Solved Examples

How to Use This Section

This is your practice powerhouse for heights and distances. Below are 30+ fully solved problems covering the whole chapter — single triangles, two triangles, elevation and depression, and real-life set-ups — roughly easy to hard.

How to read: Draw the triangle, mark the known side and angle, write the $\tan$ (or $\sin$/$\cos$) equation, then solve. Cover the solution and try each yourself first.

Keep these handy:

• $\tan 30° = \tfrac{1}{\sqrt3}$, $\tan 45° = 1$, $\tan 60° = \sqrt3$.
• Depression at the top = elevation at the bottom (alternate angles).
• Two angles ⇒ two equations sharing the common height or distance.
• Always add the observer's eye height if the question asks for height above ground.

Solved Examples

Example 1: Height at 30°

The elevation of a tower's top from a point 45 m away is 30°. Find the height.

Solution:

1. $h = 45\tan 30° = \dfrac{45}{\sqrt3} = 15\sqrt3$.

Final Answer: $15\sqrt3 \approx 25.98$ m.

Takeaway: $h = d\tan\theta$.

Example 2: Distance at 60°

A 60 m tower is seen at elevation 60°. Find the distance of the point from the foot.

Solution:

1. $\tan 60° = \dfrac{60}{d} \Rightarrow d = \dfrac{60}{\sqrt3} = 20\sqrt3$.

Final Answer: $20\sqrt3 \approx 34.64$ m.

Takeaway: $d = h/\tan\theta$.

Example 3: Find the angle

A tower is $50\sqrt3$ m high and a point is 50 m from its foot. Find the elevation of the top.

Solution:

1. $\tan\theta = \dfrac{50\sqrt3}{50} = \sqrt3 \Rightarrow \theta = 60°$.

Final Answer: 60°.

Takeaway: $\tan\theta = \sqrt3 \Rightarrow 60°$.

Example 4: Ladder height

A 12 m ladder makes 30° with the ground. How high up the wall does it reach?

Solution:

1. Height $= 12\sin 30° = 12\cdot\tfrac12 = 6$ m.

Final Answer: 6 m.

Takeaway: Height from slant uses $\sin$.

Example 5: Depression distance

From a 90 m cliff, the depression of a boat is 60°. Find its distance from the foot.

Solution:

1. $\tan 60° = \dfrac{90}{d} \Rightarrow d = \dfrac{90}{\sqrt3} = 30\sqrt3$.

Final Answer: $30\sqrt3 \approx 51.96$ m.

Takeaway: Depression = elevation at the boat.

Example 6: Shadow at 45°

Find the shadow length of a 25 m tower when the sun's elevation is 45°.

Solution:

1. $\tan 45° = \dfrac{25}{s} = 1 \Rightarrow s = 25$ m.

Final Answer: 25 m.

Takeaway: At 45°, shadow equals height.

Example 7: Two-point 30°–60°

The elevation of a tower's top from two points 20 m apart in line are 30° and 60°. Find the height.

Solution:

1. Near $x = \dfrac{h}{\sqrt3}$; far $x + 20 = h\sqrt3$.
2. $\dfrac{h}{\sqrt3} + 20 = h\sqrt3 \Rightarrow 20 = \dfrac{2h}{\sqrt3} \Rightarrow h = 10\sqrt3$.

Final Answer: $10\sqrt3 \approx 17.32$ m.

Takeaway: Two angles ⇒ two equations.

Example 8: Kite string length

A kite is at height 45 m; its string makes 45° with the ground. Find the string length.

Solution:

1. $\sin 45° = \dfrac{45}{\ell} \Rightarrow \ell = 45\sqrt2$.

Final Answer: $45\sqrt2 \approx 63.64$ m.

Takeaway: String is the hypotenuse ⇒ $\sin$.

Example 9: Add boy's height

A boy 1.7 m tall sees the top of a tower at 45° from 30 m away (horizontal). Find the tower's height.

Solution:

1. Triangle height $= 30\tan 45° = 30$ m.
2. Add eye height: $30 + 1.7 = 31.7$ m.

Final Answer: 31.7 m.

Takeaway: Add observer's eye height for total above ground.

