevation of 45°.

    Draw a vertical line AB for the tower (A at the top). Mark the student's eye at point C on the ground line through B. Join CA (the line of sight). The angle ACB between the horizontal CB and the line of sight CA is 45°. Since tan 45° = 1, the height of the tower equals the distance CB.

15. Very Important · 8 marks

    15. A 1.2 m tall girl sees a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation from her eyes is 60°. After some time it reduces to 30°. Find the distance travelled by the balloon in that interval.

    Height of the balloon above her eyes = 88.2 − 1.2 = 87 m. First position: tan 60° = 87/x₁, so x₁ = 87/√3 = 29√3 m. Second position: tan 30° = 87/x₂, so x₂ = 87√3 m. Distance travelled = x₂ − x₁ = 87√3 − 29√3 = 58√3 m ≈ 100.46 m.