Trig Problems in Three Dimensions

Real problems don’t always lie flat on a page. Finding the height of a tower you can’t reach, or the distance between two boats seen from a cliff, involves triangles in different planes. The trick is to break the 3-D situation into flat triangles and solve them one at a time. All angles are in degrees.

Key ideas

Strategy

1. Draw a clear diagram. Sketch the 3-D situation, then label every known length and angle.
2. Find the right angles. Anything vertical (a tower, a cliff, a pole) meets level ground at $90^\circ$. These give right triangles.
3. Spot the triangles. Usually one triangle lies flat on the ground and another stands up vertically. They share a side.
4. Solve in order. Find the shared side in one triangle, then use it in the next. Choose SOH CAH TOA for right triangles, and the sine law or cosine law for the others.
5. Keep full values in your calculator until the final answer.

Angles of elevation and depression

• The angle of elevation is measured up from the horizontal to an object above you.
• The angle of depression is measured down from the horizontal to an object below you.

The angle of depression from the top of a cliff to a boat equals the angle of elevation from the boat to the top (they’re alternate angles).

Worked examples

Example 1: The height of a tower

Points $A$ and $C$ are $100$ m apart on level ground. $B$ is the base of a vertical tower $BT$. In the ground triangle, $\angle BAC = 60^\circ$ and $\angle BCA = 50^\circ$. From $A$, the angle of elevation to the top $T$ is $25^\circ$. Find the height of the tower.

The ground triangle $ABC$ is flat; the triangle $ABT$ stands up, with a right angle at $B$.

Solution. Ground triangle. $\angle ABC = 180^\circ - 60^\circ - 50^\circ = 70^\circ$, opposite the $100$ m side. By the sine law:

$$AB = \frac{100\sin 50^\circ}{\sin 70^\circ} \approx 81.52$$

Vertical triangle. $\triangle ABT$ has a right angle at $B$, so:

$$h = AB\tan 25^\circ \approx 81.52\tan 25^\circ \approx 38.01$$

The tower is about $38.01$ m tall.

Example 2: The diagonal of a box

A box is $4$ cm wide, $3$ cm deep, and $12$ cm tall. Find the length of the diagonal from a bottom corner to the opposite top corner, and the angle it makes with the base.

Solution. First the diagonal of the base:

$$d_{\text{base}} = \sqrt{4^2 + 3^2} = 5$$

This diagonal and the $12$ cm height form a right triangle (standing up from the base):

$$d = \sqrt{5^2 + 12^2} = 13, \qquad \tan\theta = \frac{12}{5} \quad\Rightarrow\quad \theta \approx 67.4^\circ$$

The diagonal is $13$ cm long and makes an angle of about $67.4^\circ$ with the base.

Example 3: Two boats from a cliff

From the top of a $60$ m cliff, two boats are seen at angles of depression of $28^\circ$ and $22^\circ$. Looking down from above, the lines from the base of the cliff to the two boats make an angle of $70^\circ$. How far apart are the boats?

Solution. Two vertical right triangles give each boat’s distance from the base of the cliff:

$$d_1 = \frac{60}{\tan 28^\circ} \approx 112.84, \qquad d_2 = \frac{60}{\tan 22^\circ} \approx 148.51$$

The flat triangle on the water has sides $d_1$ and $d_2$ with $70^\circ$ between them. By the cosine law:

$$x^2 = d_1^2 + d_2^2 - 2d_1d_2\cos 70^\circ \quad\Rightarrow\quad x \approx 152.72$$

The boats are about $152.72$ m apart.

Common mistakes

Mixing up which triangle is flat and which is vertical. Shade or colour the ground triangle in your sketch.

Assuming an angle is $90^\circ$ when it isn’t. Only vertical-meets-horizontal angles are guaranteed right angles. The ground triangle is usually oblique.

Using a rounded intermediate value. In Example 1, using $AB \approx 81.5$ instead of the full value can change the final answer. Store values in your calculator.

