## Trigonometry (11th Edition) Clone

The remaining angles and sides are $$\angle A=36.54^\circ\hspace{.75cm}b\approx44.154m\hspace{.75cm}a\approx28.068m$$
$$\angle C=74.08^\circ\hspace{.75cm}\angle B=69.38^\circ\hspace{.75cm}c=45.38m$$ 1) Analysis: - Angles $\angle B$ and $\angle C$ are known. We can always calculate $\angle A$ as the sum of 3 angles in a triangle equals $180^\circ$. - Side $c$ and its opposite $\angle C$ are known. $\angle A$ and $\angle B$ are also known, so they are helpful to finding out sides $a$ and $b$. (Law of sines is to be applied) 2) Calculate the unknown angle $\angle A$ We know that the sum of 3 angles in a triangle is $180^\circ$. $$\angle A+\angle B+\angle C=180^\circ$$ $$\angle A+69.38^\circ+74.08^\circ=180^\circ$$ $$\angle A+143.46^\circ=180^\circ$$ $$\angle A=180^\circ-143.46^\circ=36.54^\circ$$ 3) Calculate the unknown sides $a$ and $b$ a) For $b$ We know the opposite angle of $b$: $\angle B=69.38^\circ$, so $\sin B=\sin69.38^\circ\approx0.936$. We also know side $c=45.38m$ and its opposite angle $\angle C=74.08^\circ$, $\sin C\approx0.962$. Therefore, using the law of sines: $$\frac{b}{\sin B}=\frac{c}{\sin C}$$ $$b=\frac{c\sin B}{\sin C}$$ $$b=\frac{45.38m\times0.936}{0.962}$$ $$b\approx44.154m$$ b) For $a$ We know the opposite angle of $a$: $\angle A=36.54^\circ$, so $\sin A=\sin36.54^\circ\approx0.595$. We also know side $c=45.38m$ and its opposite angle $\angle C=74.08^\circ$, $\sin C\approx0.962$. Therefore, using the law of sines: $$\frac{a}{\sin A}=\frac{c}{\sin C}$$ $$a=\frac{c\sin A}{\sin C}$$ $$a=\frac{45.38m\times0.595}{0.962}$$ $$a\approx28.068m$$ 4) Conclusion: The remaining angles and sides are $$\angle A=36.54^\circ\hspace{.75cm}b\approx44.154m\hspace{.75cm}a\approx28.068m$$