Finding the lengths of the sides using a given angle

[greggory]

Homework Statement

Well, here is the problem. Assuming that the shape is a triangle, find the lengths of all sides, given one angle is 51.168821565 degrees using the law of sine.

sin(51.168821565) = 0.785398161...

The Attempt at a Solution

I basically tried using the theorems, such as that all angles will add up to 180 degrees and A+B > C always.

I tried to find the lengths, but couldn't. Help would be appreciated.

EDIT:

I tried this:
d = opposite
h = hypotnues
sin 51.168821565 = d / h
d / h = 0.785398161

Does this mean there are infinite numbers that can fit side lengths?

 

[th4450]

There are infinitely many triangles with only one angle given.

 

[HallsofIvy]

Homework Statement

Well, here is the problem. Assuming that the shape is a triangle, find the lengths of all sides, given one angle is 51.168821565 degrees using the law of sine.

This is impossible. As th4450 said, there are an infinite number of different triangles having that angle. Draw the angle choose any point on one ray, any point on the other and connect them.n Go back and reread the problem. You have to be given at least one side.

sin(51.168821565) = 0.785398161...

The Attempt at a Solution

I basically tried using the theorems, such as that all angles will add up to 180 degrees and A+B > C always.

I tried to find the lengths, but couldn't. Help would be appreciated.

EDIT:

I tried this:
d = opposite
h = hypotnues
sin 51.168821565 = d / h
d / h = 0.785398161

Does this mean there are infinite numbers that can fit side lengths?

That formula works only for right triangles. You say that you are to use the "law of sine". Do you know what that is? (Not the definition of sine.) In order to use the "law of sine" to find the length of a side, you must already know two angles and one side.

 

[greggory]

Well, I accidently left out some information. The other angle is 90 degrees, which means that the other angle must be 38.831178435, right? Also, one given side is 25(if I read the problem right this time).