SOLUTION: Determine the greatest possible value for B to the nearest tenth if A = 30°, a = 5, and b = 8.

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Question 373625: Determine the greatest possible value for B to the nearest tenth if A = 30°, a = 5, and b = 8.

Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
We're given two sides of a triangle, the angle opposite one of those sides and we've been asked to find the angle opposite the other side. This is a job for the Law of Sines: sin%28A%29%2Fa+=+sin%28B%29%2Fb. Substituting the given numbers into this equation we get:
sin%2830%29%2F5+=+sin%28B%29%2F8
This being a proportion, we can use cross-multiplying:
sin(30)*8 = 5*sin(B)
Since sin(30) = 1/2 this becomes:
%281%2F2%29%2A8+=+5%2Asin%28B%29
or
4 = 5*sin(B)
Dividing by 5 we get:
4%2F5+=+sin%28B%29
or
0.8 = sin(B)
Now we have to find the largest angle (which can fit inside a triangle) whose sin is 0.8. For this we will need the inverse sin function. (On your calculator the button is probably sin or asin.) From our calculator we can find
sin-1(0.8) = 53.1301023541559808
Rounding to the nearest tenth this is 53.1
(If you don't get this number from your calculator, make sure you are using degree mode, not radian mode.)
We have to remember that this value, 53.1, is just the reference angle. Our sin is a positive 0.8. Sin is positive in the first and second quadrants. In the first quadrant the angle and the reference angle are one and the same. In the second quadrant the angle is 180 minus the reference angle: 180-53.1 = 126.9. Both of these angles are less than 150 (180 minus angle A) so they both can fit inside our triangle. Of course the larger one is 126.9.

Note: There are an infinite number of angles whose sin is 0.8. Other than the two above however, they are all either negative or too big to fit inside our triangle.