Town B is 150 km on a bearing of 050° from town A. Town C is on a bearing of 110° from town A and 150° from town B. Town D is 240 km on a bearing of 320° from town A. Calculate distances and bearings

Here is the solution to question 47.

47. Town B is 150 km on a bearing of $050^\circ$ from town A. Another town C is on a bearing of $110^\circ$ from town A and on a bearing of $150^\circ$ from town B. A fourth town D is 240 km on a bearing of $320^\circ$ from town A. Without using a scale drawing, calculate to the nearest kilometre:

First, let's determine the angles within the triangles formed by towns A, B, C, and D.
Let N be the North direction from town A.

• Bearing of B from A is $050^\circ$. So, $\angle NAB = 50^\circ$.
• Bearing of C from A is $110^\circ$. So, $\angle NAC = 110^\circ$.
• Bearing of D from A is $320^\circ$. So, $\angle NAD = 320^\circ$.

From these, we can find the angles at A:

$$\angle BAC = \angle NAC - \angle NAB = 110^\circ - 50^\circ = 60^\circ$$ $$\angle CAD = 360^\circ - (\angle NAD - \angle NAC) = 360^\circ - (320^\circ - 110^\circ) = 360^\circ - 210^\circ = 150^\circ$$

Alternatively, $\angle CAD = \angle NAD - \angle NAC = 320^\circ - 110^\circ = 210^\circ$. The interior angle is $360^\circ - 210^\circ = 150^\circ$.

Now, let's find the angle $\angle ABC$.
Draw a North line at B, parallel to the North line at A.
The alternate interior angle to $\angle NAB$ is the angle between the North line at B and the line BA (pointing towards A). So, $\angle NBA = 50^\circ$.
The bearing of C from B is $150^\circ$. This means the angle from the North line at B, clockwise to BC, is $150^\circ$. So, $\angle NBC = 150^\circ$.

$$\angle ABC = \angle NBC - \angle NBA = 150^\circ - 50^\circ = 100^\circ$$

In $\triangle ABC$, we have:
$AB = 150$ km
$\angle BAC = 60^\circ$
$\angle ABC = 100^\circ$
The third angle $\angle BCA = 180^\circ - (60^\circ + 100^\circ) = 180^\circ - 160^\circ = 20^\circ$.