Question

A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7,600 feet, and the angles of elevation are given. a. Find BC, the distance from Tower 2 to the plane, to the nearest foot. b. Find CD, the height of the plane from the ground, to the nearest foot. Please show work! It is not multiple choice.

Answer 1

Part A) Find BC, the distance from Tower 2 to the plane, to the nearest foot.

in the triangle ACD
sin16=CD/(7600+BD)--------> CD=sin16*(7600+BD)---------> equation 1 
in the triangle BCD
sin24=CD/BD-----------> CD=sin24*BD---------------> equation 2
equation 1=equation 2
sin16*(7600+BD)=sin24*BD-----> sin16*7600+sin16*BD=sin24*BD
sin24*BD-sin16*BD=sin16*7600----> BD=[sin16*7600]/[sin24-sin16]
BD=15979 ft

in the triangle BCD
cos24=BD/BC---------> BC=BD/cos24-------> 15979/cos24-------> 17491
BC=17491 ft

the answer part 1) BC is 17491 ft


Part 2) Find CD, the height of the plane from the ground, to the nearest foot.

CD=sin24*BD        ( remember equation 2)
BD=15979 ft
CD=sin24*15979 -----------> CD=6499 ft

the answer part 2)  CD is 6499 ft

Answer 2

Answer:

a) about  15,052

b) about 6,122