i need help with geometry! has to do w/triangles and angle of depression?
here's the word problem..
Two hills are 2 miles apart. The elevation of the taller hill is 2707 ft. The angle of depression from the top of the taller hill to the top of the shorter hill is 7 degrees. Find the elevation of the shorter hill to the nearest foot. (1 mile = 5280)
i have the final tommorow, so could you tell me how you got the answer, i really need to understand how to do this!!
- RackbraneLv 71 decade agoFavorite Answer
Let AB be the smaller hill with A at the top and B at the bottom.
Let CD be the taller hill with C at the top and D at the bottom.
BD is horizontal ground.
Through A, draw horizontal line AE meeting CD at E.
Through C, draw horizontal line CP with P on the same side of C as A.
The 7 deg. angle of depression is angle PCA.
Because AE and PC are parallel, angle PCA = angle EAC.
(Incidentally, angle EAC is the angle of elevation of C from A.)
In triangle ACE:
Angle AEC = 90 deg.
AE = BD = 2 miles = 2 * 5280 ft.
CE / EA = tan(7 deg)
CE = EA tan(7) = 2 * 5280tan(7) ft.
AB = ED = CD - CE
= 2707 - 2 * 5280tan(7)
= 1410 ft.
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- hawbakerLv 43 years ago
Draw the parent. it quite is a triangle with 34º, 39º and 109º angles. Then use regulation of sines (in case you have had that): x = distance to buddy x / sin 34 = 450 / sin 109 (on the grounds that a million a million/2 min = ninety sec x 5 ft/sec = 450 ft) x sin 109 = 450 sin 34 so x = (450 sin 34) ÷ sin 109; see what it is and around it off.