IGCSE Trigonometry: Angle of Depression with Inverse Tangent

A vertical lighthouse stands on a horizontal sea wall. The lamp room window $W$ is $48\text{ m}$ vertically above sea level. A lighthouse keeper at $W$ watches two fishing boats, $A$ and $B$ , which are on the sea on the same side of the lighthouse, in line with its base $F$ .

Boat $A$ is at a horizontal distance of $95\text{ m}$ from the foot of the lighthouse $F$ . Boat $B$ is closer in, at a horizontal distance of $60\text{ m}$ from $F$ .

(All distances are measured in a vertical plane. You may assume the line $FW$ is vertical and the sea surface through $F$ , $B$ , $A$ is horizontal.)

(a) Calculate the angle of depression of boat $A$ from the window $W$ . Give your answer correct to $1$ decimal place. [3]

(b) Calculate the angle of depression of boat $B$ from the window $W$ . Give your answer correct to $1$ decimal place. [2]

Setting up the problem

Draw the situation as a right-angled triangle. The window $W$ is at the top, the foot of the lighthouse $F$ is directly below it, and the boat sits out on the horizontal sea surface.

Vertical side (height): $WF = 48\text{ m}$
Horizontal side (distance along the sea): $F\!A$ or $F\!B$

The angle of depression is the angle measured downwards from the horizontal line of sight at the window $W$ . Because the keeper’s horizontal line at $W$ is parallel to the sea surface, the angle of depression at $W$ is equal (alternate angles) to the angle of elevation of $W$ as seen from the boat. So we may work inside the right-angled triangle using the angle at the boat, which has:

$\text{opposite} = WF = 48\text{ m}, \qquad \text{adjacent} = \text{horizontal distance}.$

Since we know the opposite and adjacent sides, we use tangent:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}.$

(a) Angle of depression of boat $A$

Here the adjacent side is $FA = 95\text{ m}$ .

$\tan(\theta_A) = \frac{48}{95}.$

Taking the inverse tangent:

$\theta_A = \tan^{-1}\!\left(\frac{48}{95}\right) = \tan^{-1}(0.50526\ldots) = 26.805\ldots^{\circ}.$

$\boxed{\theta_A \approx 26.8^{\circ}}$

(Make sure your calculator is in degree mode.)

(b) Angle of depression of boat $B$

Boat $B$ is nearer, so the adjacent side is $FB = 60\text{ m}$ . The height is unchanged:

$\tan(\theta_B) = \frac{48}{60} = 0.8.$

$\theta_B = \tan^{-1}(0.8) = 38.659\ldots^{\circ}.$

$\boxed{\theta_B \approx 38.7^{\circ}}$

This is sensible: the closer boat is seen at a steeper downward angle.

(c) Difference in the angles of depression

$\theta_B - \theta_A = 38.659\ldots^{\circ} - 26.805\ldots^{\circ} = 11.85\ldots^{\circ}.$

$\boxed{\approx 11.9^{\circ}}$

Tip: keep the unrounded values in your calculator when subtracting. Using the rounded figures $38.7^{\circ} - 26.8^{\circ} = 11.9^{\circ}$ happens to give the same answer here, but carrying full accuracy until the final step is the safe habit.

Alternative check for (a)

You could instead find the slant line of sight $WA$ first by Pythagoras, $WA = \sqrt{48^2 + 95^2} = \sqrt{2304 + 9025} = \sqrt{11329} \approx 106.4\text{ m},$ and then use sine: $\sin(\theta_A) = \dfrac{48}{106.4} = 0.4511\ldots$ , giving $\theta_A = \sin^{-1}(0.4511) \approx 26.8^{\circ}$ , the same result, confirming the answer.