+185 at point the angle elevation of a mountain peak is 40 degrees 20 minutes. at a point 9560 feet farther away in the same horizontal plane, its angle of elevation is 29 degrees 50 minutes. find the distance of the peak above the horizontal plane.
+128598 Let y be the distance from the first observation to the base of the mountain
So we have
tan [40 +1/3]° = h / y where h is the height of the mountain
Solving for y, we have
y = h / tan [40 +1/3]°
And.....from the second observation pt, we have that
tan [ 29 + 5/6] ° = h / [ 9560 + h / tan [ 40 + 1/3]°]
Multiply both sides by [ 9560 + h / tan [ 40 + 1/3]°]
[ 9560 + h / tan [ 40 + 1/3]°] * tan [ 29 + 5/6]° = h simplify
9560 tan[29 + 5/6]° + h * tan [ 29 + 5/6]° / tan [ 40 + 1/3]° = h
Subtract h * tan [ 29 + 5/6]° / tan [ 40 + 1/3]° from both sides
9560 tan [29 + 5/6]° = h - h * tan [ 29 + 5/6]° / tan [ 40 + 1/3]°
Factor out h on the right
9560 tan [29 + 5/6]° = h [ 1 - tan [ 29 + 5/6]° / tan [ 40 + 1/3]° ]
Divide both sides by [ 1 - tan [ 29 + 5/6]° / tan [ 40 + 1/3]° ]
9560 tan [29 + 5/6] / [ 1 - tan [ 29 + 5/6] / tan [ 40 + 1/3] ] = h = about 16,891 ft
