one side of triangelar lot has a length of 100 feet; the angle opposite that side is 55 degrees. another angle is 63 degrees. how much fence will be needed to enclose the lot? i am having trouble figuring out which angle goes where and if i should use Cos or Tan

- stapel_eliz
**Posts:**1738**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

theshadow wrote:one side of triangelar lot has a length of 100 feet; the angle opposite that side is 55 degrees. another angle is 63 degrees. how much fence will be needed to enclose the lot?

You know that the angle sum for

theshadow wrote:i am having trouble figuring out which angle goes where and if i should use Cos or Tan

When you have the angles and the length of the side opposite one of those angles, it's usually a safe bet that you'll need to use the

The actual placement, in your drawing, of the various angle measures is irrelevant. Just draw a triangle, label the angles with their measures, and label the one side with the given length. Then work from the picture to set up your ratios, and solve for the lengths of the other two sides. The sum of their lengths is the amount of fencing needed.

Hope that helps!

i did 180-55-63=62 then 100/sin55=x/sin63=y/sin62 so x=108.8 y=107.8

is it ok that their about the same?

is it ok that their about the same?

- stapel_eliz
**Posts:**1738**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**