Question

The bearing of Q from P is 150

0 and the bearing of P from R is 015
0
. If Qand Rare 24kmand 32km respectively from P:
(i) represent this information in a diagram;
(ii) Calculate the distance between Q and R, correct to two decimal places;
(iii) Find the bearing of R from Q, correct to the nearest degree.

Answer 1

(i) see first attached diagram

(ii) QR = 22.67 km (2 dp)

(iii) 243°

Explanation:

A bearing is the angle in degrees measured clockwise from north.

Part (i)

see first attached diagram

Part (ii)

The angles marked in blue on the second attached diagram are Consecutive Interior Angles. In this case they add to 180° as the North lines are parallel.

The sum of angles around a point is 360°

⇒ m∠QPR (shown in red on the second attached diagram) = 360 - 150 - (180 - 15) = 45°

To calculate the distance between Q and R (marked in red on the attached diagram), use the cosine rule.

⇒ QR² = PR² + QP² - 2(PR)(QP)cos(QPR)

⇒ QR² = 32² + 24² - 2(32)(24)cos(45)

⇒ QR² = 513.8839841...

⇒ QR = √513.8839841...

⇒ QR = 22.67 km (2 dp)

Part (iii)

We need to find the angle marked in green on the third attached diagram.

To do this, we need to find the angle marked in pink and the angle marked in orange, then subtract them from 360°

Pink angle = 180 - 150 = 30° (using the same consecutive angle theorem as before)

Orange angle (o) using the sine rule:

⇒ sin(o)/32 = sin(45)/QR

⇒ sin(o) = 32sin(45)/22.67

⇒ sin(o) = 0.9981652337...

⇒ o = 86.52868176...°

Therefore, the green angle = 360 - 30 - 86.52868176... = 243° (nearest degree)