Example 3
Completion requirements
| Example 3 |
Solve \(\triangle ABC\) given that \(AB = 17.6\), \(BC = 14.2\), and \(\angle A = 37^\circ\).
Step 1: Draw a possible diagram of the triangle.
Step 2: Determine how many triangles are possible with the information.
Step 3: Solve the triangles.
Since two sides and an acute angle not contained within those two sides are given, the ambiguous case of the sine law should be considered.
Step 2: Determine how many triangles are possible with the information.
\(\begin{align}
c \sin A &= 17.6 \sin 37^\circ \\
&= 10.591...\end{align}\)
Because side \(a\) is between \(10.591...\) and \(17.6\), there are two possible triangles.
The second possible triangle is:
c \sin A &= 17.6 \sin 37^\circ \\
&= 10.591...\end{align}\)
Because side \(a\) is between \(10.591...\) and \(17.6\), there are two possible triangles.
The second possible triangle is:
Step 3: Solve the triangles.
|
Case 1: \(\triangle ABC'\) |
| Find \(\angle C'\). \(\begin{align} \frac{a}{{\sin A}} &= \frac{c}{{\sin C'}} \\ \frac{{14.2}}{{\sin 37^\circ }} &= \frac{{17.6}}{{\sin C'}} \\ \sin C' &= \frac{{17.6\sin 37^\circ }}{{14.2}} \\ \sin C' &= 0.745... \\ \angle C' &= \sin ^{ - 1} \left( {0.745...} \right) \\ \angle C' &= 48.237... \\ \angle C' &\doteq 48.2^\circ \\ \end{align}\) |
| Find \(\angle B\). \(\begin{align} \angle B &\doteq 180^\circ - 37^\circ - 48.2^\circ \\ &\doteq 94.8^\circ \end{align}\) |
| Find side \(AC'\). \(\begin{align} \frac{b}{{\sin B}} &= \frac{a}{{\sin A}} \\ \frac{{AC'}}{{\sin 94.8^\circ }} &\doteq \frac{{14.2}}{{\sin 37^\circ }} \\ AC' &\doteq \frac{{14.2\sin 94.8^\circ }}{{\sin 37^\circ }} \\ AC' &\doteq 23.512... \\ AC' &\doteq 23.5 \\ \end{align}\) |
|
Case 2: \(\triangle ABC''\) |
| Find \(\angle C''\). This is the same step as in Case 1. Using the information from Case 1, you know that the reference angle for Case 2 is \(48.2^\circ\). In Case 2, \(\angle C''\) will be obtuse, and located in Quadrant II. \(\begin{align} \angle C'' &\doteq 180^\circ - 48.2^\circ \\ &\doteq 131.8^\circ \end{align}\) |
| Find \(\angle B\). \(\begin{align} \angle B &\doteq 180^\circ - 37^\circ - 131.8^\circ \\ &\doteq 11.2^\circ \end{align}\) |
| Find side \(AC''\). \(\begin{align} \frac{b}{{\sin B}} &= \frac{a}{{\sin A}} \\ \frac{{AC''}}{{\sin 11.2^\circ }} &\doteq\frac{{14.2}}{{\sin 37^\circ }} \\ AC'' &\doteq \frac{{14.2\sin 11.2^\circ }}{{\sin 37^\circ }} \\ AC'' &\doteq 4.583... \\ AC'' &\doteq 4.6 \\ \end{align}\) |