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Completion requirements
Solve \(\triangle QRS\) given \(RS = 10\thinspace \rm{m}\), \(QR =
14\thinspace \rm{m}\), and \(\angle Q = 39^\circ\). If two solutions are
possible, give both.
First, check to see how many answers there are:
\(\begin{align}
b\sin A &= s\sin Q \\
&= 14\sin 39^\circ \\
&= 8.810... \\
\end{align}\)
Because side \(RS\) is greater than \(8.810…\), but less than side \(QR\), there are two possible triangles.
Case 1: \(\triangle QRS''\)
Find \(\angle S''\).
Find \(\angle R\).
\(\begin{align}
\angle R &\doteq 180^\circ - 39^\circ - 61.8^\circ \\
&\doteq 79.2^\circ \end{align}\)
Find side \(QS''\).
Case 2: \(\triangle QRS'\)
Find \(\angle S'\).
The reference angle is the same as \(\angle S''\).
\(\begin{align}
\angle S' &= 180^\circ - \theta _R \\
\angle S' &\doteq 180^\circ - 61.8^\circ \\
\angle S' &\doteq 118.2^\circ \\
\end{align}\)
Find \(\angle R\).
\(\begin{align}
\angle R &\doteq 180^\circ - 39^\circ - 118.2^\circ \\
&\doteq 22.8^\circ \end{align}\)
Find side \(QS'\).
\(\begin{align}
\frac{b}{{\sin B}} &= \frac{a}{{\sin A}} \\
\frac{{QS'}}{{\sin 22.8^\circ }} &\doteq \frac{{10}}{{\sin 39^\circ }} \\
QS' &\doteq \frac{{10\sin 22.8^\circ }}{{\sin 39^\circ }} \\
QS' &\doteq 6.157... \\
QS' &\doteq 6.2\thinspace {\rm {m}} \\
\end{align}\)
\(\begin{align}
b\sin A &= s\sin Q \\
&= 14\sin 39^\circ \\
&= 8.810... \\
\end{align}\)
Because side \(RS\) is greater than \(8.810…\), but less than side \(QR\), there are two possible triangles.
Case 1: \(\triangle QRS''\)
Find \(\angle S''\).
\[\begin{align}
\frac{a}{{\sin A}} &= \frac{c}{{\sin C}} \\
\frac{{10}}{{\sin 39^\circ }} &= \frac{{14}}{{\sin S''}} \\
\sin S'' &= \frac{{14\sin 39^\circ }}{{10}} \\
\sin S'' &= 0.881... \\
\angle S'' &= \sin ^{ - 1} \left( {0.881...} \right) \\
\angle S'' &= 61.769... \\
\angle S'' &\doteq 61.8^\circ \\
\end{align}\]
\frac{a}{{\sin A}} &= \frac{c}{{\sin C}} \\
\frac{{10}}{{\sin 39^\circ }} &= \frac{{14}}{{\sin S''}} \\
\sin S'' &= \frac{{14\sin 39^\circ }}{{10}} \\
\sin S'' &= 0.881... \\
\angle S'' &= \sin ^{ - 1} \left( {0.881...} \right) \\
\angle S'' &= 61.769... \\
\angle S'' &\doteq 61.8^\circ \\
\end{align}\]
Find \(\angle R\).
\(\begin{align}
\angle R &\doteq 180^\circ - 39^\circ - 61.8^\circ \\
&\doteq 79.2^\circ \end{align}\)
Find side \(QS''\).
\[\begin{align}
\frac{b}{{\sin B}} &= \frac{a}{{\sin A}} \\
\frac{{QS''}}{{\sin 79.2^\circ }} &\doteq \frac{{10}}{{\sin 39^\circ }} \\
QS'' &\doteq \frac{{10\sin 79.2^\circ }}{{\sin 39^\circ }} \\
QS'' &\doteq 15.608... \\
QS'' &\doteq 15.6{\rm{ m}} \\
\end{align}\]
\frac{b}{{\sin B}} &= \frac{a}{{\sin A}} \\
\frac{{QS''}}{{\sin 79.2^\circ }} &\doteq \frac{{10}}{{\sin 39^\circ }} \\
QS'' &\doteq \frac{{10\sin 79.2^\circ }}{{\sin 39^\circ }} \\
QS'' &\doteq 15.608... \\
QS'' &\doteq 15.6{\rm{ m}} \\
\end{align}\]
Case 2: \(\triangle QRS'\)
Find \(\angle S'\).
The reference angle is the same as \(\angle S''\).
\(\begin{align}
\angle S' &= 180^\circ - \theta _R \\
\angle S' &\doteq 180^\circ - 61.8^\circ \\
\angle S' &\doteq 118.2^\circ \\
\end{align}\)
Find \(\angle R\).
\(\begin{align}
\angle R &\doteq 180^\circ - 39^\circ - 118.2^\circ \\
&\doteq 22.8^\circ \end{align}\)
Find side \(QS'\).
\(\begin{align}
\frac{b}{{\sin B}} &= \frac{a}{{\sin A}} \\
\frac{{QS'}}{{\sin 22.8^\circ }} &\doteq \frac{{10}}{{\sin 39^\circ }} \\
QS' &\doteq \frac{{10\sin 22.8^\circ }}{{\sin 39^\circ }} \\
QS' &\doteq 6.157... \\
QS' &\doteq 6.2\thinspace {\rm {m}} \\
\end{align}\)
| For further information about the ambiguous case see pp. 104 to 107 of Pre-Calculus 11. |