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?

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 [tex]\dfrac{a}{sinA}=\dfrac{b}{sin B}=\dfrac{c}{sinC}[/tex] 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 [tex]7.4^o[/tex] north-west on a ship traveling at [tex]300^o[/tex] bearing.
According to the given information in the question, the following diagram has been prepared.
[tex]\angle A+\angle B+\angle C=180^o\\\angle A=180-(90+60)-7.4\\\angle A= 22.6^o[/tex]
Now, from sine law-
[tex]\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[/tex]
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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