Asked by someonestupid
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.
Answers
Answered by
Reiny
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
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
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