Posted by **Jennifer** on Wednesday, November 11, 2009 at 12:48pm.

From a point A on the ground, the angle of elevation to the top of a tall building is 24.1 degrees. From a point B, which is 600 ft closer to the building, the angle of elevation is measured to be 30.2 degrees. Find the height of the building. PLEASE HELP ME!

- Trig -
**drwls**, Wednesday, November 11, 2009 at 1:15pm
Get out a piece of paper and draw the situation. You have two right triangles. Each have two points that are the same (the top and bottom of the bulding), but the third points (where the observer is located) are different. Let H be the height of the building, in feet. Point A is X ft away and point B is A - 600 feet away.

Solve those two simultaneous equations:

H/X = tan 24.1

H/(600-X) = tan 30.2

As a first step in solving them, you can solve for X first, using

(600-X)/X = tan 30.2/tan 24.1 = 1.3011

(I used a calculator for that)

Rewrite as

780.67 = 2.3011 X

X = 339.3 ft

Now use either of the first two equations to solve for H.

- Trig -
**Reiny**, Wednesday, November 11, 2009 at 1:45pm
or

look at the non-right-angles triangle, with its top angle at the top of the building.

That top angle can be easily found to be 6.1 degrees. We can find the side coming up from the 30.2 degree angle, call it y

by sine law

y/sin24.1 = 600/sin6.1

y = 2305.56

then in the small right-angled triangle

sin 30.2 = H/y

H = 2305.56sin30.2

= 1159.74

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