Question

from the top of a vertical cliff 125 metres high, the angle of depression of a boat is 18°50. find the distance of the boat from the cliff. select an answer: 360.5 m 266.5 m 361.3 m 366.5 m

Explanation:

Step1: Convert angle of depression to decimal degrees

The angle of depression is $18^{\circ}50'$. Since $1^{\circ} = 60'$, then $50'=\frac{50}{60}\approx0.833^{\circ}$. So the angle $\theta = 18 + 0.833=18.833^{\circ}$.

Step2: Set up tangent - ratio equation

Let the distance of the boat from the cliff be $x$. We know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Here, the height of the cliff (opposite side) is $h = 125$ m and the distance from the cliff to the boat (adjacent side) is $x$. So $\tan(18.833^{\circ})=\frac{125}{x}$.

Step3: Solve for $x$

We can rewrite the equation as $x=\frac{125}{\tan(18.833^{\circ})}$. Since $\tan(18.833^{\circ})\approx0.346$, then $x=\frac{125}{0.346}\approx361.3$ m.

Answer:

361.3 m