## 2 buildings 150' apart...find height of taller building

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
maria1983
Posts: 9
Joined: Sat Mar 07, 2009 10:00 pm

### 2 buildings 150' apart...find height of taller building

two apartment buildings are 150 feet apart. from a window of a shorter building, the angle od elevation to the top of the other building is 50 degrees and the angle of depression to the base is 38 degrees. find the height of the taller building.

stapel_eliz
Posts: 1743
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
maria1983 wrote:two apartment buildings are 150 feet apart. from a window of a shorter building, the angle od elevation to the top of the other building is 50 degrees and the angle of depression to the base is 38 degrees. find the height of the taller building.

Draw two rectangles, both taller than wide, and one taller than the other.

From somewhere in the middle of the shorter rectangle, draw a horizontal line across to the other rectangle, and label this as "150".

Draw a slantly line from where the horizontal leaves the shorter rectangle to the top of the taller rectangle, and another from the same starting point to the bottom of the taller rectangle. This will form two right triangles.

Label the angles at the shorter building with the given angles values.

Use the tangent ratio, the given distance between the buildings, and the given angles to find the heights of the two right triangles. Add the two heights to find the total height of the taller building.

If you get stuck, please reply showing your work. Thank you!

maria1983
Posts: 9
Joined: Sat Mar 07, 2009 10:00 pm

### Re: 2 buildings 150' apart...find height of taller building

is it 150tan50+150tan38=295.9558829?

maria1983
Posts: 9
Joined: Sat Mar 07, 2009 10:00 pm

### Re: 2 buildings 150' apart...find height of taller building

nvm i got it thnx

morris2710
Posts: 1
Joined: Fri Apr 10, 2009 1:25 am
Contact:

### Re: 2 buildings 150' apart...find height of taller building

It always helps to draw a picture first, so for this problem start by drawing two triangles. One triangle with theta=50 degrees and one with theta=38 degrees. Since the distance between the buildings is 150 ft, the adjacent side is 150 for both triangles. Next, use tangent=opposite/adjacent for each triangle and solve for the opposite side which would look like 150tan50=opposite and 150tan38=opposite. Add those values together and that will give you the height of the building.

Return to “Trigonometry”