Question 204821
Two students standing 3000M apart see the same airplane flying in the sky between them. 
Student A measures the angle of elevation to be 37 degrees.
Student B measures the angle of elevation to be 30 degrees. 
How high is the airplane.
:
Find the 3rd angle of the triangle formed by the plane and the two observers:
180 - 37 - 30 = 113 degrees
:
Find the side (a) opposite the 37 degree angle using the law of sines
{{{sin(113)/3000}}} + {{{sin(37)/a}}}
Find the sines and cross multiply
.9502a = 3000 * .6018
.9502a = 1805.445
a = {{{1805.445/.9502}}}
a = 1961.364 
:
Drop a line from the 113 degree angle and perpendicular to the base,
 this is the altitude of the plane
;
This forms two right angle triangles, use the one containing the angle of 30 degrees
Find the altitude (h) using the sine of 30 degrees:
Sin(30) = {{{h/1961.364}}}}
.5 = {{{h/1961.364}}}}
h = .5 * 1961.346
h = 980.673  meters is the height of the plane