Question

A radar operator on a ship discovers a large sunken vessel lying parallel to the ocean surface, 120 m directly below the ship. The length of the vessel is a clue to which wreck has been found. The radar operator measures the angles of depression to the front and back of the sunken vessel to be 55° and 46°. How long, to the nearest tenth of a metre, is the sunken vessel?

Answer 1

Answer:

The length of the sunken vessel, to the nearest tenth is 200 m

Step-by-step explanation:

The given information are;

The depth of the sunken vessel below the ship = 120 m

The angle of depression to the front of the sunken vessel = 55°

The angle of depression to the back of the sunken vessel = 46°

Therefore, we have;

The length, A, of the from directly below the ship to the front of the sunken vessel given as follows;

Tan(55°) = ((120 m)/A)

A = 120 / Tan(55°) ≈ 84.025 m

The length, B, of the from directly below the ship to the back of the sunken vessel given as follows;

tan(46°) = ((120 m)/B)

B = 120 / tan(46°) ≈ 115.88 m

The length, L,  of the sunken vessel = A + B

L = A + B = 120 / Tan(55°) + 120 / tan(46°) ≈ 84.025 m + 115.88 m ≈  199.91 m

∴ The length, L, of the sunken vessel, to the nearest tenth = 200 m.