Use the Law of Sines to Solve, if Possible, the Triangle or Triangles in the Ambiguous Case
-$$\mathrm { A } = 30 ^ { \circ } , \mathrm { a } = 11 , \mathrm {~b} = 22$$
A) $\mathrm { B } = 90 ^ { \circ } , \mathrm { C } = 60 ^ { \circ } , \mathrm { c } = 19.1$
B) $B = 60 ^ { \circ } , C = 90 ^ { \circ } , c = 19.1$
C) $B = 60 ^ { \circ } , C = 60 ^ { \circ } , \mathrm { C } = 19.1$
D) no triangle

Question

Quizplus · Accepted Answer

Given $A = 30^\circ, a = 11, b = 22$, using the Law of Sines: $\frac{\sin A}{a} = \frac{\sin B}{b}$. Substituting the given values and solving for $B$ shows that $B = 90^\circ$, which makes $C = 60^\circ$ since the sum of angles in a triangle is $180^\circ$. The side $c$ can be found using the Law of Sines again or the Pythagorean theorem due to the right angle at $B$, leading to $c \approx 19.1$.

Use the Law of Sines to Solve, If Possible, the Triangle