A plane travelled for 600 km from airport A, on a bearing of 210°, before changing direction and travelling a further 250 km on a bearing of 300° to land at airport B.
Calculate the three-figure bearing of airport B from airport A.
Hi, this is a topic about bearings and trigonometry.
I would like some assistance, this is the working I have so far:
link to image
Thanks
Best Answer
To answer your question, it might be helpful to refer to a diagram, as shown below:
Since the plane travelled from airport A on a bearing of $210$°, the angle formed by the line taken by the plane and the first South line is $210 \text {°} - 180 \text {°} = 30$°.
Likewise, after changing direction, since the plane travelled on a bearing of $300$° to airport B, the angle formed by the line taken by the plane and the second West line is $300 \text {°} - 270 \text {°} = 30$°.
Let us now calculate the angle formed between the first line taken by the plane (from A) and the second line (to B).
Notice the relationship between angle $q$ and our calculated angle $30$°. Since both North lines are parallel to each other, the first line taken by the plane forms a transversal. Therefore, angle $q$ and our calculated angle are alternate angles. Thus, angle $q = 30$°.
Let us now find angle $p$. Since angle $p$ and our second calculated angle are complementary, angle $p = 90 \text {°} - 30 \text {°} = 60$°.
Thus, the angle formed between the first and second line taken by the plane is $90$° i.e. a right-angle triangle is formed.
We can now use trigonometry to calculate angle $\theta$. Since angle $p + q$ forms a right-angle, the distance $250$ kilometres represents the opposite and the distance $600$ kilometres represents the adjacent. Therefore, we can make use of the tangent formula to calculate angle $\theta$, as follows:
Thus, the three-figure bearing of airport B from airport A is simply the sum of angle $\theta$ and the first bearing of $210$°, as follows:
I hope that helps!