Measuring an angle of depression from the vertical. It’s measured from the horizontal line of sight.

Practice

1. (Warm-up) A $20$ m flagpole is seen from $35$ m away across level ground. Find the angle of elevation to the top.

Solution

$\tan\theta = \tfrac{20}{35}$, so $\theta \approx 29.7^\circ$.

2. (Warm-up) Find the length of the space diagonal of a box measuring $6$ cm by $8$ cm by $10$ cm.

Solution

$$\sqrt{6^2 + 8^2 + 10^2} = \sqrt{200} \approx 14.14 \text{ cm}$$

3. (Core) From the top of a $45$ m tower, the angle of depression to a car is $18^\circ$. How far is the car from the base of the tower?

Solution

The angle of elevation from the car to the top is also $18^\circ$, so:

$$d = \frac{45}{\tan 18^\circ} \approx 138.50 \text{ m}$$

4. (Core) Points $A$ and $B$ are $40$ m apart on level ground, and $F$ is the base of a tree. $\angle FAB = 48^\circ$ and $\angle FBA = 62^\circ$. From $B$, the angle of elevation to the top of the tree is $35^\circ$. How tall is the tree?

Solution

Ground triangle: $\angle AFB = 180^\circ - 48^\circ - 62^\circ = 70^\circ$.

$$FB = \frac{40\sin 48^\circ}{\sin 70^\circ} \approx 31.63$$

Vertical right triangle at $F$: height $= FB\tan 35^\circ \approx 22.15$ m.

5. (Core) A cube has edges of $5$ cm. Find its space diagonal, and the angle the diagonal makes with the base.

Solution

Base diagonal: $5\sqrt{2}$. Space diagonal: $\sqrt{(5\sqrt{2})^2 + 5^2} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ cm.

$\tan\theta = \dfrac{5}{5\sqrt{2}} = \dfrac{1}{\sqrt{2}}$, so $\theta \approx 35.3^\circ$.

6. (Core) From the top of an $80$ m cliff, two boats are seen at angles of depression of $25^\circ$ and $18^\circ$. The angle between the lines from the base of the cliff to the boats is $50^\circ$. How far apart are the boats?

Solution

$$d_1 = \frac{80}{\tan 25^\circ} \approx 171.56, \qquad d_2 = \frac{80}{\tan 18^\circ} \approx 246.21$$$$x^2 = d_1^2 + d_2^2 - 2d_1d_2\cos 50^\circ \quad\Rightarrow\quad x \approx 189.08$$

The boats are about $189.08$ m apart.

7. (Challenge) A pyramid has a square base with sides of $10$ m, and each slanted edge (from a base corner to the top) is $13$ m long. Find the height of the pyramid and the angle each slanted edge makes with the base.

Solution

The top is directly above the centre of the base. Half the base diagonal is $\tfrac{1}{2}(10\sqrt{2}) = 5\sqrt{2} \approx 7.07$ m.

The height, half-diagonal, and slanted edge form a right triangle:

$$h = \sqrt{13^2 - (5\sqrt{2})^2} = \sqrt{169 - 50} = \sqrt{119} \approx 10.91 \text{ m}$$

$\cos\theta = \dfrac{5\sqrt{2}}{13}$, so $\theta \approx 57.0^\circ$.

8. (Challenge) A drone hovers $120$ m above point $D$ on level ground. Observer $A$ sees it at an angle of elevation of $40^\circ$, and observer $B$ sees it at $30^\circ$. On the ground, $\angle ADB = 90^\circ$. How far apart are the observers?

Solution

$$AD = \frac{120}{\tan 40^\circ} \approx 143.01, \qquad BD = \frac{120}{\tan 30^\circ} \approx 207.85$$

The ground triangle $ADB$ has a right angle at $D$:

$$AB = \sqrt{AD^2 + BD^2} \approx 252.29 \text{ m}$$