A small plane is 150 km south of the equator. The plane is flying at 150 km/h at a heading of 30∘ to the west of north. how many minutes will the plane cross the equator?

Question

A small plane is  150 km  south of the equator. The plane is flying at  150 km/h  at a heading of 30∘ to the west of north. how many minutes will the plane cross the equator?

MathMate · Accepted Answer

We have a right triangle ABC, CA lies on the equator, where C is the destination (where plane intercepts the equator).
B is the initial position of the plane, and is immediately south of A.
So AB=150 km, ∠ABC=30°.
We need length of BC.
Using definition of cosine,
AB/BC=cos(30)=AB/BC
BC=AB/cos(30)=150/(sqrt(3)/2)=173.21 km
Time required = distance/speed
=150/(sqrt(3)/2) km / 150 km/hr
=2/sqrt(3) hours
=1.155 hours.

Note:
Real life situation is that all distances are on the surface of a sphere of radius 6471 km. The actual calculations used by navigators require spherical trigonometry. In this case, the "spherical triangle" is relatively small compared to the radius of the earth, and the difference in distance amounts to only about 0.01 km.