Question

A submarine traveling from the east, fires a torpedo at a ship traveling at 300° bearing. The torpedo is fired 7.4° north of west. How much faster must the torpedo be traveling than the ship in order to hit it?​

Answer 1

The answer is 3.88

The explanation is below!

I hope this helps (:

Answer 2

The speed of the torpedo fired must be equal to 3.88 times of the speed of the ship.

Sine Law

The Sine Law is  $\dfrac{a}{sinA}=\dfrac{b}{sin B}=\dfrac{c}{sinC}$ where, a, b, c are the length of the sides of the triangle and A, B, C are the angles of the triangle.

How to apply sine law?

Submarine is heading towards east and fired a torpedo at $7.4^o$ north-west on a ship traveling at $300^o$ bearing.

According to the given information in the question, the following diagram has been prepared.

$\angle A+\angle B+\angle C=180^o\\\angle A=180-(90+60)-7.4\\\angle A= 22.6^o$

Now, from sine law-

$\dfrac{AC}{sinB}=\dfrac{AB}{sinC}\\\dfrac{AC}{sin150}=\dfrac{AB}{sin7.4}\\\dfrac{AC}{AB}=\dfrac{0.1288}{0.5}\\\dfrac{AC}{AB}=0.2576\\AB=3.88AC$

As distance is directly proportional to the time so, the speed of the torpedo fired must be equal to 3.88 times of the speed of the ship.

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