+9479 \(\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \\~\\ \frac{\sin 56}{17} = \frac{\sin C}{14} \\~\\ \frac{14\sin 56}{17} = \sin C \\~\\ \arcsin{(\frac{14\sin 56}{17})} = C \\~\\ C \approx 43.058^{\circ}\)
Since there are 180 degrees in every triangle,
180 = A + B + C
A = 180 - B - C
A ≈ 180 - 56 - 43.058
A ≈ 80.942º
All that's left to find is side a.
\(\frac{\sin 56}{17} \approx \frac{\sin 80.942}{a} \\~\\ (a)\frac{\sin 56}{17} \approx \sin 80.942 \\~\\ a \approx \frac{17\sin80.942}{sin56} \\~\\ a \approx 20.250\)
Recap:
A ≈ 80.942º a ≈ 20.25
B = 56º b = 17
C ≈ 43.058º c = 14
*edit* I put c = 4 instead of c = 14
+9479 \(\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \\~\\ \frac{\sin 56}{17} = \frac{\sin C}{14} \\~\\ \frac{14\sin 56}{17} = \sin C \\~\\ \arcsin{(\frac{14\sin 56}{17})} = C \\~\\ C \approx 43.058^{\circ}\)
Since there are 180 degrees in every triangle,
180 = A + B + C
A = 180 - B - C
A ≈ 180 - 56 - 43.058
A ≈ 80.942º
All that's left to find is side a.
\(\frac{\sin 56}{17} \approx \frac{\sin 80.942}{a} \\~\\ (a)\frac{\sin 56}{17} \approx \sin 80.942 \\~\\ a \approx \frac{17\sin80.942}{sin56} \\~\\ a \approx 20.250\)
Recap:
A ≈ 80.942º a ≈ 20.25
B = 56º b = 17
C ≈ 43.058º c = 14
*edit* I put c = 4 instead of c = 14
+129852 triangle was A,B,C angle B 56 and side length little c is 14 little b is 17 solve triangle using Law of sines (SSA case)
Solving for angle C we have
sin C / 14 = sin 56 / 17
sin C = [14 * sin 56] / 17
arcsin [ (14 * sin 56) / 17 ] ≈ 43°
And angle A = 180 - 56 - 43 ≈ 81°
And side a =
a / sin 81 = 17/ sin 56 → a = 17sin 81/sin 56 ≈ 20.25
To see if we have a second triangle, C might also be 180 - 43 = 137°
However......when we add this to the known angle of 56°, we get more than 180°....so....only one triangle is possible
a = 20.25 b = 17 c = 14
A = 81 B = 56 C = 43
