Question

A ships bearing is 240(deg). If a submarine due East of the ship, shoots a torpedo traveling 5 times as fast as the ship. What angle should the submarine fire the torpedo in order to hit the ship?

Answer 1

5.739°

Explanation:

For the ship new angle bearing (β), since we are dealing with the second quarter and part of the third quater; we have β to be:

β= 240° - 90°

β= 150°

The question goes further by stating that the submarine shoots a torpedo traveling 5 times as fast as the ship.

i.e if the ship is = y

the submarine = 5y

∴ by sine rule which state that
`(SinA)/(A) =(Sin B)/(B)` ; we can determine the angle that the submarine should fire the torpedo in order to hit the ship

∴ substituting our parameters; we have:

`(Sin(150))/(5y)=(Sin\alpha )/(y)`

(Sin ∝) 5y = y × sin (150)

(Sin ∝) =
`(y*sin150 )/(5y)`

(Sin ∝) =
`(sin150)/(5)`

(Sin ∝) =
`(0.5)/(5)`

(Sin ∝) = 0.1

∝ = Sin⁻¹ 0.1

∝ = 5.739°

Thus, the angle that the submarine should fire the torpedo in order to hit the ship = 5.739°