6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP and 324 from Q. How far is the ship from each of the coastguard stations?

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = -

     = 0

Sum of angles in a triangle =

<P + <Q + <S =

048° + 0 + <S =

+ <S =

<S  = -  

    =

<S =

Applying the Sine rule,

= =

=

=

=

⇒ y =

       = 12.703

y = 12.70 km

=

=

=

⇒ x =

      = 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.