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?

5.739°

Step-by-step 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 ; we can determine the angle that the submarine should fire the torpedo in order to hit the ship

∴ substituting our parameters; we have:

(Sin ∝) 5y  = y × sin (150)

(Sin ∝) =

(Sin ∝) =

(Sin ∝) =

(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°

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step-by-step explanation:

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =