Question 429711
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Consider the triangle with vertices at the aircraft, the car with the *[tex \Large 35^\circ] angle of depression, and the point on the ground directly below the aircraft.


Given the angle of depression of *[tex \Large 35^\circ], the angle of the triangle at the aircraft vertex must be the complement of the depression angle, namely *[tex \Large 55^\circ]


Let *[tex \Large y] represent the distance from the point on the ground immediately below the aircraft and the car with the *[tex \Large 55^\circ] angle of depression.


Then *[tex \Large \tan(55^\circ)\ =\ \frac{y}{5150}]


hence *[tex \Large y\ =\ 5150\cdot\tan(55^\circ)]


Using your calculator you can easily calculate the value of *[tex \Large y]


If you let *[tex \Large x] represent the distance from the point on the ground immediately below the aircraft and the OTHER car, you can use the same process above to derive *[tex \Large x\ =\ 5150\cdot\tan(33^\circ)]


The sum *[tex \Large x\ +\ y] rounded to the nearest foot, is the desired distance.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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