1. ## Height of Aircraft

Two sensors are spaced 700 feet apart along the approach to a small airport. When an aircraft is nearing the airport, the angle of elevation from the first sensor to the aircraft is 20 degrees and from the second sensor to the aircraft it is 15 degrees. Determine how high the aircraft is at this time.

2. Originally Posted by magentarita
Two sensors are spaced 700 feet apart along the approach to a small airport. When an aircraft is nearing the airport, the angle of elevation from the first sensor to the aircraft is 20 degrees and from the second sensor to the aircraft it is 15 degrees. Determine how high the aircraft is at this time.
Pardon my artwork, but this one needs a picture. See attachment.

$\displaystyle \tan 20=\frac{A}{X}$

$\displaystyle X\tan20=A$ [1]

$\displaystyle \tan 15=\frac{A}{X+700}$

$\displaystyle (X+700)\tan 15=A$ [2]

$\displaystyle X\tan20=(X+700)\tan 15$

$\displaystyle X\tan20=X\tan15+700\tan15$

$\displaystyle X\tan20-X\tan15=700\tan 15$

$\displaystyle X(\tan20-\tan15)=700\tan15$

$\displaystyle X=\frac{700\tan15}{\tan20-\tan15}$

$\displaystyle X\approx 1953.37$ ft.

Now substitute back into one of the first equations for A.

$\displaystyle A=X\tan20$ ft.

$\displaystyle A\approx710.9678247$

3. ## wonderfully done!

Originally Posted by masters
Pardon my artwork, but this one needs a picture. See attachment.

$\displaystyle \tan 20=\frac{A}{X}$

$\displaystyle X\tan20=A$ [1]

$\displaystyle \tan 15=\frac{A}{X+700}$

$\displaystyle (X+700)\tan 15=A$ [2]

$\displaystyle X\tan20=(X+700)\tan 15$

$\displaystyle X\tan20=X\tan15+700\tan15$

$\displaystyle X\tan20-X\tan15=700\tan 15$

$\displaystyle X(\tan20-\tan15)=700\tan15$

$\displaystyle X=\frac{700\tan15}{\tan20-\tan15}$

$\displaystyle X\approx 1953.37$ ft.

Now substitute back into one of the first equations for A.

$\displaystyle A=X\tan20$ ft.

$\displaystyle A\approx710.9678247$
What can I say? You did fabulous!