Bearing problems & Airplane problems

Question

A woman looks at a plane flying directly at her at a constant height and a speed. The plane is traveling 400mph. She first sees the plane at an angle of elevation of 6 degrees with the horizontal. Six minutes later, the angle of elevation is 15 degrees. What is that height of the plane's path? 

Answer 1

Construct a trapezium, ABCD to represent this question;

 

AD and CB are parallel and horizontal, BD is vertical and represents the height of the plane, A and D are on the ground with the woman positioned at A.

 

angle DAB =ABC (Z angles) is 6 degrees, and DAC is 15 degrees, making BAC = 15-6 =9 degrees

 

distance travelled by the plane in 6 minutes BC, is 400*6/60 = 40 miles

 

using the sine formula AC/sine6 = CB/sine9  ; AC = 40 *0.1045/0.1564 = 26.73 miles

 

drop a vertical line from C to a point X on AD; CX = BD (opposite sides of a rectangle)

 

CX/AC = sine15; therefore CX = 26.73 * 0.2588 = 6.92 miles = BD, or the height of the plane.

 

Hence the height of the plane is 6.92 miles