Angle of elevation/inclination problem help!

• Jan 19th 2010, 03:51 AM
purplesky828
Angle of elevation/inclination problem help!
Here's the problem:

1. At a certain point, the angle of elevation of a mountain peak is 40 degrees and 20 minutes. At a point 9560 feet farther away on the same horizontal plane, it's angle of elevation is 29 degrees and 50 minutes. Find the perpendicular distance of the peak from the horizontal.

2. From the top of a tower 60 feet tall the angle of elevation of the top of a building is 14 degrees. From the base of the tower, the angle of elevation of the tops of the building is 28 degrees. How far is the tower from the building and how high is the building?
• Jan 19th 2010, 04:41 AM
Prove It
Quote:

Originally Posted by purplesky828
Here's the problem:

1. At a certain point, the angle of elevation of a mountain peak is 40 degrees and 20 minutes. At a point 9560 feet farther away on the same horizontal plane, it's angle of elevation is 29 degrees and 50 minutes. Find the perpendicular distance of the peak from the horizontal.

2. From the top of a tower 60 feet tall the angle of elevation of the top of a building is 14 degrees. From the base of the tower, the angle of elevation of the tops of the building is 28 degrees. How far is the tower from the building and how high is the building?

1. Draw the situation. You should end up with two triangles, both of which have the same vertical height. Call this height $h$.

If we call the distance from the first point to the peak $x$, the the distance from the second point to the peak will be $x + 9560$.

From triangle 1:

$\tan{40^{\circ}20'} = \frac{h}{x}$

Therefore $x = \frac{h}{\tan{40^{\circ}20'}}$.

From triangle 2:

$\tan{29^{\circ}50'} = \frac{h}{x + 9560}$

Therefore $x = \frac{h}{\tan{29^{\circ}50'}} - 9560$.

So $\frac{h}{\tan{40^{\circ}20'}} = \frac{h}{\tan{29^{\circ}50'}} - 9560$.

Solve for $h$.
• Jan 19th 2010, 05:22 AM
Hello purplesky828

Welcome to Math Help Forum!
Quote:

Originally Posted by purplesky828
...2. From the top of a tower 60 feet tall the angle of elevation of the top of a building is 14 degrees. From the base of the tower, the angle of elevation of the tops of the building is 28 degrees. How far is the tower from the building and how high is the building?

This is very similar to number 1. Let the distance from the foot of the tower to the foot of the building be $d$ feet, and let the height of the tower be $h$ feet.

Draw a diagram, marking in all the distances and the two angles. (Can you do that?)

Then you have two right angled triangles. From one you get:
$\tan14^o = \frac{h-60}{d}$
and from the other, you get:
$\tan28^o = \frac{h}{d}$
Then you need to:

• eliminate $d$ between these two equations;

• solve the resulting equation for $h$;

• substitute back to find $d$.

Can you complete it now?