Question

13 a shark is swimming 28ft below sea level. if the angle of depression from a boat on the water to the shark is 19°, what is the horizontal distance (in feet) between the boat and the shark?

Explanation:

Step1: Define right triangle sides

Let $x$ = horizontal distance (adjacent side), $28$ ft = vertical depth (opposite side), angle of depression = $19^\circ$ (equal to the angle inside the triangle at the boat).

Step2: Use tangent trigonometric ratio

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$
$\tan(19^\circ) = \frac{28}{x}$

Step3: Solve for $x$

Rearrange to isolate $x$:
$x = \frac{28}{\tan(19^\circ)}$
Calculate $\tan(19^\circ) \approx 0.3443$, so:
$x \approx \frac{28}{0.3443}$

Step4: Compute final value

$x \approx 81.3$

Answer:

$\approx 81.3$ feet