## Trigonometry (11th Edition) Clone

The remaining angles and sides needed to find are $$\angle B=18.5^\circ\hspace{.75cm}a\approx239.193yd\hspace{.75cm}c\approx230.573yd$$
$$\angle A=87.2^\circ\hspace{.75cm}\angle C=74.3^\circ\hspace{.75cm}b=75.9yd$$ 1) Analysis: - Angles $\angle A$ and $\angle C$ are known. We can always calculate $\angle B$ as the sum of 3 angles in a triangle equals $180^\circ$. - Side $b$ and its opposite $\angle B$ are known. $\angle A$ and $\angle C$ are also known, so they are helpful to finding out sides $a$ and $c$. (Law of sines is to be applied) 2) Calculate the unknown angle $\angle B$ We know that the sum of 3 angles in a triangle is $180^\circ$. $$\angle A+\angle B+\angle C=180^\circ$$ $$87.2^\circ+\angle B+74.3^\circ=180^\circ$$ $$\angle B+161.5^\circ=180^\circ$$ $$\angle B=180^\circ-161.5^\circ=18.5^\circ$$ 3) Calculate the unknown sides $a$ and $c$ a) For $a$ We know the opposite angle of $a$: $\angle A=87.2^\circ$, so $\sin A=\sin87.2^\circ\approx0.999$. We also know side $b=75.9yd$ and its opposite angle $\angle B=18.5^\circ$, $\sin B\approx0.317$. Therefore, using the law of sines: $$\frac{a}{\sin A}=\frac{b}{\sin B}$$ $$a=\frac{b\sin A}{\sin B}$$ $$a=\frac{75.9yd\times0.999}{0.317}$$ $$a\approx239.193yd$$ b) For $c$ We know the opposite angle of $c$: $\angle C=74.3^\circ$, so $\sin C=\sin74.3^\circ\approx0.963$. We also know side $b=75.9yd$ and its opposite angle $\angle B=18.5^\circ$, $\sin B\approx0.317$. Therefore, using the law of sines: $$\frac{c}{\sin C}=\frac{b}{\sin B}$$ $$c=\frac{b\sin C}{\sin B}$$ $$c=\frac{75.9yd\times0.963}{0.317}$$ $$c\approx230.573yd$$ 4) Conclusion: The remaining angles and sides are $$\angle B=18.5^\circ\hspace{.75cm}a\approx239.193yd\hspace{.75cm}c\approx230.573yd$$