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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

User Kenny Yap
by
5.5k points

2 Answers

5 votes

Answer:

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

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

A salvage ship’s sonar locates wreckage at a 12 degree angle of depression. A diver-example-1
User Sunilr
by
5.8k points
0 votes

Answer: First option is correct.

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

Explanation:

Since we have given from the figure shown below:


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

Consider ΔABC,


tan\theta=(AB)/(BC)\\\\tan12\textdegree=(40)/(BC)\\\\BC=(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.


A salvage ship’s sonar locates wreckage at a 12 degree angle of depression. A diver-example-1
User Adiana
by
5.8k points