Key Concepts & Formulae at a Glance
- Line of sight:The line from the observer's eye to the point being viewed.
- Angle of elevation:The angle between the line of sight and the horizontal, when the object is above eye level (looking up).
- Angle of depression:The angle between the line of sight and the horizontal, when the object is below eye level (looking down). By alternate angles, it equals the angle of elevation seen from the object looking back.
- Choosing the ratio:If the two knownunknown sides are the opposite and adjacent sides of the angle, use tan or cot. If one of them is the hypotenuse (a rope, string, ladder, spiral), use sin or cos.
- Two right triangles:Problems with two different angles from two different points, or a structure on top of another, give two right triangles sharing a common side — set up both equations and solve together.
- Observer height:When a person (not the ground) is the observer, subtract their height first to get the true vertical side, then add it back if the final answer needs height above the ground.
Exercise 9.1
15 Questions
1A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).
Let AB be the pole and AC the rope, with ∠ACB = 30° and AC = 20 m.
In right ΔABC, sin C = ABAC (AB is opposite the angle, AC is the hypotenuse)
sin 30° = AB20
AB = 20 × 12 = 10
In right ΔABC, sin C = ABAC (AB is opposite the angle, AC is the hypotenuse)
sin 30° = AB20
AB = 20 × 12 = 10
The height of the pole is 10 m.
2A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Let D be the foot of the tree, B the point where it breaks, and C the point where the top touches the ground, with DC = 8 m and ∠DCB = 30°.
In right ΔBDC, tan C = BDDC (the standing stub BD is opposite the angle)
tan 30° = BD8 → BD = 8√3 m
cos C = DCBC (BC is the broken part — the hypotenuse)
cos 30° = 8BC → BC = 8cos 30° = 16√3 m
Total height = BD + BC = 8√3 + 16√3 = 24√3 = 8√3
In right ΔBDC, tan C = BDDC (the standing stub BD is opposite the angle)
tan 30° = BD8 → BD = 8√3 m
cos C = DCBC (BC is the broken part — the hypotenuse)
cos 30° = 8BC → BC = 8cos 30° = 16√3 m
Total height = BD + BC = 8√3 + 16√3 = 24√3 = 8√3
The height of the tree is 8√3 m ≈ 13.86 m.
3A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Slide for younger children
Height = 1.5 m, angle = 30°. If L is the slide length (hypotenuse),
sin 30° = 1.5L → L = 1.5/(12) = 3
sin 30° = 1.5L → L = 1.5/(12) = 3
Slide length = 3 m.
Slide for elder children
Height = 3 m, angle = 60°.
sin 60° = 3L → L = 3/(√32) = 6√3 = 2√3
sin 60° = 3L → L = 3/(√32) = 6√3 = 2√3
Slide length = 2√3 m ≈ 3.46 m.
4The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Let AB be the tower and C the point on the ground, with BC = 30 m and ∠ACB = 30°.
tan C = ABBC (opposite over adjacent)
tan 30° = AB30 → AB = 30√3 = 10√3
tan C = ABBC (opposite over adjacent)
tan 30° = AB30 → AB = 30√3 = 10√3
The height of the tower is 10√3 m ≈ 17.32 m.
5A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Let the string length be L, height of kite = 60 m, angle with ground = 60°.
sin 60° = 60L (height is opposite, string is the hypotenuse)
L = 60/(√32) = 120√3 = 40√3
sin 60° = 60L (height is opposite, string is the hypotenuse)
L = 60/(√32) = 120√3 = 40√3
The length of the string is 40√3 m ≈ 69.28 m.
6A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
The boy's eyes are at 1.5 m, so the effective height to consider is 30 − 1.5 = 28.5 m (subtract observer height first).
