Question

Solve triangle abc which have angle=25.25, angleb=60.51 and a=3.82 find c

Answer 1

Answer on Question #58987 – Math – Trigonometry

Question

Solve triangle ABC which has angle=25.25, angle B=60.51 and a=3.82. Find c

Solution

It is not known which of angles A or C has the measure of $25.25{}^{\circ}$.

If $\angle A = 25.25{}^{\circ}$, then $\angle C = 180{}^{\circ} - \angle A - \angle B = 180{}^{\circ} - 25.25{}^{\circ} - 60.51{}^{\circ} = 94.24{}^{\circ}$ and using the theorem of sines

$$\frac{a}{\sin \angle A} = \frac{c}{\sin \angle C}, \text{ hence } c = a \frac{\sin \angle C}{\sin \angle A} = 3.82 \frac{\sin(94.24{}^{\circ})}{\sin(25.25{}^{\circ})} = 3.82 \frac{0.9972631}{0.426569} \approx 8.93.$$

If $\angle C = 25.25{}^{\circ}$, then $\angle A = 180{}^{\circ} - \angle B - \angle C = 180{}^{\circ} - 60.51{}^{\circ} - 25.25{}^{\circ} = 94.24{}^{\circ}$ using the theorem of sines

$$\frac{a}{\sin \angle A} = \frac{c}{\sin \angle C}, \text{ hence } c = a \frac{\sin \angle C}{\sin \angle A} = 3.82 \frac{\sin(25.25{}^{\circ})}{\sin(94.24{}^{\circ})} = 3.82 \frac{0.426569}{0.9972631} \approx 1.63.$$

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