Question 1
A 20 m rope reaches from the ground to the top of a vertical pole and makes a 30° angle with the ground. Find the pole height.
Steps only
sin 30° = h/20
h = 20 × 1/2 = 10
Answer
10 m.
CLASS 10 · MATHEMATICS · CHAPTER 9
Some Applications of Trigonometry
Turn angles of elevation and depression into right triangles, then find heights and distances with familiar trigonometric ratios.
15 worked answersAll 1 exercise sets
What you will learn
- Draw a right-triangle model from a word problem.
- Choose sine, cosine or tangent from the known sides.
- Keep eye level and object height separate when needed.
15 worked answers · short steps first, explanations when you need them
Exercise 9.1
Textbook pages 141–143
Question 2
A broken tree touches the ground 8 m from its foot and the broken part makes a 30° angle with the ground. Find the original tree height.
Steps only
Stump = 8 tan 30° = 8/√3
Broken part = 8/cos 30° = 16/√3
Total = 24/√3 = 8√3
Answer
8√3 m, about 13.86 m.
Question 3
Find the slide lengths for heights 1.5 m at 30° and 3 m at 60° to the ground.
Steps only
First: L = 1.5/sin30° = 3 m
Second: L = 3/sin60° = 2√3 m
Answer
3 m and 2√3 m (about 3.46 m).
Question 4
A point is 30 m from a tower and the angle of elevation to its top is 30°. Find the height.
Steps only
tan30° = h/30
h = 30/√3 = 10√3
Answer
10√3 m, about 17.32 m.
Question 5
A kite is 60 m above the ground and its taut string is inclined at 60°. Find the string length.
Steps only
sin60° = 60/L
L = 60 × 2/√3 = 40√3
Answer
40√3 m, about 69.28 m.
Question 6
A 1.5 m tall boy views the top of a 30 m building first at 30° and then at 60°. Find how far he walks toward it.
Steps only
Height above eyes = 30 − 1.5 = 28.5 m
First distance = 28.5/tan30° = 28.5√3
Second distance = 28.5/tan60° = 9.5√3
Walked = 19√3
Answer
19√3 m, about 32.91 m.
Question 7
The angles of elevation of the bottom and top of a tower on a 20 m building are 45° and 60°. Find the tower height.
Steps only
tan45° = 20/x, so x = 20
Total height = x tan60° = 20√3
Tower = 20√3 − 20 = 20(√3 − 1)
Answer
20(√3 − 1) m, about 14.64 m.
Question 8
A 1.6 m statue stands on a pedestal. From one point, the top angles are 60° for the statue and 45° for the pedestal. Find the pedestal height.
Steps only
Let pedestal height = h. Since tan45° = h/x, x = h
tan60° = (h + 1.6)/h = √3
h = 1.6/(√3 − 1) = 0.8(√3 + 1)
Answer
0.8(√3 + 1) m, about 2.19 m.
Question 9
From the feet of a tower and building, the elevation angles to the opposite tops are 30° and 60°. If the tower is 50 m high, find the building height.
Steps only
tan60° = 50/d, so d = 50/√3
Building height = d tan30° = 50/3
Answer
50/3 m, about 16.67 m.
Question 10
Equal poles stand on opposite sides of an 80 m road. From a point between them, their elevation angles are 60° and 30°. Find the pole height and the two distances.
Steps only
Let distances be x and y. x + y = 80
h = x tan60° = y tan30°, so y = 3x
x = 20, y = 60, h = 20√3
Answer
Height 20√3 m; distances 20 m and 60 m.
Question 11
From the point directly opposite a canal-side tower its elevation is 60°. From a point 20 m farther away it is 30°. Find tower height and canal width.
Steps only
Let canal width = x and height = h
h = x√3 and h = (x + 20)/√3
3x = x + 20, so x = 10 and h = 10√3
Answer
Canal width 10 m; tower height 10√3 m.
Question 12
From a 7 m building top, the elevation to a cable-tower top is 60° and depression to its foot is 45°. Find the tower height.
Steps only
Horizontal distance = 7/tan45° = 7 m
Height above building = 7 tan60° = 7√3
Tower height = 7 + 7√3
Answer
7(1 + √3) m, about 19.12 m.
Question 13
From a 75 m lighthouse, two ships on the same side have depression angles 30° and 45°. Find the distance between them.
Steps only
Far ship distance = 75/tan30° = 75√3
Near ship distance = 75/tan45° = 75
Separation = 75(√3 − 1)
Answer
75(√3 − 1) m, about 54.90 m.
Question 14
A 1.2 m girl observes a balloon at height 88.2 m. Its elevation changes from 60° to 30°. Find the horizontal distance travelled.
Steps only
Height above eyes = 88.2 − 1.2 = 87 m
Distances = 87/tan60° = 29√3 and 87/tan30° = 87√3
Travel = 58√3
Answer
58√3 m, about 100.46 m.
Question 15
A car approaches a tower. Its depression angle from the top changes from 30° to 60° in 6 seconds. Find the additional time needed to reach the tower foot at the same speed.
Steps only
Let tower height = h. Distances are h√3 and h/√3
Distance in 6 s = h√3 − h/√3 = 2h/√3
Remaining distance h/√3 is half as much, so time = 3 s
Answer
3 seconds.