Pleasee Help!! If a triangle with a = 10.9 m, c = 30.2 m, and C = 50° exists. If it does, determine A to the nearest degree.?

Use the Sine Rule
Which is
SinA/a = SinB/b = SinC/c
Permutate any two of the above out of the three.
Hence
SinA/a = SinC / c =>
SinA = a*SinC/c
SinA = 10.9 (Sin50) / 30.2
SinA = 10.9(0.766044443) / 30.2
SinA = 0.276486239...
A = Sin^-1(0.2764...)
A = 16.05 degrees.
A = 16 degrees ( nearest degree).

a/sin A = b/sin B = c/sin C
10.9/sin A = c/sin(50)
=> sin A = 10.9 sin(50) /30.2
sin A = 0.276
A = sin^-1(0.276)
A = 16.05°
B = 180 - (50+16.05) = 113.95°
b/sin B = c/sin C
b/sin(113.95) = 30.2/sin(50)
b = sin(113.95) *30.2 /sin(50)
= 36.03 m

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