Ambiguous case — two triangles from two sides and an angle

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A triangle $ABC$ has $\overline{AB}=5$, $\overline{BC}=3$ and $\widehat{A}=20°$.

(a) Explain why two possible triangles exist, and sketch them.

(b) Find the missing sides and angles of both triangles.

Solution

Side $\overline{BC}=a=3$ is opposite $\widehat{A}=20°$, while side $\overline{AB}=c=5$ is opposite $\widehat{C}$. By the sine rule: $\sin\widehat{C}=\frac{c\sin\widehat{A}}{a}=\frac{5\sin20°}{3}\approx0.570.$ Since $\sin\widehat{C}<1$ and the known side opposite $\widehat A$ ($a=3$) is shorter than side $c=5$, angle $\widehat C$ can be either acute or obtuse: this is the ambiguous case (SSA), with two solutions: $\widehat{C}_1\approx34.75°\qquad\text{or}\qquad\widehat{C}_2\approx180°-34.75°=145.25°.$

Triangle 1. $\widehat{C}_1\approx34.75°$, so $\widehat{B}_1=180°-\widehat A-\widehat C_1\approx125.25°$, and by the sine rule $\overline{AC}=b_1=\dfrac{a\sin\widehat B_1}{\sin\widehat A}\approx7.16$.

Triangle 2. $\widehat{C}_2\approx145.25°$, so $\widehat{B}_2=180°-\widehat A-\widehat C_2\approx14.75°$, and $b_2=\dfrac{a\sin\widehat B_2}{\sin\widehat A}\approx2.23$.

$\boxed{\widehat C_1\approx34.75°,\ \widehat B_1\approx125.25°,\ b_1\approx7.16;\quad \widehat C_2\approx145.25°,\ \widehat B_2\approx14.75°,\ b_2\approx2.23}$