Use the Law of Sines to solve for all passible triangles tha

Use the Law of Sines to solve for all passible triangles that satisfy the given conditions. a = 20, b = 45,

a) a =20 , b =45 , A = 125 deg

sine rule: a/sinA = b/sinB

sinB = b*sinA/a = 45*sin125/20 = 1.84

since value of sin cannot be greater than zero.

So, no triangles are possible

b) a = 75 ; b= 100 A = 30 deg

sine rule : a/sinA = b/sinB

75/sin30 = 100/sinB

sinB = 100*0.5/75 = 0.67 ----> B1 = 41.81 deg

B2 = 180 -B1= 138.20 deg

Angle C1 = 180-30 -B1 = 108.19 deg

C2 = 180 -30 - B2 = 41.81 deg

Two traingle are possible with angles AB1C1 and AB2C2