A jet A is flying above 2 ships B and C which are 18km apart. If the angles of depression of B and C from are 60 and 32 degrees respectively, find the height of the jet above the sea level, correct to the nearest m.

Answer 1

How good are you at sketching 3-D diagrams?
I let P be the point directly below plane A,
and I now have 4 triangles, each in their own plane.
Redraw each with a side view:
right-triangle ABP , right angle at P
right-triangle ACP, right angle at P
triangle BCP , (on ocean floor)
triangle ABC
let AP, the height of the plane be h km

in triangle ABP , sin60 = h/BA
h = ABsin60

in triangle ACP , sin 32 = h/AC
h = ACsin32
therefore, ABsin60 = ACsin32
√3/2AB = .5299AC
AB = .6119 AC

See if that gets you anywhere.
I hope I am not overthinking this problem