Answer
The remaining angles and sides are
$$\angle A=36.54^\circ\hspace{.75cm}b\approx44.154m\hspace{.75cm}a\approx28.068m$$
Work Step by Step
$$\angle C=74.08^\circ\hspace{.75cm}\angle B=69.38^\circ\hspace{.75cm}c=45.38m$$
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
$$\angle A=36.54^\circ\hspace{.75cm}b\approx44.154m\hspace{.75cm}a\approx28.068m$$