How Far Away Is That Boat? Angles From a Lighthouse Tell You

A tower, a tree or a lighthouse forms a right angle with the level ground. If you know one distance and one angle, a trigonometric ratio gives the missing side of that right triangle — no climbing or wading needed.

The ratio used most is

$$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$$

because heights and horizontal distances are the two shorter sides. This chapter covers the two key angles, one-triangle and two-triangle problems, and rivers, boats and poles. Values use $\sqrt{3}\approx 1.732$.

The angle of elevation is the angle between the horizontal and the line of sight when you look up at an object; the angle of depression is the angle between the horizontal and the line of sight when you look down.

Marking them on a diagram:

- Draw a horizontal line through the observer's eye
- Draw the line of sight to the object
- Elevation is measured up from the horizontal; depression is measured down from it

A key fact. If $A$ looks down at $B$ with angle of depression $\theta$, then $B$ looks up at $A$ with angle of elevation $\theta$. The two horizontal lines are parallel, so the angles are alternate angles.

Worked example. From the top of a building, a boat is seen at an angle of depression of $30^{\circ }$. From the boat, the angle of elevation of the top is also $30^{\circ }$.

An everyday example. Looking up at a kite gives an angle of elevation; looking down from a balcony at a scooter on the street gives an angle of depression.

The trap. The angle of depression is measured from the horizontal, not from the vertical wall of the building.

Sketch the right triangle, mark the known side and angle, choose the ratio linking them to the unknown, and solve.

Worked example 1 — tower. From a point $50$ m from the foot of a tower, the angle of elevation of the top is $60^{\circ }$.

$$\tan 60^{\circ }=\frac{h}{50}h=50\sqrt{3}\approx 86.6\text{ m}$$

Worked example 2 — kite. A kite string is $120$ m long and makes $30^{\circ }$ with the ground.

$$\sin 30^{\circ }=\frac{h}{120}h=60\text{ m}$$

Worked example 3 — shadow. A pole is $10$ m tall and the sun's elevation is $30^{\circ }$.

$$\tan 30^{\circ }=\frac{10}{s}s=10\sqrt{3}\approx 17.32\text{ m}$$

Worked example 4 — eye height. A person whose eyes are $1.5$ m above the ground stands $30$ m from a tree and sees its top at $45^{\circ }$.

$$\text{height above eye}=30\tan 45^{\circ }=30\text{ m}\text{tree}=30+1.5=31.5\text{ m}$$

An everyday example. A kite flyer on a rooftop can estimate the kite's height this way.

The substance. Add the observer's eye height when the angle is measured from eye level.

Draw both triangles sharing a common side, write an equation from each, and solve them together to find the unknowns.

Worked example 1 — moving observer. The top of a tower is seen at $60^{\circ }$. After walking $20$ m farther away, it is seen at $30^{\circ }$. Let the first distance be $x$ and the height $h$.

$$h=x\sqrt{3}h=\frac{x+20}{\sqrt{3}}$$
$$3x=x+20x=10\text{ m}h=10\sqrt{3}\approx 17.32\text{ m}$$

Worked example 2 — top and bottom of a tower. From the top of a $60$ m building, the angles of depression of the top and bottom of a tower are $30^{\circ }$ and $60^{\circ }$.

$$\text{distance}=\frac{60}{\tan 60^{\circ }}=20\sqrt{3}\text{ m}\text{drop to tower top}=20\sqrt{3}\times \tan 30^{\circ }=20\text{ m}$$
$$\text{tower}=60-20=40\text{ m}$$

An everyday example. Two angles to a temple gopuram, taken from two spots on one road, fix its height.

The substance. The shared side — usually the horizontal distance — links the two triangles, so find it first.

Set up right triangles from the observer's position, find the horizontal distances or heights in each, then add or subtract them depending on how the objects are placed.

Worked example 1 — river. A tree on the far bank is seen at $60^{\circ }$ from the near bank. Moving $40$ m straight back, it is seen at $30^{\circ }$.

$$x\sqrt{3}=\frac{x+40}{\sqrt{3}}3x=x+40x=20\text{ m}$$

**The river is $20$ m wide** and the tree is $20\sqrt{3}\approx 34.64$ m tall.

Worked example 2 — two boats. From a lighthouse $100$ m high, two boats are seen at angles of depression of $30^{\circ }$ and $45^{\circ }$.

$$\text{distances}=\frac{100}{\tan 30^{\circ }}=173.2\text{ m}\text{and}\frac{100}{\tan 45^{\circ }}=100\text{ m}$$

- On opposite sides: $173.2+100=273.2$ m apart
- On the same side: $173.2-100=73.2$ m apart

Worked example 3 — pole on a building. From a point on the ground, the bottom and top of a flagpole on a $20$ m building are seen at $45^{\circ }$ and $60^{\circ }$.

$$\text{distance}=20\text{ m}\text{top}=20\sqrt{3}\approx 34.64\text{ m}\text{pole}\approx 14.64\text{ m}$$

An everyday example. A lighthouse keeper can judge how far fishing boats are from their angles.

The substance. Same side means subtract; opposite sides means add.