At P₁ (angle 30°): tan 30° = 28.5d₁ → d₁ = 28.5√3
At P₂ (angle 60°): tan 60° = 28.5d₂ → d₂ = 28.5√3
Distance walked = d₁ − d₂ = 28.5√3 − 28.5√3 = 28.53−1√3 = 57√3 = 19√3
At P₁ (angle 30°): tan 30° = 28.5d₁ → d₁ = 28.5√3
At P₂ (angle 60°): tan 60° = 28.5d₂ → d₂ = 28.5√3
Distance walked = d₁ − d₂ = 28.5√3 − 28.5√3 = 28.53−1√3 = 57√3 = 19√3
The boy walked 19√3 m ≈ 32.91 m.
7From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Let P be the point on the ground, d the horizontal distance to the building, and h the height of the tower (above the building).
For the bottom of the tower (top of building, 20 m): tan 45° = 20d → d = 20
For the top of the tower (20+h): tan 60° = 20+hd → 20+h = 20√3
h = 20√3 − 20 = 20(√3−1)
For the bottom of the tower (top of building, 20 m): tan 45° = 20d → d = 20
For the top of the tower (20+h): tan 60° = 20+hd → 20+h = 20√3
h = 20√3 − 20 = 20(√3−1)
The height of the tower is 20(√3−1) m ≈ 14.64 m.
8A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Let h be the pedestal height and d the horizontal distance from the observation point.
Top of pedestal: tan 45° = hd → d = h
Top of statue: tan 60° = h+1.6d → h+1.6 = h√3 (substituting d = h)
h(√3−1) = 1.6 → h = 1.6√3−1 = 1.6√3+12 = 0.8(√3+1)
Top of pedestal: tan 45° = hd → d = h
Top of statue: tan 60° = h+1.6d → h+1.6 = h√3 (substituting d = h)
h(√3−1) = 1.6 → h = 1.6√3−1 = 1.6√3+12 = 0.8(√3+1)
The height of the pedestal is 0.8(√3+1) m ≈ 2.19 m.
9The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Let the building and tower stand distance d apart on the same ground, tower height = 50 m, building height = h.
From foot of building, tower's top: tan 60° = 50d → d = 50√3
From foot of tower, building's top: tan 30° = hd → h = d√3 = 503
From foot of building, tower's top: tan 60° = 50d → d = 50√3
From foot of tower, building's top: tan 30° = hd → h = d√3 = 503
The height of the building is 503 m ≈ 16.67 m.
10Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Let the point P be at distance x from the pole seen at 60°, so it is (80−x) from the other pole (seen at 30°). Let h be the common height.
tan 60° = hx → h = x√3
tan 30° = h80−x → h = 80−x√3
Equating: x√3 = 80−x√3 → 3x = 80−x → x = 20
h = 20√3
tan 60° = hx → h = x√3
tan 30° = h80−x → h = 80−x√3
Equating: x√3 = 80−x√3 → 3x = 80−x → x = 20
h = 20√3
The poles are 20√3 m ≈ 34.64 m tall; the point is 20 m from one pole and 60 m from the other.
11A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.
Let B be the foot of the tower, C the point directly opposite on the other bank (canal width BC = x), D a point 20 m further from C, and A the top of the tower (height h).
From C: tan 60° = hx → h = x√3
From D: tan 30° = hx+20 → h = x+20√3
Equating: x√3 = x+20√3 → 3x = x+20 → x = 10
h = 10√3
From C: tan 60° = hx → h = x√3
From D: tan 30° = hx+20 → h = x+20√3
Equating: x√3 = x+20√3 → 3x = x+20 → x = 10
h = 10√3
The tower is 10√3 m ≈ 17.32 m tall; the canal is 10 m wide.
12From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Let P be the top of the building (height 7 m) and d the horizontal distance to the tower.
Angle of depression to the tower's foot = 45° (alternate angle to the elevation from the foot): tan 45° = 7d → d = 7
Let the tower's height above the building level be h′. Angle of elevation to the tower's top: tan 60° = h′d → h′ = 7√3
Total tower height = 7 + 7√3 = 7(1+√3)
Angle of depression to the tower's foot = 45° (alternate angle to the elevation from the foot): tan 45° = 7d → d = 7
Let the tower's height above the building level be h′. Angle of elevation to the tower's top: tan 60° = h′d → h′ = 7√3
Total tower height = 7 + 7√3 = 7(1+√3)
The height of the tower is 7(1+√3) m ≈ 19.12 m.
