# Trig

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Two observers 1 km apart spot an airplane. The angles of elevation of the airplane from the two observers are 22 degrees 50 minutes and 24 degrees 40 minutes. If the observers and the airplane are in the some vertical plnae, compute the altitude of the airplane.

ok i don\'t know how to do this if you could show me how to do this

Also I think i may be reading this wrong i don\'t know

they put 22 then raise it to a circle which is used for degrees but than after that they do something like 50\' which i believe are minutes

a2 = unkown horizontal distance from observer to verticle component that the line of sight makes at the air plane

a2 tan 24.7 = opposite

(1 km + a2) tan 22.8 = x

a2 tan 24.7 = (1 km + a2) tan 22.8

not sure how to solve for a2

• Trig -

let's make a diagram.
(I think your's is probably close)

draw a straight line and mark two points A and B and label AB = 1 km

Place the plane above the line AB and to the right of it, call it P
Extend AB, and from P draw a perpendicular down to the extended line AB and let Q be the point where it meets AB extended.
You should now have two right-angled triangles, APQ and BPQ, with the right angle at Q.
Label BQ as x, and PQ as y

now in the inside triangle,
tan 24.7º = y/x
y = xtan24.7

in the larger triangle,
tan 22.8 = y/(x+1)
y = (x+1)tan22.8

so (x+1)tan22.8 = xtan24.7
expanding
xtan22.8 + tan22.8 = xtan24.7
tan22.8 = xtan24.7 - xtan22.8
tan22.8 = x(tan24.7 - tan22.8)
x = tan22.8/(tan24.7-tan22.8)

now use your calculator to find x (I don't mine handy)
once you have x, sub it back into
y = xtan24.7 to find the height y

• Trig -

0.2

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