Question

A radar operator on a ship discovers a large sunken vessel lying parallel to the ocean surface, 165 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 40° and 62°.
How long, to the nearest tenth of a metre, is the sunken vessel?

Answer 1

It is hard to explain but here : 

Answer 2

Sunken vessel is 273.6 meters long.

Explanation:

We are given that,

Depth of the sunken vessel = 165 meter.

Angle of depression to the front = 40°

Angle of depression to the front = 65°

Since, we know that,

'The measure of angle of depression is equal to the measure of the angle of elevation'.

So, the corresponding measures of angle of elevations are shown as in the figure.

Now, using trigonometric form for right triangles, we will find the values of x and y.

That is,
`\tan x=(Perpendicular)/(Base)`

So,
`\tan 40=(165)/(x)`

i.e.
`x=(165)/(\tan 40)`

i.e.
`x=(165)/(0.8391)`

i.e. x= 196.64 meter

Also,
`\tan 65=(165)/(y)`

i.e.
`y=(165)/(\tan 65)`

i.e.
`y=(165)/(2.145)`

i.e. y= 76.92 meter

Thus, the length of the vessel = x + y = 196.64 + 76.92 = 273.6 meter

Hence, the sunken vessel is 273.6 meters long.