Example 10: Two ships depression

From a $60\sqrt3$ m lighthouse, the depressions of two ships in line are 30° and 60°. Find the distance between them.

Solution:

1. Near $x_1 = \dfrac{60\sqrt3}{\sqrt3} = 60$. Far $x_2 = 60\sqrt3\cdot\sqrt3 = 180$.
2. Distance $= 180 - 60 = 120$ m.

Final Answer: 120 m.

Takeaway: Larger depression ⇒ nearer ship.

Example 11: Building from elevation/depression

From the top of a 15 m building, the elevation of a tower's top is 30° and depression of its foot is 60°. Find the tower's height.

Solution:

1. Depression 60°: $\tan 60° = \dfrac{15}{d} \Rightarrow d = \dfrac{15}{\sqrt3} = 5\sqrt3$.
2. Elevation 30°: rise $b = d\tan 30° = 5\sqrt3\cdot\dfrac{1}{\sqrt3} = 5$.
3. Tower height $= 15 + 5 = 20$ m.

Final Answer: 20 m.

Takeaway: Depression gives $d$; elevation gives the rise above eye level.

Example 12: Moving towards tower

The elevation of a tower's top is 30°; after walking 40 m towards it, it is 60°. Find the height.

Solution:

1. Near $x = \dfrac{h}{\sqrt3}$; far $x + 40 = h\sqrt3$.
2. $\dfrac{h}{\sqrt3} + 40 = h\sqrt3 \Rightarrow 40 = \dfrac{2h}{\sqrt3} \Rightarrow h = 20\sqrt3$.

Final Answer: $20\sqrt3 \approx 34.64$ m.

Takeaway: The 30°–60° pair gives $d = 2h/\sqrt3$.

Example 13: River width with 60°

A tower 30 m tall stands on the far bank. From the near bank, its top has elevation 60°. Find the river's width.

Solution:

1. $\tan 60° = \dfrac{30}{d} \Rightarrow d = \dfrac{30}{\sqrt3} = 10\sqrt3$.

Final Answer: $10\sqrt3 \approx 17.32$ m.

Takeaway: Width from height uses $d = h/\tan\theta$.

Example 14: Tower on a hill

From a point on the ground, the elevation of the bottom of a tower (top of a 40 m hill) is 30° and of the tower's top is 45°. Find the tower's height.

Solution:

1. $\tan 30° = \dfrac{40}{d} \Rightarrow d = 40\sqrt3$.
2. $\tan 45° = \dfrac{40 + t}{d} = 1 \Rightarrow 40 + t = 40\sqrt3$.
3. $t = 40\sqrt3 - 40 = 40(\sqrt3 - 1)$.

Final Answer: $40(\sqrt3 - 1) \approx 29.28$ m.

Takeaway: Lower angle fixes $d$; upper angle gives total height.

Example 15: Speed of a boat

From a 100 m cliff, a boat's depression changes from 30° to 45° in 1 minute. Find the boat's speed.

Solution:

1. Far $x_2 = \dfrac{100}{\tan 30°} = 100\sqrt3$. Near $x_1 = \dfrac{100}{\tan 45°} = 100$.
2. Distance $= 100\sqrt3 - 100 = 100(\sqrt3 - 1) \approx 73.2$ m.
3. Speed $= \dfrac{100(\sqrt3 - 1)}{1} \approx 73.2$ m/min.

Final Answer: $100(\sqrt3 - 1) \approx 73.2$ m/min.

Takeaway: Distance ÷ time = speed.

Example 16: Pole and its shadow

A pole casts a 6 m shadow when the sun's elevation is 60°. Find the pole's height.

Solution:

1. $\tan 60° = \dfrac{h}{6} \Rightarrow h = 6\sqrt3$.

Final Answer: $6\sqrt3 \approx 10.39$ m.

Takeaway: Height $=$ shadow $\times \tan(\text{elevation})$.

Example 17: Building between two points

The elevations of a building's top from two points 60 m apart on opposite sides are 30° and 60°. Find the height.

