Example  1
Solve the triangle from Case 4.




Information given:

\(\angle A = 35^\circ\)
\(a = 8\)
\(b = 10\)

The sine law may be used to solve for the first case, which will be the base where \(\angle B\) is an acute angle.

\[\begin{align}
 \frac{a}{{\sin A}} &= \frac{b}{{\sin B}} \\
 \frac{8}{{\sin 35^\circ }} &= \frac{{10}}{{\sin B''}} \\
 \sin B'' &= \frac{{10\sin 35^\circ }}{8} \\
 \sin B'' &= 0.716... \\
 \angle B'' &= \sin ^{ - 1} \left( {0.716...} \right) \\
 \angle B'' &= 45.804...^\circ  \\
 \angle B'' &\doteq 45.8^\circ  \\
 \end{align}\]
Use the definition of a triangle to find \(\angle C\).

\(\begin{align}
 \angle C &\doteq 180^\circ - 35^\circ - 45.8^\circ \\
 &\doteq 99.2^\circ \end{align}\)

The sine law can be used to find the remaining side, \(c\).

\[\begin{align}
 \frac{a}{{\sin A}} &= \frac{c}{{\sin C}} \\
 \frac{c}{{\sin 99.2^\circ }} &\doteq \frac{8}{{\sin 35^\circ }} \\
 c &\doteq \frac{{8\sin 99.2^\circ }}{{\sin 35^\circ }} \\
 c &\doteq 13.768... \\
 c &\doteq 13.8 \\
 \end{align}\]
The question is not done yet; recall there are two solutions to this question. Think back to reference angles, and note that sine is positive in Quadrants I and II; therefore, \(\angle B'' \doteq 45.8^\circ\) in Quadrant I and \(\angle B' \doteq 180^\circ - 45.8^\circ \doteq 134.2^\circ\) in Quadrant II.



In this second case, \(\angle C \doteq 180^\circ - 35^\circ - 134.2^\circ \doteq 10.8^\circ\). This angle is part of \(\triangle AB'C\), as shown in the diagram below.



Because a pair is known, this triangle can be solved using the sine law.

\[\begin{align}
 \frac{a}{{\sin A}} &= \frac{c}{{\sin C}} \\
 \frac{8}{{\sin 35^\circ }} &\doteq \frac{c}{{\sin 10.8^\circ }} \\
 \frac{{8\sin 10.8^\circ }}{{\sin 35^\circ }} &\doteq c \\
 2.613... &\doteq c \\
 2.6 &\doteq c \\
 \end{align}\]