How do you solve the triangle of A=76^circ, a=34, b=21A=76,a=34,b=21?

1 Answer
Oct 23, 2017

See the following :)

Explanation:

By the Law of Sines, (sinA)/a = (sinB)/b = (sinC)/csinAa=sinBb=sinCc
For this question, we know that A= 76^@, a= 34 and b=21A=76,a=34andb=21, and we can simply plug in the numbers and get B:

[sin(76^@)]/34 = (sin B)/21sin(76)34=sinB21

B= 36.81980185^@B=36.81980185

A+B+C=180^@A+B+C=180 (angle sum of triangle)
C=67.18019815^@C=67.18019815

[sin(76^@)]/34 = (sin C)/csin(76)34=sinCc
[sin(76^@)]/34 = (sin67.18019815^@ )/csin(76)34=sin67.18019815c
c=32.29818587c=32.29818587

Hope this can help you :)