Solution:

1. $x = \dfrac{h}{\sqrt3}$ (60° side); $60 - x = h\sqrt3$ (30° side).
2. Add: $60 = \dfrac{h}{\sqrt3} + h\sqrt3 = \dfrac{4h}{\sqrt3} \Rightarrow h = \dfrac{60\sqrt3}{4} = 15\sqrt3$.

Final Answer: $15\sqrt3 \approx 25.98$ m.

Takeaway: Opposite sides ⇒ distances add to the separation.

Example 18: Cloud over a lake (preview)

From a point 20 m above a lake, the elevation of a cloud is 30°. Set up $\tan 30° = \dfrac{H - 20}{d}$ where $H$ is the cloud's height above the lake and $d$ the horizontal distance. If $d = 60\sqrt3$, find $H$.

Solution:

1. $\tan 30° = \dfrac{H - 20}{60\sqrt3} \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{H - 20}{60\sqrt3}$.
2. $H - 20 = 60 \Rightarrow H = 80$ m.

Final Answer: $H = 80$ m above the lake.

Takeaway: Measure the cloud's height relative to the observation level, then adjust.

Example 19: 45°-then-60°

The elevation of a tower from a point is 45°; moving 10 m closer it becomes 60°. Find the height.

Solution:

1. Near $x = \dfrac{h}{\sqrt3}$; far $x + 10 = h$ (since $\tan 45° = 1$).
2. $\dfrac{h}{\sqrt3} + 10 = h \Rightarrow 10 = h\cdot\dfrac{\sqrt3 - 1}{\sqrt3} \Rightarrow h = \dfrac{10\sqrt3}{\sqrt3 - 1} = 5\sqrt3(\sqrt3 + 1) = 15 + 5\sqrt3$.

Final Answer: $15 + 5\sqrt3 \approx 23.66$ m.

Takeaway: Rationalise $\sqrt3 - 1$ with its conjugate.

Example 20: Aeroplane speed

An aeroplane flying at $3000$ m has an elevation of 60° from a point; 10 s later it is 30°. Find its speed.

Solution:

1. Near $x_1 = \dfrac{3000}{\tan 60°} = \dfrac{3000}{\sqrt3} = 1000\sqrt3$. Far $x_2 = \dfrac{3000}{\tan 30°} = 3000\sqrt3$.
2. Distance $= 3000\sqrt3 - 1000\sqrt3 = 2000\sqrt3$ m in 10 s.
3. Speed $= 200\sqrt3 \approx 346.4$ m/s.

Final Answer: $200\sqrt3 \approx 346.4$ m/s.

Takeaway: Constant-height flight ⇒ two triangles.

Example 21: Find the height of a balloon

A balloon's elevation from a point is 30°. Walking 50 m towards the point below it, the elevation is 60°. Find the balloon's height.

Solution:

1. Far $x + 50 = h\sqrt3$; near $x = \dfrac{h}{\sqrt3}$.
2. $\dfrac{h}{\sqrt3} + 50 = h\sqrt3 \Rightarrow 50 = \dfrac{2h}{\sqrt3} \Rightarrow h = 25\sqrt3$.

Final Answer: $25\sqrt3 \approx 43.3$ m.

Takeaway: Standard 30°–60° two-triangle template.

Example 22: Angle of elevation of the sun

The ratio of a vertical pole's height to its shadow is $1 : \sqrt3$. Find the sun's elevation.

Solution:

1. $\tan\theta = \dfrac{\text{height}}{\text{shadow}} = \dfrac{1}{\sqrt3}$.
2. $\theta = 30°$.

Final Answer: 30°.

Takeaway: $\tan\theta = \tfrac{1}{\sqrt3} \Rightarrow 30°$.

Example 23: Two towers

Two towers of equal height stand 80 m apart. From the midpoint between them, the elevation of each top is 45°. Find their height.

Solution:

1. Midpoint is 40 m from each. $\tan 45° = \dfrac{h}{40} = 1 \Rightarrow h = 40$ m.

Final Answer: 40 m.

Takeaway: Symmetry: the midpoint is equidistant, so one equation suffices.

Example 24: Find distance between cars

From the top of a $50\sqrt3$ m tower, two cars on a straight road have depressions 30° and 60°. Find the distance between them.

