SOLUTION: How do I solve triangle ABC, given A = 43 degrees, a = 13 and c = 16? Not sure how to figure out side b. The triangle is not a right angled triangle. Thanks. Matt

Algebra ->  Trigonometry-basics -> SOLUTION: How do I solve triangle ABC, given A = 43 degrees, a = 13 and c = 16? Not sure how to figure out side b. The triangle is not a right angled triangle. Thanks. Matt      Log On


   

Question 698781: How do I solve triangle ABC, given A = 43 degrees, a = 13 and c = 16? Not sure how to figure out side b. The triangle is not a right angled triangle.
Thanks.
Matt
Answer by Simnepi(216) About Me  (Show Source):
You can put this solution on YOUR website!
You need to use the cosine rule, which is
a%5E2+=+b%5E2+%2B+c%5E2+-+2%2Ab%2Ac%2ACosA
Substitute in the values for the variables to give
13%5E2+=+b%5E2+%2B+16%5E2+-+2%2Ab%2A16%2ACos43
to give
169 = b%5E2 + 256 - 23.403*b
rearranging gives
b%5E2 - 23.403b + 87 = 0
now if we let b = x we get
x%5E2 - 23.304x + 87 = 0
Now we can solve this equation using the formula for quadratic solutions eg
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ so
x+=+%28-23.403+%2B-+sqrt+%28-23.403%5E2+-+4%2A1%2A87%29%29%2F%282%2A1%29
calculating this gives us 2 values for x(b)
these are x = 18.767 or x = 4.636