1. ## length of pole

Im suppose to find how high a building is [ not literally, lol ].

A flagpole 40ft stands on top of a building. The angle of elavation for the top of the pole is: 54 degrees 54' and the angle of elevation for the bottom of the pole is 47 degrees 30'. How high is the building?

I did:

tan 54 54' = 40
x

x= 28.11

and then did tan 47 30' = 40
x

x= 36.65

and subtract that and get a difference of 8.54.

Is that how you find the height of the building?

2. $\tan(54^{\circ}54') = \frac{40+h}{x}$

$\tan(47^{\circ}30') = \frac{h}{x}$

where x is the unknown horizontal distance to the building and h is the height of the building.

using the 2nd equation ...

$x = \frac{h}{\tan(47^{\circ}30')}$

substitute for x in the 1st equation and solve for h.

3. how do you go about solving for h?

4. $\tan(54^{\circ}54') = \frac{40+h}{x}$

$x = \frac{h}{\tan(47^{\circ}30')}$

substitute for x in the 1st equation and solve for h ...

$\tan(54^{\circ}54') = \frac{(40+h)\tan(47^{\circ}30')}{h}$

$h\tan(54^{\circ}54') = (40+h)\tan(47^{\circ}30')$

$h\tan(54^{\circ}54') = 40\tan(47^{\circ}30') + h\tan(47^{\circ}30')$

$h\tan(54^{\circ}54') - htan(47^{\circ}30') = 40\tan(47^{\circ}30')$

$h[\tan(54^{\circ}54') - tan(47^{\circ}30')] = 40\tan(47^{\circ}30')$

$h = \frac{40\tan(47^{\circ}30')}{\tan(54^{\circ}54') - tan(47^{\circ}30')}$