In a triangle ABC, AB=7, BC=10, Angle A = 80. To the nearest tenth, what is Angle C?

We need to be careful here... we have a SSA situation.....using the Law of Sines we have

sinA / 10  = sin C / 7

7*sin 80 / 10 = sin C

sin^-1( 7* sin 80 / 10)  = about 43.6°

And angle B = (180 - 80 - 43.8) = 56.4°

We need to check to see if B could be  the supplemental to 56.4°

But. this is impossible since, if A is 80°, the angle supplemental to 56.4° would be 123.6° and these two angles added together would be > 180°

So....we only have one triangle.....

i dont know if this is right but i think:

a=80

b=-73

c=83

Draw the triangle first then use the sine rule

Thanks Chris, I forget about that quite often.

Chris is saying that      $$sin\theta = sin(180-\theta)$$

So C could be obtuse or acute and you need to consider this when using the sine rule.

Sometimes there are 2 answers.

The acute angle here is    about 44 degrees.

the obtuse equivalent is  about  180-44=136 degrees.

We know hat angle A = 80 degrees 

80+136>180   so there is no way this triangle can have an angle of 136degrees because the angle sum of any triangle has to be equal to 180 degrees.  (the sum of any 2 angles must be less than 180 degrees)

so

C=44 degrees to the nearest degree.