Solution:

1. Near (60°): $x_1 = \dfrac{50\sqrt3}{\sqrt3} = 50$. Far (30°): $x_2 = 50\sqrt3\cdot\sqrt3 = 150$.
2. Distance $= 150 - 50 = 100$ m.

Final Answer: 100 m.

Takeaway: Both cars on the same side ⇒ subtract the distances.

Example 25: Height with eye level 1.5 m

An observer 1.5 m tall is 28.5 m from a chimney. The elevation of the top is 45°. Find the chimney's height.

Solution:

1. Triangle height $= 28.5\tan 45° = 28.5$ m.
2. Total $= 28.5 + 1.5 = 30$ m.

Final Answer: 30 m.

Takeaway: Always add the eye height for total above ground.

Example 26: Width of a road

From the top of a 10 m building, the angles of depression of two points on the same side of a road are 45° and 30°. Find the width of the road between the points.

Solution:

1. Near (45°): $x_1 = 10$. Far (30°): $x_2 = 10\sqrt3$.
2. Width $= 10\sqrt3 - 10 = 10(\sqrt3 - 1)$.

Final Answer: $10(\sqrt3 - 1) \approx 7.32$ m.

Takeaway: The two points lie on the same side; subtract the distances.

Example 27: Height of a chimney from two angles

The elevation of the top of a chimney from two points 50 m apart in line are 45° and 30°. Find the height.

Solution:

1. Near $x = h$ (45°); far $x + 50 = h\sqrt3$ (30°).
2. $h + 50 = h\sqrt3 \Rightarrow 50 = h(\sqrt3 - 1) \Rightarrow h = \dfrac{50}{\sqrt3 - 1} = 25(\sqrt3 + 1)$.

Final Answer: $25(\sqrt3 + 1) \approx 68.3$ m.

Takeaway: 45°–30° pair gives a conjugate-rationalised answer.

Example 28: Slant of a hill path

A straight path up a hill rises to a height of 50 m at an inclination of 30° to the horizontal. Find the length of the path.

Solution:

1. $\sin 30° = \dfrac{50}{\ell} = \tfrac12 \Rightarrow \ell = 100$ m.

Final Answer: 100 m.

Takeaway: The path is the hypotenuse ⇒ $\sin$ links it to the height.

Example 29: Flagstaff on a tower

A flagstaff stands on a 20 m tower. From a point on the ground, the elevation of the bottom of the flagstaff is 45° and of its top is 60°. Find the flagstaff's height.

Solution:

1. $\tan 45° = \dfrac{20}{d} \Rightarrow d = 20$.
2. $\tan 60° = \dfrac{20 + f}{20} = \sqrt3 \Rightarrow 20 + f = 20\sqrt3$.
3. $f = 20\sqrt3 - 20 = 20(\sqrt3 - 1)$.

Final Answer: $20(\sqrt3 - 1) \approx 14.64$ m.

Takeaway: Lower angle fixes $d$; difference of the two verticals gives the flagstaff.

Example 30: Distance walked

At a point the elevation of a 100 m tower's top is 30°. How far must one walk towards it to make the elevation 45°?

Solution:

1. Far $x_2 = \dfrac{100}{\tan 30°} = 100\sqrt3$. Near $x_1 = \dfrac{100}{\tan 45°} = 100$.
2. Distance walked $= 100\sqrt3 - 100 = 100(\sqrt3 - 1)$.

Final Answer: $100(\sqrt3 - 1) \approx 73.2$ m.

Takeaway: Distance walked = difference of the two horizontal distances.

Example 31: Two boats opposite sides

From the top of a $30$ m lighthouse, two boats on opposite sides have depressions 45° and 30°. Find the distance between the boats.

Solution:

1. One side (45°): $x_1 = \dfrac{30}{\tan 45°} = 30$.
2. Other side (30°): $x_2 = \dfrac{30}{\tan 30°} = 30\sqrt3$.
3. Distance $= x_1 + x_2 = 30 + 30\sqrt3 = 30(1 + \sqrt3)$.

Final Answer: $30(1 + \sqrt3) \approx 81.96$ m.

Takeaway: Opposite sides ⇒ ADD the two horizontal distances.