13As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Let the lighthouse height be 75 m, and d₁, d₂ the distances of the nearer (45°) and farther (30°) ships from its base.
tan 45° = 75d₁ → d₁ = 75
tan 30° = 75d₂ → d₂ = 75√3
Distance between ships = d₂ − d₁ = 75√3 − 75 = 75(√3−1)
tan 45° = 75d₁ → d₁ = 75
tan 30° = 75d₂ → d₂ = 75√3
Distance between ships = d₂ − d₁ = 75√3 − 75 = 75(√3−1)
The distance between the two ships is 75(√3−1) m ≈ 54.90 m.
14A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.
The girl's eyes are at 1.2 m, so the effective height of the balloon above eye level is 88.2 − 1.2 = 87 m (subtract observer height).
At 60°: tan 60° = 87d₁ → d₁ = 87√3 = 29√3
At 30°: tan 30° = 87d₂ → d₂ = 87√3
Distance travelled = d₂ − d₁ = 87√3 − 29√3 = 58√3
At 60°: tan 60° = 87d₁ → d₁ = 87√3 = 29√3
At 30°: tan 30° = 87d₂ → d₂ = 87√3
Distance travelled = d₂ − d₁ = 87√3 − 29√3 = 58√3
The balloon travelled 58√3 m ≈ 100.46 m.
15A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Let the tower height be h, and d₁, d₂ the car's distance from the foot at the first and second sightings.
tan 30° = hd₁ → d₁ = h√3
tan 60° = hd₂ → d₂ = h√3
Distance covered in 6 s = d₁−d₂ = h√3−h√3 = 2h√3, so speed = (2h√3)/6 = h3√3
Time to cover the remaining distance d₂ = h√3 at this speed = (h√3) ÷ (h3√3) = 3 (h cancels — the answer doesn't depend on the tower's actual height)
tan 30° = hd₁ → d₁ = h√3
tan 60° = hd₂ → d₂ = h√3
Distance covered in 6 s = d₁−d₂ = h√3−h√3 = 2h√3, so speed = (2h√3)/6 = h3√3
Time to cover the remaining distance d₂ = h√3 at this speed = (h√3) ÷ (h3√3) = 3 (h cancels — the answer doesn't depend on the tower's actual height)
The car takes a further 3 seconds to reach the foot of the tower.
Frequently Asked Questions
Both are measured from the horizontal line of sight. The angle of elevation is used when you raise your head to look up at something above you (e.g. a person on the ground looking at the top of a tower). The angle of depression is used when you lower your head to look down at something below you (e.g. a person on a tower looking at a car on the ground). By alternate angles, the angle of depression from the top equals the angle of elevation from the bottom, for the same line of sight.
Almost every question in this chapter reduces to one right triangle where you know two of {opposite side, adjacent side, angle} and need the third. If the opposite and adjacent sides are involved, use tan (or cot). If the hypotenuse is involved (like a rope, string, or ladder), use sin or cos instead.
Whenever a problem gives two different angles of elevationdepression from two different positions, or asks about a structure sitting on top of another (like a flagstaff on a building, or a tower on a building), you get two right triangles that share a common side. Set up the trigonometric equation for each triangle separately, then solve the two equations together — usually by substitution.
When a person is looking at an object, the angle is measured from their eye level, not from the ground. So before applying any trigonometric ratio, subtract the observer's height from the object's height to get the true vertical side of the right triangle — then add the observer's height back on at the end if the question asks for the object's height above the ground.
Yes — a printable Formula Card covering all 7 units, including heights and distances, is available for ₹99.
Continue to the Next Chapter
Chapter 10 (Circles) moves on to tangents and their properties.