Question

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?

Answer 1

Answer:

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 = $360^{o}$ - $324^{o}$

     = 0$36^{o}$

Sum of angles in a triangle = $180^{o}$

<P + <Q + <S = $180^{o}$

048° + 0$36^{o}$ + <S = $180^{o}$

$84^{o}$ + <S = $180^{o}$

<S  = $180^{o}$ -  $84^{o}$

    = $96^{o}$

<S = $96^{o}$

Applying the Sine rule,

$\frac{y}{Sin P}$ = $\frac{x}{Sin Q}$ = $\frac{z}{Sin S}$

$\frac{y}{Sin P}$ = $\frac{z}{Sin S}$

$\frac{y}{Sin 48^{o} }$ = $\frac{17}{Sin 96^{o} }$

$\frac{y}{0.74314}$ = $\frac{17}{0.99452}$

⇒ y = $\frac{12.63338}{0.99452}$

       = 12.703

y = 12.70 km

$\frac{x}{Sin Q}$ = $\frac{z}{Sin S}$

$\frac{x}{Sin 36^{o} }$ = $\frac{17}{Sin 96^{o} }$

$\frac{x}{0.58779}$ = $\frac{17}{0.99452}$

⇒ x = $\frac{9.992430}{0.99452}$

      = 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.