Right Triangle Problems

Surveyors, pilots, builders and navigators all use right triangles to find distances they can’t measure with a tape: the height of a cliff, the distance to a boat, the angle of a ramp. Once you know the primary trigonometric ratios, the hard part of these problems is turning the words into a clear diagram. This page shows you how. All angles are in degrees.

Key ideas

Draw a diagram first

For every word problem:

1. Sketch the situation and find the right angle (a wall meets the ground, a tower stands upright, north is perpendicular to east).
2. Label what you know and use a letter for what you want.
3. Mark the angle, and label the sides as opposite, adjacent or hypotenuse relative to it.
4. Choose SOH, CAH or TOA, solve, and answer in a sentence with units.
5. Ask whether the answer is reasonable.

Angles of elevation and depression

• The angle of elevation is the angle up from the horizontal to your line of sight, when you look up at something.
• The angle of depression is the angle down from the horizontal to your line of sight, when you look down at something.

Both are always measured from a horizontal line, never from a vertical one.

The angle of elevation from $P$ equals the angle of depression from $T$.

The horizontal line at $T$ is parallel to the ground, so the angle of depression from $T$ to $P$ and the angle of elevation from $P$ to $T$ are alternate angles (a Z pattern), and they are equal. That means you can put the angle of depression inside the triangle at the bottom, where it’s easier to work with.

Clinometers and eye height

A clinometer is a simple tool that measures an angle of elevation. You look along it at the top of an object and read the angle. But the angle is measured from your eyes, not from the ground, so the triangle gives the height above eye level. Add your eye height at the end.

Problems with two right triangles

Some problems involve two right triangles that share a side (often a height or a horizontal distance). The plan:

• solve the triangle that has enough information first, and use the shared side to get into the second triangle, or
• if neither triangle can be solved alone, write an expression for the shared side from each triangle and set them equal.

Worked examples

Example 1: A ladder against a wall

A $6.0$ m ladder leans against a wall and makes an angle of $72^\circ$ with the ground. How high up the wall does it reach, and how far is its foot from the wall?

Solution. The wall meets the ground at a right angle. The ladder is the hypotenuse. Relative to the $72^\circ$ angle at the ground, the height $h$ on the wall is opposite and the distance $d$ along the ground is adjacent.

$$\sin 72^\circ = \frac{h}{6.0} \quad\Rightarrow\quad h = 6.0 \sin 72^\circ \approx 5.7 \text{ m}$$ $$\cos 72^\circ = \frac{d}{6.0} \quad\Rightarrow\quad d = 6.0 \cos 72^\circ \approx 1.9 \text{ m}$$

The ladder reaches about $5.7$ m up the wall, and its foot is about $1.9$ m from the wall.

Check: $\sqrt{5.71^2 + 1.85^2} \approx 6.0$. ✓

Example 2: Using a clinometer

Kai stands $25$ m from the base of a flagpole on level ground. Using a clinometer, he measures the angle of elevation to the top as $38^\circ$. His eyes are $1.6$ m above the ground. How tall is the flagpole?

Solution. Draw a horizontal line from Kai’s eyes to the pole. That makes a right triangle: the horizontal side is $25$ m (adjacent to $38^\circ$), and the vertical side $x$ is the part of the pole above eye level (opposite).

$$\tan 38^\circ = \frac{x}{25} \quad\Rightarrow\quad x = 25 \tan 38^\circ \approx 19.53 \text{ m}$$

Now add Kai’s eye height:

$$\text{height} \approx 19.53 + 1.6 = 21.13 \approx 21.1 \text{ m}$$

The flagpole is about $21.1$ m tall.

Example 3: An angle of depression

From the top of a lighthouse, $42$ m above the water, a keeper sees a sailboat at an angle of depression of $9^\circ$. How far is the boat from the base of the lighthouse?

Solution. The angle of depression from the top equals the angle of elevation from the boat (alternate angles). So in the right triangle formed by the lighthouse, the water and the line of sight, the angle at the boat is $9^\circ$.

Relative to that angle, the $42$ m height is opposite and the distance $d$ is adjacent:

$$\begin{aligned} \tan 9^\circ &= \frac{42}{d} \\ d \tan 9^\circ &= 42 \\ d &= \frac{42}{\tan 9^\circ} \approx 265.2 \end{aligned}$$

The boat is about $265.2$ m from the base of the lighthouse. A small angle of depression means the boat is far away compared with the height, so a large answer makes sense.

Example 4: Two angles of elevation

From point $A$ on level ground, the angle of elevation to the top of a tower is $30^\circ$. From point $B$, $40$ m closer to the tower in a straight line, the angle of elevation is $50^\circ$. How tall is the tower?

