Thursday

July 24, 2014

July 24, 2014

Posted by **Trig** on Thursday, September 10, 2009 at 5:29pm.

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 -
**Reiny**, Thursday, September 10, 2009 at 6:05pmlet'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

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