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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?​

A Submarine Traveling From The East Fires A Torpedo At A Ship Traveling At 300 Bearing The Torpedo Is Fired 74 North Of West How Much Faster Must The Torpedo Be class=

Sagot :

The answer is 3.88

The explanation is below!

I hope this helps (:
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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.

Learn more: https://brainly.com/question/4372174

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