Problem:

A person standing 30 m from the base of a vertical tower observes its top at an angle of elevation of 30°. Find the height of the tower.

Solution:

In right triangle ABC, AB = height (h), BC = 30 m, ∠ACB = 30°

tan 30° = AB / BC = h / 30

1/√3 = h / 30

h = 30/√3 = 30√3/3 = 10√3 m ≈ 17.32 m

Problem:

An electrician needs to reach a point 3.7 m above the ground on a vertical pole. If a ladder is inclined at 60° to the horizontal, find (a) the length of the ladder and (b) how far from the pole its foot must be placed. (Use √3 = 1.73)

Solution:

BD = 3.7 m (height to reach), BC = ladder length, ∠BCD = 60°

(a) Length of ladder: sin 60° = BD/BC → √3/2 = 3.7/BC → BC = (3.7 × 2)/√3 = 7.4/1.73 ≈ 4.28 m

(b) Distance from pole: cos 60° = DC/BC → 1/2 = DC/4.28 → DC ≈ 2.14 m

Problem:

An observer 1.5 m tall stands 28.5 m from the base of a chimney. The angle of elevation of the top of the chimney from the observer’s eyes is 45°. Find the total height of the chimney.

Solution:

Let the chimney be AB. The observer CD is 1.5 m tall. Eye level DE is parallel to CB.

AE (height above eye level): tan 45° = AE/DE = AE/28.5 → 1 = AE/28.5 → AE = 28.5 m

Total height AB = AE + BE = AE + CD = 28.5 + 1.5 = 30 m

Problem:

From point P on the ground, the angle of elevation of the top of a 10 m building is 30°. A flag is hoisted at the top of the building. The angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and distance of the building from P. (√3 = 1.732)

Solution:

Step 1 — Find distance PA: tan 30° = AB/PA → 1/√3 = 10/PA → PA = 10√3 ≈ 17.32 m

Step 2 — Find flagstaff DB = x: AD = (10 + x) m

tan 45° = AD/PA → 1 = (10 + x)/(10√3) → 10 + x = 10√3 → x = 10(√3 − 1) = 10(1.732 − 1) = 7.32 m

Problem:

The shadow of a tower is 40 m longer when the sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Solution:

Let height = h m, shorter shadow (60° case) = x m → longer shadow (30° case) = (x + 40) m

At 60°: tan 60° = h/x → √3 = h/x → h = x√3  …(1)

At 30°: tan 30° = h/(x+40) → 1/√3 = h/(x+40) → h = (x+40)/√3  …(2)

From (1) and (2): x√3 = (x+40)/√3 → 3x = x + 40 → x = 20 m

h = 20√3 m

Problem:

The angles of depression of the top and 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 them.

Solution:

Let PC = multi-storeyed building height, AB = 8 m (shorter building), AC = horizontal distance.

Using alternate angles: ∠PAC = 45° and ∠PBD = 30°

From △PAC: tan 45° = PC/AC → PC = AC → AC = BD

Let PD = h. From △PBD: tan 30° = PD/BD → 1/√3 = h/BD → BD = h√3

Since PC = PD + DC = PD + 8 = AC = BD = h√3 → h + 8 = h√3 → h(√3−1) = 8 → h = 8/(√3−1) = 4(√3+1) m

Height of multi-storeyed building = h + 8 = 4(√3+1) + 8 = 4(3+√3) m. Distance = 4(3+√3) m.

Problem:

From a point P on a bridge 3 m above the water, the angles of depression of the two banks are 30° and 45°. Find the width of the river.

Solution:

DP = 3 m (bridge height). In △APD: tan 30° = DP/AD → AD = 3√3 m

In △BPD: tan 45° = DP/DB → DB = DP = 3 m

Width AB = AD + DB = 3√3 + 3 = 3(√3 + 1) m ≈ 8.20 m

Problem:

From the top of a 75 m lighthouse, the angles of depression of two ships on the same side are 45° and 30°. Find the distance between the two ships.

Solution:

Lighthouse height = 75 m. Let Ship 1 be closer (45°) and Ship 2 farther (30°).

Distance to Ship 1: tan 45° = 75/d₁ → d₁ = 75 m

Distance to Ship 2: tan 30° = 75/d₂ → 1/√3 = 75/d₂ → d₂ = 75√3 m

Distance between ships = d₂ − d₁ = 75√3 − 75 = 75(√3 − 1) m ≈ 54.9 m

Problem:

A 1.2 m tall girl observes a balloon moving horizontally at 88.2 m height. The angle of elevation is initially 60°, then reduces to 30°. Find the distance the balloon travelled.

Solution:

Height of balloon above girl’s eyes = 88.2 − 1.2 = 87 m

Position 1 (60°): tan 60° = 87/d₁ → d₁ = 87/√3 = 29√3 m

Position 2 (30°): tan 30° = 87/d₂ → d₂ = 87√3 m

Distance travelled = d₂ − d₁ = 87√3 − 29√3 = 58√3 m ≈ 100.5 m

Problem:

A man at the top of a tower observes a car at 30° depression. After 6 seconds the angle of depression becomes 60°. At uniform speed, find the time for the car to reach the base of the tower.

Solution:

Let tower height = h. Car positions: C (far) and D (closer).

BC = h·cot 30° = h√3  |  BD = h·cot 60° = h/√3

Distance CD = BC − BD = h√3 − h/√3 = h(3−1)/√3 = 2h/√3

Speed = CD/6 = (2h/√3)/6 = h/(3√3)

Time to travel BD = h/√3 from D to base: t = BD/speed = (h/√3)/(h/3√3) = (h/√3)×(3√3/h) = 3 seconds

Problem:

A TV tower stands on a canal bank. From the opposite bank directly opposite, the angle of elevation is 60°. From a point 20 m farther along that bank, the angle of elevation is 30°. Find the tower height and width of the canal.

Solution:

Let tower height = h m, canal width = BC = d m. Point D is 20 m from C.

From C: tan 60° = h/d → h = d√3  …(1)

From D: tan 30° = h/(d+20) → h = (d+20)/√3  …(2)

From (1)=(2): d√3 = (d+20)/√3 → 3d = d + 20 → d = 10 m

h = 10√3 m