Question

A salvage ship’s sonar locates wreckage at a 12 degree angle of depression. A diver is lowered 40 meters to the ocean floor. How far does the diver need to walk along the ocean floor to the wreckage?
188.2 units
8.5 m
192.4 m
41.0 m

Answer 1

Answer: First option is correct.

The diver need to walk 188.2 units along the ocean floor to the wreckage.

Step-by-step explanation:

Since we have given from the figure shown below:

$AB=40\text{ meters}\\\\\theta=\angle {ACB}=12\textdegree$

Consider ΔABC,

$tan\theta=\frac{AB}{BC}\\\\tan12\textdegree=\frac{40}{BC}\\\\BC=\frac{40}{tan12\textdegree}\\\\BC=\frac{40}{tan{12\textdegree}\\\\BC=188.2\text{ units}$

So, the diver need to walk 188.2 units along the ocean floor to the wreckage.

Hence, First option is correct.

Answer 2

Answer:

The diver need to walk 188.2m along the ocean floor to the wreckage.

Step-by-step explanation:

Check the attachment for diagram.

Using SOH, CAH TOA to get the distance the diver need to walk along the ocean floor to the wreckage,

Note that the angle is a right angled triangle with the height of the triangle being the opposite side and the distance needed as the adjacent side.

Using TOA

Tan(theta) = Opposite/Adjacent

Given theta = 12°

Opposite = 40m

Adjacent side = x

Tan12° = 40/x

x = 40/tan12°

x = 188.2m