# Find Angle of Elevation

• Sep 4th 2008, 01:38 PM
magentarita
Find Angle of Elevation
A carpenter is preparing to put a roof on a garage that is 20 feet in height by 40 feet long by 20 feet wide. A steel support beam 46 feet in length is positioned in the center of the garage. To support the roof, another beam will be attached to the top of the center beam. At what angle of elevation is the new beam?
In other words, what is the pitch of the roof?

• Sep 4th 2008, 01:41 PM
Matt Westwood
Is the garage 40 feet tall or 40 feet wide / long? I think we need to be told.
• Sep 4th 2008, 05:23 PM
magentarita
Here it is...
Quote:

Originally Posted by Matt Westwood
Is the garage 40 feet tall or 40 feet wide / long? I think we need to be told.

Yes, you are right.

It is 20 feet in height, 40 feet long and 20 feet wide.

Now, can you show me how to find the answer?
• Sep 4th 2008, 10:24 PM
Matt Westwood
Again, to solve these problems you have to draw a picture. And again, the question seems to be incomplete.

The beam at the centre, I presume it's pointing straight up. The second beam though, I expect it leads from the top of this centre beam and goes to the top of the side of the garage. But which side - the one 20 feet away or 10 feet away? The angle will of course be different depending on which side it's talking about. Or is it the corner?

Let's assume it's the corner. The walls are 20 feet high. So you're looking at a right-angled triangle whose points are at:

a) The top of the centre pole (46 feet up)
b) The top of the corner of the garage (20 feet up)
c) The point 20 feet up the centre pole.

The vertical line of the triangle is 46-20 feet = 26 feet.
The horizontal line of the triangle is the distance from the corner to the centre which you can find by pythagoras: $\sqrt{20^2 + 10^2}$. Now you can use the approprate trigonometrical formula (I suggest the one for tangent) to get the angle of elevation, which is the slope of the hypotenuse (the long slopey line) of this right-angled triangle.

I can't emphasise the importance of drawing a picture to clarify things.

Also, write down the assumptions you make (i.e. the ones I made above) when doing these problems if stuff is not clear, then at least the examiner knows you know what you're talking about.
• Sep 5th 2008, 03:13 AM
magentarita
great...
Quote:

Originally Posted by Matt Westwood
Again, to solve these problems you have to draw a picture. And again, the question seems to be incomplete.

The beam at the centre, I presume it's pointing straight up. The second beam though, I expect it leads from the top of this centre beam and goes to the top of the side of the garage. But which side - the one 20 feet away or 10 feet away? The angle will of course be different depending on which side it's talking about. Or is it the corner?

Let's assume it's the corner. The walls are 20 feet high. So you're looking at a right-angled triangle whose points are at:

a) The top of the centre pole (46 feet up)
b) The top of the corner of the garage (20 feet up)
c) The point 20 feet up the centre pole.

The vertical line of the triangle is 46-20 feet = 26 feet.
The horizontal line of the triangle is the distance from the corner to the centre which you can find by pythagoras: $\sqrt{20^2 + 10^2}$. Now you can use the approprate trigonometrical formula (I suggest the one for tangent) to get the angle of elevation, which is the slope of the hypotenuse (the long slopey line) of this right-angled triangle.

I can't emphasise the importance of drawing a picture to clarify things.

Also, write down the assumptions you make (i.e. the ones I made above) when doing these problems if stuff is not clear, then at least the examiner knows you know what you're talking about.

I thank you so much Matt.