I got the stuff down cold but I cant seem to figure out this problem

Side 1=7.21

Side 2=7.21

Side 3=10

The question states: Using law of sines or law of cosines find all 3 angles.

Its so easy but i keep blanking.

- Apr 21st 2009, 07:40 AMjohn504I feel like an idiot for not being able to figure out this basic trig proplem, help?
I got the stuff down cold but I cant seem to figure out this problem

Side 1=7.21

Side 2=7.21

Side 3=10

The question states: Using law of sines or law of cosines find all 3 angles.

Its so easy but i keep blanking. - Apr 21st 2009, 07:47 AMe^(i*pi)
I changed 1,2 and 3 to a,b and c respectively to make notation easier. In a triangle side a is opposite angle A and so on.

Use the cos rule to find angle C:

$\displaystyle cos(C)=\frac{a^2+b^2-c^2}{2ab}$

Then you can use the sine rule rule to find A and/or B:

$\displaystyle \frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)}$

To find the last angle either use the sine rule again or (easier) take the other two angles from pi or 180 degrees

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edit: Since this is an isosceles triangle side a = side b and so angle A = angle B

Once you've found angle C you can do $\displaystyle \frac{\pi - C}{2}$ - Apr 21st 2009, 07:57 AMjohn504
Thanks, I got it now. I had to switch up angles b and c but i got it.