Question

A shark spotter is standing at lookout point 25m above the waters edge. He spots a shark in the water at an angle of depression of 50. If the swimmer that is also in the water is sim from the foot of the lookout spot, how far is the shark from the swimmer?​

Answer 1

The shark is 29.78 from the swimmer.

What is angle of depression?

Angle of depression is the angle between the line of sight and the vertical or perpendicular to the earth surface.

Analysis:

The height of the lookout, the distance of shark from swimmer and line of sight form a right angle triangle.

So that, tan 50 = distance of shark from swimmer/25

Distance of swimmer = 25 tan 50

Distance of swimmer from shark = 25(1.1910 = 29.78m

In conclusion, the distance of the shark from the swimmer is 29.78m

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Answer 2

The distance of the shark from the swimmer is 21m

Trigonometry

From the question, we are to determine the distance the shark is from the swimmer

In the diagram,

S is the position of the shark

M is where the shark spotter is standing

and B is the position of the swimmer

The distance between the shark and the swimmer is /BS/

Using SOH CAH TOA

$tan 50^\circ = \frac{25}{/BS/}$

$1.19175 = \frac{25}{/BS/}$

∴ $/BS/ = \frac{25}{1.19175}$

/BS/ = 20.97755

/BS/ ≅ 21 m

Hence, the distance of the shark from the swimmer is 21m.

Learn more on Trigonometry here: https://brainly.com/question/2004882

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