# final question on bearings

• Oct 13th 2007, 06:10 PM
cocoknny
final question on bearings
The angle of elevation from the top of the building 200 feet high to the top of a taller building 100 yards away is 10 degree. Find the height of the taller building correct to the nearest tenth of a foot.

100yards=300ft

200 is the short building. 300 is the adjacent
• Oct 13th 2007, 06:27 PM
Jhevon
Quote:

Originally Posted by cocoknny
The angle of elevation from the top of the building 200 feet high to the top of a taller building 100 yards away is 10 degree. Find the height of the taller building correct to the nearest tenth of a foot.

100yards=300ft

200 is the short building. 300 is the adjacent

see the diagram below. the black rectangle is the short building, the gray rectangle is the tall building.

they are 300 feet apart. and the angle of elevation is 10 degrees. thus we can use the tangent ratio to find BC.

Now, $\displaystyle \tan \left( 10^o \right) = \frac {BC}{300}$

$\displaystyle \Rightarrow BC = 300 \tan \left( 10^o \right)$

So, the height, h, of the taller building is given by:

$\displaystyle h = 200 + BC$

i suppose you can take it from here
• Oct 13th 2007, 06:50 PM
cocoknny
hello
can u explain why 200ft is added even though its the shorter building?
• Oct 13th 2007, 07:02 PM
Jhevon
Quote:

Originally Posted by cocoknny
can u explain why 200ft is added even though its the shorter building?

look at the diagram. the taller building is taller by the length BC. so its height is the height of the short building plus BC. so it is 200 + BC