1. The line drawn from the eye of an observer to the point in the object being viewed is called:
(a) angle of elevation (b) angle of depression (c) line of sight (d) horizontal level
2. When an object is above the horizontal level, the angle formed by the line of sight with the horizontal is called:
(a) angle of depression (b) angle of elevation (c) right angle (d) obtuse angle
3. In Example 1 of the chapter, if the distance from the foot of the tower is 15 m and angle of elevation is 60°, the height of the tower is:
(a) 15 m (b) 15√3 m (c) 15/√3 m (d) 30 m
4. Assertion (A): In a right triangle formed by line of sight, the tangent ratio is used when the opposite side (height) and adjacent side (distance) are involved.
Reason (R): tan θ = opposite side / adjacent side.
Options: (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true.
5. The angle of depression is formed when:
(a) the line of sight is above the horizontal (b) the line of sight is below the horizontal (c) the observer looks straight (d) the object is at the same level
6. In the method to find height of a minar (Fig. 9.1), BD equals:
(a) height of student AE (b) distance DE (c) total height CD (d) length of line of sight
7. Assertion (A): When the angle of elevation changes from 30° to 60°, the height of an object can be found using two right triangles sharing a common side.
Reason (R): tan 30° = 1/√3 and tan 60° = √3 are used to set up equations for the unknown height.
Options: (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true.
8. Which trigonometric ratio is used in Example 2 to find the length of the ladder?
(a) cos 60° (b) sin 60° (c) tan 60° (d) cot 30°
9. If the angle of elevation is 45°, then the height of the object equals:
(a) the distance from the observer (b) twice the distance (c) half the distance (d) zero
10. In Fig. 9.3, the angle formed by the line of sight with the horizontal when viewing downward is:
(a) angle of elevation (b) angle of depression (c) right angle (d) complementary angle
11. Define angle of elevation and angle of depression with reference to the line of sight and horizontal level.
12. In a right triangle, if tan A = BC/AB, name the sides opposite and adjacent to angle A.
13. From the chapter, state the three pieces of information needed to find the height of a minar without measuring it directly.
14. In Example 3, why is AE added to 1.5 m to obtain the height of the chimney?
15. If the angle of elevation is 30° and distance is 20 m, write the expression for height using tan 30°.
16. State the relationship between angles of elevation and depression when the line of sight is a transversal to parallel lines (as in Example 6).
17. A tower stands vertically. From a point 20 m away, the angle of elevation of the top is 60°. Find the height of the tower.
18. An observer 1.6 m tall is 20 m away from a pole. The angle of elevation of the top of the pole from her eyes is 45°. Find the height of the pole.
19. From a point on the ground, the angle of elevation of the top of a 12 m tall building is 30°. Find the distance of the point from the building.
20. The shadow of a tower is 30 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
21. A 1.5 m tall boy observes the top of a building. The angle of elevation increases from 30° to 60° as he walks 20 m towards the building. Verify the height of the building using the given data.
22. From a point P on the ground, the angle of elevation of the top of a 15 m tall building is 30°. A flagstaff is fixed at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from P. (Use √3 = 1.732)
23. The angles of depression of the top and bottom of a 10 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.
24. From a point on a bridge 4 m above the banks of a river, the angles of depression of the opposite banks are 30° and 45°. Find the width of the river.
25. A student standing on the ground observes a vertical tower. The angle of elevation of the top of the tower from a point 25 m away from the foot is 60°. Later, the student moves closer and the angle becomes 45°.
(a) Draw a labelled diagram showing the two positions and angles of elevation.
(b) Find the height of the tower using the 60° position.
(c) Find the new distance of the student from the foot of the tower at 45°.
(d) Calculate the distance walked by the student between the two positions.
26. From the top of a lighthouse 60 m high, the angles of depression of two ships on the same side are observed as 30° and 45°.
(a) Draw a labelled diagram showing the lighthouse, ships and angles of depression.
(b) Find the distance of the nearer ship from the base of the lighthouse.
(c) Find the distance of the farther ship from the base of the lighthouse.
(d) Calculate the distance between the two ships.
1. (c) line of sight — 1 mark
2. (b) angle of elevation — 1 mark
3. (b) 15√3 m — 1 mark
4. (i) Both true and R explains A — 1 mark
5. (b) line of sight below horizontal — 1 mark
6. (a) height of student AE — 1 mark
7. (i) Both true and R explains A — 1 mark
8. (b) sin 60° — 1 mark
9. (a) equals the distance — 1 mark
10. (b) angle of depression — 1 mark
11. Angle of elevation: angle between line of sight and horizontal when object is above horizontal level (1 mark). Angle of depression: angle between line of sight and horizontal when object is below horizontal level (1 mark).
12. Opposite side = BC, adjacent side = AB (2 marks).
13. Distance DE from foot, angle of elevation ∠BAC, height AE of observer (2 marks).
14. AB = AE + BE; BE = observer height 1.5 m, so total chimney height = AE + 1.5 m (2 marks).
15. Height = 20 × tan 30° = 20/√3 m (2 marks).
16. Angles of depression equal alternate interior angles formed by transversal with parallel lines (2 marks).
17. Let height = h. tan 60° = h/20 → √3 = h/20 → h = 20√3 m (correct formula 1 mark, substitution 1 mark, answer 1 mark).
18. tan 45° = h/20 → 1 = h/20 → h = 20 + 1.6 = 21.6 m (correct ratio 1 mark, addition of observer height 1 mark, answer with unit 1 mark).
19. tan 30° = 12/d → 1/√3 = 12/d → d = 12√3 m (formula 1 mark, solving 1 mark, answer 1 mark).
20. Let height = h, shorter shadow = x. tan 60° = h/x → h = x√3. tan 30° = h/(x+30) → 1/√3 = x√3/(x+30) → x = 15 m, h = 15√3 m (two equations 2 marks, solving 1 mark).
21. Let distance at 30° = d+20. tan 30° = h/(d+20), tan 60° = h/d. Solving gives h = 20√3 m (setup 2 marks, solution 1 mark).
22. For building: tan 30° = 15/AP → AP = 15√3 m. Let flagstaff = x. tan 45° = (15+x)/(15√3) → x = 15(√3−1) = 7.32 m (building distance 2 marks, flagstaff 3 marks).
23. Let height of multi-storeyed building = h, distance = d. tan 30° gives d = (h−10)/√3; tan 45° gives h = d. Solving: h = 5(3+√3) m, d = 5(3+√3) m (equations 2 marks, solution 3 marks).
24. tan 30° gives one part = 4√3 m; tan 45° gives other part = 4 m. Width = 4(1+√3) m (two triangles 2 marks, addition 3 marks).
25. (a) Diagram with two positions and angles (1 mark). (b) Height = 25√3 m (1 mark). (c) Distance at 45° = 25 m (1 mark). (d) Distance walked = 25(√3−1) m (1 mark).
26. (a) Diagram with lighthouse and two ships (1 mark). (b) Nearer distance = 60 m (1 mark). (c) Farther distance = 60√3 m (1 mark). (d) Distance between ships = 60(√3−1) m (1 mark).
All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.