Single-angle problems

The simplest heights-and-distances problems involve exactly one angle and one unknown length. They reduce to a single right triangle, and the entire problem-solving process is: draw, label, pick a trig ratio, solve. If you make this routine automatic, you have already secured a large slice of the marks this chapter offers.

The reason these problems appear so often in board examinations is twofold. First, they cleanly test whether you can translate a sentence into a diagram. Second, they reward neat, organised work , the student who labels the diagram thoughtfully almost never makes an error. A messy diagram, on the other hand, leads to mismatched sides and the wrong ratio.

Definitions and conventions

A single-angle problem features one observer and one object. The line connecting them is the line of sight; together with the horizontal and the vertical it forms a right triangle. The given information is some combination of: the angle of elevation or depression, a horizontal distance, a vertical height, or the line-of-sight length.

For brevity we will write:

• $h$ for the vertical leg (the height we typically want),
• $b$ for the horizontal leg (the distance from the observer to the foot of the object),
• $L$ for the line-of-sight (the hypotenuse).

Then for a measured angle $\theta$,

$\sin\theta = \frac{h}{L}, \qquad \cos\theta = \frac{b}{L}, \qquad \tan\theta = \frac{h}{b}.$

Which ratio to choose

A useful rule: pick the ratio that contains exactly one unknown. If you know $b$ and want $h$, use $\tan\theta$. If you know $h$ and want $L$, use $\sin\theta$. If you know $b$ and want $L$, use $\cos\theta$. The wrong ratio leads to two unknowns and unnecessary algebra.

The most frequently used ratio is $\tan$, because most problems give either the horizontal distance or the height (the two legs), not the line of sight (the hypotenuse). Train your eye to spot when $\tan$ is the natural choice.

Standard angle values

Almost every single-angle problem uses one of $30^\circ, 45^\circ, 60^\circ$. Memorise these to instant recall:

$\tan 30^\circ = \frac{1}{\sqrt 3}, \quad \tan 45^\circ = 1, \quad \tan 60^\circ = \sqrt 3.$

$\sin 30^\circ = \frac{1}{2}, \quad \sin 45^\circ = \frac{1}{\sqrt 2}, \quad \sin 60^\circ = \frac{\sqrt 3}{2}.$

If the angle does not appear in this list, the problem will provide the value (or you will use complementary-angle relations).

Worked examples

Example 1. A vertical tower stands on level ground. From a point $50$ m from its foot, the angle of elevation of the top is $60^\circ$. Find the height.

Let $h$ be the height. Then $\tan 60^\circ = \frac{h}{50} \Rightarrow h = 50\sqrt 3 \approx 86.6 \text{ m}.$

Example 2. The angle of elevation of the top of a pole from a point $24$ m away is $45^\circ$. Find the height of the pole.

$\tan 45^\circ = h/24 = 1 \Rightarrow h = 24$ m. (When the elevation is $45^\circ$, height equals horizontal distance.)

Example 3. A kite is flying at a height of $60$ m above the ground. The string attached to it is tied at a point on the ground and makes an angle of $60^\circ$ with the ground. Assuming there is no slack, find the length of the string.

Here $h = 60$, angle $= 60^\circ$, find $L$: $\sin 60^\circ = \frac{60}{L} \Rightarrow L = \frac{60}{\sqrt 3/2} = \frac{120}{\sqrt 3} = 40\sqrt 3 \approx 69.3 \text{ m}.$

Example 4. A $1.5$-m-tall observer stands $28.5$ m from the foot of a tower. The angle of elevation of the top of the tower from the observer's eye is $45^\circ$. Find the height of the tower.

The triangle is formed by the eye, the top of the tower, and a point on the tower at eye-level (height $1.5$ m). Let the vertical leg = $x$. $\tan 45^\circ = \frac{x}{28.5} \Rightarrow x = 28.5.$ The total tower height $= x + 1.5 = 28.5 + 1.5 = 30$ m. (Never forget to add the observer's height when it is given.)

Example 5. From a point on the ground, the angle of elevation of the top of a $15$-m tall tower is $30^\circ$. How far is the observer from the foot of the tower?

$\tan 30^\circ = 15/b \Rightarrow b = 15/\tan 30^\circ = 15\sqrt 3 \approx 25.98 \text{ m}.$

Try it yourself

1. A tower casts a shadow $30$ m long when the sun's elevation is $60^\circ$. Find the tower's height.
2. The angle of elevation of a balloon from a point on the ground is $45^\circ$. The balloon is at $50$ m. How far is the observer from the point directly below the balloon?
3. A ladder $10$ m long rests against a vertical wall making a $60^\circ$ angle with the ground. How high up does it reach?
4. From a window $12$ m high, the angle of depression of a point on the road is $30^\circ$. Find the distance of the point from the foot of the building.
5. The string of a kite makes a $30^\circ$ angle with the ground. The string is $200$ m long and there is no slack. Find the height of the kite.
6. A pole stands vertically on the ground. From a distance of $50$ m, its top is seen at an angle of elevation of $45^\circ$. Find the pole's height.
7. An observer of height $1.6$ m stands $40$ m from a chimney. The top of the chimney is seen at $30^\circ$ from the observer's eyes. Find the chimney's height.
8. A flagpole $9$ m high stands on the ground. Find the length of the shadow when the sun's elevation is $30^\circ$.
9. The angle of depression of a boat from the top of a $75$-m cliff is $45^\circ$. Find the boat's distance from the foot of the cliff.
10. A vertical tower is $50$ m tall. From a point on the ground the elevation of its top is $60^\circ$. Find the distance from that point to the foot.

Pitfalls / Insight

The two most common mistakes are (a) forgetting to add the observer's height when it is non-zero and (b) confusing which side is opposite and which is adjacent to the marked angle. The cure for both is a clean diagram with the angle clearly marked at the observer's eye and the right-angle clearly marked at the foot.

Always cross-check: if your computed height is comically large (e.g., $300$ m for a kite at $30^\circ$ elevation, $50$ m string), retrace.