Heights and Distances with One Triangle

The Single-Triangle Method

Many problems involve just one right triangle. The recipe:

1. Draw the triangle and mark the right angle (usually where the vertical meets the ground).
2. Label the known angle and the known side.
3. Choose the ratio that connects the known side with the unknown side.
4. Solve using the value of the standard angle.

Key Point: Decide which ratio to use by what you know and what you want: opposite & adjacent ⇒ $\tan$; opposite & hypotenuse ⇒ $\sin$; adjacent & hypotenuse ⇒ $\cos$.

[Board Important] Most height-distance problems use $\tan$, because the vertical (height) and horizontal (distance) are the opposite and adjacent sides.

Useful Standard-Angle Facts

Since the angles are almost always 30°, 45° or 60°, keep these handy:

• $\tan 30° = \dfrac{1}{\sqrt3}$, $\tan 45° = 1$, $\tan 60° = \sqrt3$.
• $\sin 30° = \tfrac12$, $\sin 45° = \tfrac{1}{\sqrt2}$, $\sin 60° = \tfrac{\sqrt3}{2}$.
• $\cos 30° = \tfrac{\sqrt3}{2}$, $\cos 45° = \tfrac{1}{\sqrt2}$, $\cos 60° = \tfrac12$.

Key Point: At 45° the height equals the base distance; at 60° the height is $\sqrt3$ times the base; at 30° the base is $\sqrt3$ times the height.

[JEE/NEET Tip] Keeping answers in surd form (e.g. $10\sqrt3$) is exact; use $\sqrt3 \approx 1.732$ only if a decimal is asked.

Finding a Height

If you know the horizontal distance $d$ and the angle of elevation $\theta$, the height is $$h = d\tan\theta.$$ If instead you know the slant (line of sight) $\ell$, then $h = \ell\sin\theta$.

Key Point: Height $= d\tan\theta$ (from horizontal distance) or $\ell\sin\theta$ (from the slant).

[Board Important] Read carefully whether the given length is the horizontal distance or the slant line of sight — they call for different ratios.

Finding a Distance or an Angle

To find the horizontal distance when the height $h$ and elevation $\theta$ are known: $$d = \frac{h}{\tan\theta}.$$ To find the angle when both height and distance are known, identify $\tan\theta = \dfrac{h}{d}$ and recognise the standard angle.

Key Point: Rearrange $\tan\theta = h/d$ for whichever quantity is unknown.

[Board Important] If $\tan\theta$ comes out as $1$, $\sqrt3$, or $\tfrac{1}{\sqrt3}$, the angle is $45°$, $60°$, or $30°$ respectively.

Solved Examples

Example 1: Height from distance

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

Solution:

1. $h = 30\tan 30° = 30\cdot\dfrac{1}{\sqrt3} = \dfrac{30}{\sqrt3} = 10\sqrt3$.

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

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

Example 2: Distance from height

A tower is $50$ m high. The angle of elevation of its top from a point on the ground is 45°. How far is the point from the foot?

Solution:

1. $\tan 45° = \dfrac{50}{d} \Rightarrow 1 = \dfrac{50}{d} \Rightarrow d = 50$ m.

Final Answer: 50 m.

Takeaway: At 45°, distance equals height.

Example 3: Find the angle

A 10 m pole casts a $10\sqrt3$ m shadow. Find the sun's angle of elevation.

Solution:

1. $\tan\theta = \dfrac{10}{10\sqrt3} = \dfrac{1}{\sqrt3}$.
2. $\theta = 30°$.

Final Answer: 30°.

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

Example 4: Ladder against a wall

A ladder 10 m long reaches a window. It makes 60° with the ground. How high is the window?

Solution:

1. Height $= 10\sin 60° = 10\cdot\dfrac{\sqrt3}{2} = 5\sqrt3$.

Final Answer: $5\sqrt3 \approx 8.66$ m.

Takeaway: Height from the slant uses $\sin$.

Example 5: Foot of the ladder

For the same 10 m ladder at 60°, how far is the foot from the wall?

Solution:

1. Distance $= 10\cos 60° = 10\cdot\dfrac12 = 5$ m.

Final Answer: 5 m.

Takeaway: Adjacent from the slant uses $\cos$.

Example 6: River width

From a point on one bank, the angle of elevation of the top of a tree on the opposite bank (height 20 m) is 45°. Find the river's width.

Solution:

1. $\tan 45° = \dfrac{20}{d} \Rightarrow d = 20$ m.

Final Answer: 20 m.

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

Example 7: Kite string

A kite is flying at a height of 60 m. The string makes 60° with the ground. Find the length of the string (assume it is straight).

Solution:

1. $\sin 60° = \dfrac{60}{\ell} \Rightarrow \ell = \dfrac{60}{\sqrt3/2} = \dfrac{120}{\sqrt3} = 40\sqrt3$.

Final Answer: $40\sqrt3 \approx 69.28$ m.

Takeaway: The string is the hypotenuse, so use $\sin$.

Example 8: Tower height with 60°

The angle of elevation of the top of a tower from a point 15 m away is 60°. Find the height.

Solution:

1. $h = 15\tan 60° = 15\sqrt3$.

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

Takeaway: At 60°, height is $\sqrt3$ times the base distance.

Example 9: Add observer's height

A boy 1.5 m tall stands 28.5 m from a tower. The elevation of the top from his eyes is 45°. Find the tower's height.

Solution:

1. Triangle height above eyes $= 28.5\tan 45° = 28.5$ m.
2. Add eye height: $28.5 + 1.5 = 30$ m.

Final Answer: 30 m.

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

Example 10: Shadow length

A tower is $30$ m high. Find the length of its shadow when the sun's elevation is 60°.

Solution:

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

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

Takeaway: Shadow $= \dfrac{\text{height}}{\tan(\text{elevation})}$.