# given triangle ABC.....

• Jun 26th 2010, 08:32 PM
ninobrn99
given triangle ABC.....
The question is:
Given the triangle ABC with AB=5 and BC= 5sqrt3/3

The measure of the angle A is 30 degrees. How many choices are there for the measure of angle C?

I've hit a problem with the final solution.

sin(c)/5= sin(30degrees)/(5sqrt3/3)
multiply both sides by 5
=>sin(c)=(1/2)/(sqrt3/3)
=(1/2)(3/sqrt3)
=sqrt3/2

since Im dealing with sin, I reference the unit circle and see that sin is sqrt3/2 at pi/3 and 2pi/3.

That means I have 2 choices for angle C?
• Jun 26th 2010, 09:28 PM
dwsmith
Quote:

Originally Posted by ninobrn99
The question is:
Given the triangle ABC with AB=5 and BC= 5sqrt3/3

The measure of the angle A is 30 degrees. How many choices are there for the measure of angle C?

I've hit a problem with the final solution.

sin(c)/5= sin(30degrees)/(5sqrt3/3)
multiply both sides by 5
=>sin(c)=(1/2)/(sqrt3/3)
=(1/2)(3/sqrt3)
=sqrt3/2

since Im dealing with sin, I reference the unit circle and see that sin is sqrt3/2 at pi/3 and 2pi/3.

That means I have 2 choices for angle C?

As long as the two angles aren't greater than 180, you should be fine since all 3 angles need to equal 180.
• Jun 26th 2010, 09:31 PM
ninobrn99
Thank you dw!