Question

A navy cruiser detects a nuclear submarine that is 2500 feet from the cruiser. The angle between the water level and the submarine is 15.5 degrees. How deep is the submarine?

Answer 1

Answer:

Depth of the submarine = 668.09 feet

Step-by-step explanation:

From the figure attached,

In triangle ABC,

Distance between a navy cruiser and a nuclear submarine = 2500 feet

Angle between the water level and the submarine = 15.5°

By applying sine rule in the given triangle,

sin(∠BAC)° = $\frac{BC}{AC}$

sin(15.5°) = $\frac{BC}{2500}$

BC = 2500[sin(15.5)]

BC = 668.09 feet

Therefore, submarine is 668.09 feet deep in the sea.