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

B is 156° because 180 minus 24°. C (for the left triangle) is 8° because 180-156-16. 180-90-24= 66° for the other side of C. So now that you have all of the angles labeled, you’re gonna do some SOH, CAH, TOA. Going from angle A, 16°, you have the 7600ft as an adjacent measurement, label it as “A”. then you have BC as an opposite measurement, label it as “O”. You are going to use the TOA method. The “O” is going to be the numerator and the “A” denominator. so it will look like: O/7600. then take the 7600 and do this: 7600(T16). which is the denominator and the Tan option in the calculator and the angle you used, and you will get 2284.8. so BC equals 2285. I hope that sets you up. i’m not sure how to do the second part.

the answer is 31.27mph

5/12, or 5: 12

step-by-step explanation: