Question

An airplane flies due north from ikeja airport for 500km.It then flies on a bearing of 060 from a further distance of 300km before over flying a road junction. Calculate A. Distance of airplane from ikeja airport when it was directly above the road junction B. The bearing of the airplane from ikeja airport at this instant

Answer 1

Answer:

482 km

63.94 degrees

Step-by-step explanation:

to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.

Cosine rule states that

C = $\sqrt{A^{2} + B^{2} -2AB cos(c) }$

where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060

solving now, we have

C = $\sqrt{500^{2} + 300^{2} -2 * 500 * 300 cos(60) }$

C = $\sqrt{250000 + 90000 - [215000 cos(60) }]$

C = $\sqrt{340000 - [215000 * 0.5 }]$

C = $\sqrt{340000 - [107500 }]$

C =$\sqrt{232500}$

C = 482 km

The bearing can be gotten by using the Sine Rule.

$\frac{sina}{A}$ = $\frac{sinc}{C}$

sina/500 = sin60/482

482 sina = 500 sin60

sina = $\frac{500 sin60}{482}$

sina = 0.8983

a = sin^-1(0.8983)

a = 63.94 degrees