Two right triangles, $\triangle ACT$ and $\triangle BCT$, share the height $h$.

Solution. Let $h$ be the height and $x$ be the distance $BC$. Neither triangle can be solved alone: each has only one known length or none. So write $h$ from each triangle.

In $\triangle BCT$: $\quad \tan 50^\circ = \dfrac{h}{x}$, so $h = x \tan 50^\circ$.

In $\triangle ACT$: $\quad AC = x + 40$, and $\tan 30^\circ = \dfrac{h}{x + 40}$, so $h = (x + 40)\tan 30^\circ$.

Both expressions equal $h$, so set them equal and solve for $x$:

$$\begin{aligned} x \tan 50^\circ &= (x + 40)\tan 30^\circ \\ x \tan 50^\circ &= x \tan 30^\circ + 40 \tan 30^\circ && \text{expand} \\ x \tan 50^\circ - x \tan 30^\circ &= 40 \tan 30^\circ && \text{collect the } x \text{ terms} \\ x(\tan 50^\circ - \tan 30^\circ) &= 40 \tan 30^\circ && \text{common factor} \\ x &= \frac{40 \tan 30^\circ}{\tan 50^\circ - \tan 30^\circ} \approx 37.59 \end{aligned}$$

Then

$$h = x \tan 50^\circ \approx 37.59 \tan 50^\circ \approx 44.8$$

The tower is about $44.8$ m tall.

Check with the other triangle: $(37.59 + 40)\tan 30^\circ \approx 44.8$. ✓

(Later in this unit you’ll see a second way to do this, using the sine law in $\triangle ABT$.)

Common mistakes

Measuring the angle of depression from the vertical. Angles of elevation and depression are always measured from the horizontal. If a problem gives an angle of depression of $9^\circ$, the angle between the line of sight and the vertical lighthouse is $81^\circ$, not $9^\circ$.

Putting the angle of depression in the wrong place. The angle of depression sits outside the triangle, at the top, between the horizontal and the line of sight. Use alternate angles to move it to the bottom of the triangle, as in Example 3.

Forgetting eye height. With a clinometer, the triangle starts at your eyes. Add the eye height to get the full height of the object.

Skipping the diagram. Most wrong answers in these problems come from labelling the wrong side as opposite or adjacent. A quick sketch, with the right angle and the given angle marked, prevents this.

Using the wrong distance in two-triangle problems. In Example 4, the base of the big triangle is $x + 40$, not $40$. Label each triangle’s sides separately.

Rounding in the middle. In multi-step problems, keep full calculator values (or at least two extra decimal places) until the final answer.

Practice

1. (Warm-up) A kite is flying on a string $50$ m long. The string makes an angle of elevation of $40^\circ$ with the ground. Assuming the string is straight and is held at ground level, how high is the kite?

Solution

The string is the hypotenuse, and the height $h$ is opposite the $40^\circ$ angle: SOH.

$$\sin 40^\circ = \frac{h}{50} \quad\Rightarrow\quad h = 50 \sin 40^\circ \approx 32.1 \text{ m}$$

The kite is about $32.1$ m high.

2. (Warm-up) From the roof of a $30$ m building, the angle of depression to a parked car is $25^\circ$.

• (a) What is the angle of elevation from the car to the roof?
• (b) How far is the car from the base of the building?

Solution

(a) $25^\circ$. The angles of elevation and depression between the same two points are equal (alternate angles).

(b) At the car, the $30$ m height is opposite and the distance $d$ is adjacent: TOA.

$$\tan 25^\circ = \frac{30}{d} \quad\Rightarrow\quad d = \frac{30}{\tan 25^\circ} \approx 64.3 \text{ m}$$

3. (Warm-up) A $5.5$ m ladder leans against a wall with its foot $1.4$ m from the wall. What angle does the ladder make with the ground?

Solution

The ladder is the hypotenuse, and the $1.4$ m distance is adjacent to the angle at the ground: CAH.

$$\cos \theta = \frac{1.4}{5.5} \quad\Rightarrow\quad \theta = \cos^{-1}\left(\frac{1.4}{5.5}\right) \approx 75^\circ$$

4. (Core) Priya stands $18.0$ m from the base of a tree. Her clinometer shows an angle of elevation of $41^\circ$ to the top, and her eyes are $1.5$ m above the ground. How tall is the tree?

Solution

Height above eye level (opposite) from the horizontal distance (adjacent): TOA.

$$x = 18.0 \tan 41^\circ \approx 15.65 \text{ m}$$

Add the eye height: $15.65 + 1.5 \approx 17.1$ m. The tree is about $17.1$ m tall.

