ASA, ASS, SAS's 2nd triangle(pre calc)?

Y

YUJIN

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1. A=40 B=20 a=2 (givens)
2. A=110 C=30 c=3 (givens)
3. a=2 c=1 C=25 (givens)
i need to solve for first triangle which i understand by using ths law of sine.
but i also need to find the second triangle.
how can i come up with it???
please help!!!!!!!!!!! i have a test on it in about 40 minutes..
 
2.) It's done the same way.

A + B + C = 180
B = 180 - A - C
B = 180 - 110 - 30
B = 40

sine rule

sinA / a = sinB / b = sinC / c

sinC / c = sinA / a
a / sinA = c / sinC
a = c * sinA / sinC
a = 3 * sin(110) / sin(30)
a = 6 * sin(110)

b / sinB = c / sinC
b = sinB * c / sinC
b = sin(40) * 3 / sin(30)
b = 6 * sin(40)
 
2.) It's done the same way.

A + B + C = 180
B = 180 - A - C
B = 180 - 110 - 30
B = 40

sine rule

sinA / a = sinB / b = sinC / c

sinC / c = sinA / a
a / sinA = c / sinC
a = c * sinA / sinC
a = 3 * sin(110) / sin(30)
a = 6 * sin(110)

b / sinB = c / sinC
b = sinB * c / sinC
b = sin(40) * 3 / sin(30)
b = 6 * sin(40)
 
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