URGENT! PLEASE HELP.

A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7600 feet and 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

SHOW YOUR WORK

a) BC = 1386.8 ft

b) CD = 565.8 ft

Step-by-step explanation:

Looking at the triangle,

AD = BD + 7600

BD = AD - 7600

Considering triangle BCD, we would apply the the tangent trigonometric ratio.

Tan θ = opposite side/adjacent side. Therefore,

Tan 24 = CD/BD = CD/(AD - 700)

0.445 = CD/(AD - 700)

CD = 0.445(AD - 700)

CD = 0.445AD - 311.5 - - - - - - - -1

Considering triangle ADC,

Tan 16 = CD/AD

CD = ADtan16 = 0.287AD

Substituting CD = 0.287AD into equation 1, it becomes

CD = 0.445AD - 311.5

0.287AD = 0.445AD - 311.5

0.445AD - 0.287AD = 311.5

0.158AD = 311.5

AD = 311.5/0.158

AD = 1971.52

CD = 0.287AD = 0.287 × 1971.52

CD = 565.8 ft

To determine BC, we would apply the Sine trigonometric ratio which is expressed as

Sin θ = opposite side/hypotenuse

Sin 24 = CD/BC

BC = CD/Sin24 = 565.8/0.408

BC = 1386.8 ft

idk

step-by-step explanation: i just don't