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

[tex]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}[/tex]

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

Hence, First option is correct.

abidemiokin abidemiokin · Answer 2 · 2020-04-26T03:50:38+02:00

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

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

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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