Question

14 An aeroplane leaves an airport A and to 2 decimal place. (WAEC) flies on a bearing 035° for 1 hours at 600 km/h to an airport B. It then flies on a bearing 130° for 1 hours at 400 km/h to an airport C. Calculate the correct (a) distance from C to A, correct to the nearest kilometre. (b) bearings romC from A, correct to the nearest degree. (WAEC)​

Answer 1

So the bearing from C to A is approximately 87° to the nearest degree.

Explanation:

(a) To find the distance from C to A, we can use the cosine rule. Let's call the distance from A to B "d1" and the distance from B to C "d2". Then:

d1² = (600 km/h * 1 hour)² = 360,000 km²

d2² = (400 km/h * 1 hour)² = 160,000 km²

Now, let's find the angle between d1 and d2. We know that the angle between the two bearings is 130° - 35° = 95°. We also know that the angle between d1 and d2 is 180° - 95° = 85° (because the three angles in a triangle add up to 180°).

Using the cosine rule:

d² = d1² + d2² - 2d1d2cos(85°)

d² = 360,000 + 160,000 - 2(600)(400)cos(85°)

d ≈ 1,013 km

So the distance from C to A is approximately 1,013 km to the nearest kilometre.

(b) To find the bearings from C to A, we can use trigonometry. Let's call the bearing from C to A "x". Then:

cos(x) = (d2/d)

cos(x) = 160,000/1,013,000

x ≈ 87°

So the bearing from C to A is approximately 87° to the nearest degree.