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.
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