5. (Core) A wheelchair ramp rises $0.5$ m over a horizontal distance of $6.0$ m.

• (a) What angle does the ramp make with the ground?
• (b) How long is the sloped surface of the ramp?

Solution

(a) The rise is opposite the angle and the horizontal distance is adjacent: TOA.

$$\tan \theta = \frac{0.5}{6.0} \quad\Rightarrow\quad \theta = \tan^{-1}\left(\frac{0.5}{6.0}\right) \approx 5^\circ$$

(b) The sloped surface is the hypotenuse. By the Pythagorean theorem:

$$L = \sqrt{6.0^2 + 0.5^2} = \sqrt{36.25} \approx 6.0 \text{ m}$$

(To two decimal places it’s $6.02$ m, just a little longer than the horizontal distance, since the ramp is so gentle.)

6. (Core) A boat leaves a dock on Georgian Bay and sails $12$ km due north, then $7$ km due east. How far is it from the dock, and in what direction (as an angle east of north) is it from the dock?

Solution

North and east are perpendicular, so the path makes a right triangle with legs $12$ km and $7$ km. The direct distance $d$ is the hypotenuse:

$$d = \sqrt{12^2 + 7^2} = \sqrt{193} \approx 13.9 \text{ km}$$

At the dock, the angle $\theta$ between north and the direct line has the $7$ km leg opposite and the $12$ km leg adjacent:

$$\tan \theta = \frac{7}{12} \quad\Rightarrow\quad \theta = \tan^{-1}\left(\frac{7}{12}\right) \approx 30^\circ$$

The boat is about $13.9$ km from the dock, in a direction about $30^\circ$ east of north.

7. (Core) From a window $12$ m above the ground, the angle of elevation to the top of a building across the street is $35^\circ$, and the angle of depression to the bottom of that building is $22^\circ$. How tall is the building across the street?

Solution

Draw a horizontal line from the window to the other building, of length $d$. It splits the problem into two right triangles that share $d$.

Lower triangle (angle of depression $22^\circ$, opposite side $12$ m):

$$\tan 22^\circ = \frac{12}{d} \quad\Rightarrow\quad d = \frac{12}{\tan 22^\circ} \approx 29.70 \text{ m}$$

Upper triangle (angle of elevation $35^\circ$, adjacent side $d$):

$$y = d \tan 35^\circ \approx 29.70 \tan 35^\circ \approx 20.80 \text{ m}$$

The building’s height is the part below the window line plus the part above it:

$$12 + 20.80 \approx 32.8 \text{ m}$$

8. (Challenge) A plane is flying level at an altitude of $2500$ m toward a lake. The pilot sees the lake at an angle of depression of $18^\circ$. A little later, the angle of depression is $35^\circ$. How far did the plane fly between the two sightings? Give your answer in kilometres to one decimal place.

Solution

At each sighting, the altitude $2500$ m is opposite the angle (moved down to the lake using alternate angles), and the horizontal distance to the lake is adjacent.

$$d_1 = \frac{2500}{\tan 18^\circ} \approx 7694.2 \text{ m}, \qquad d_2 = \frac{2500}{\tan 35^\circ} \approx 3570.4 \text{ m}$$

The plane flew $d_1 - d_2 \approx 7694.2 - 3570.4 = 4123.8$ m, or about $4.1$ km.

9. (Challenge) Two people stand on opposite sides of a radio tower, on level ground, $100$ m apart and in line with its base. From one person, the angle of elevation to the top of the tower is $40^\circ$; from the other, it is $28^\circ$. How tall is the tower?

Solution

Let $h$ be the height, and let $x$ be the distance from the first person (the $40^\circ$ one) to the base. Then the second person is $100 - x$ from the base.

$$h = x \tan 40^\circ \qquad \text{and} \qquad h = (100 - x)\tan 28^\circ$$

Set them equal and solve:

$$\begin{aligned} x \tan 40^\circ &= 100 \tan 28^\circ - x \tan 28^\circ \\ x \tan 40^\circ + x \tan 28^\circ &= 100 \tan 28^\circ \\ x(\tan 40^\circ + \tan 28^\circ) &= 100 \tan 28^\circ \\ x &= \frac{100 \tan 28^\circ}{\tan 40^\circ + \tan 28^\circ} \approx 38.79 \end{aligned}$$$$h = x \tan 40^\circ \approx 38.79 \tan 40^\circ \approx 32.5 \text{ m}$$

Check: $(100 - 38.79)\tan 28^\circ \approx 32.5$. ✓ The tower is about $32.5$ m tall.