Expert: Sherman D. - 6/12/2005

Question
triangle ABC is a right triangle with right angle C.

1) if a=2 and b=7, find angle A.
2) if angle A=40 degrees, and c=10, find b.

Answer
1.)
a = 2
b = 7
A = ?

a^2 + b^2 = c^2
2^2 + 7^2 = c^2
4 + 49 = c^2
c^2 = 53
c = sqrt(53)

2/sinA = (sqrt(53))/(sin(90))
as we already know sin(90) = 1, so this would be

2/sinA = sqrt(53)
sin(A) = 2/(sqrt(53))
A = 15.9453959°

ANS : about 16°

Even if you were to first find angle B, then subtract that from 90, you will still get the same answer.

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2.)
a/(sin(40)) = 10
a = 10sin(40)
a = 6.427876097

B = 90 - 40
B = 50

6.427876097/(sin(40)) = b/(sin(50))
b = (6.427876097sin(50))/(sin(40))
b = 7.660444431

ANS : b = about 7.66, or 7.7

even if you were to use a^2 + b^2 = c^2, you will still get it to work out.