Answer
The remaining angles and sides are
$$\angle A=37.2^\circ\hspace{.75cm}AB\approx244.015m\hspace{.75cm}BC\approx178.296m$$
Work Step by Step
$$\angle B=18.7^\circ\hspace{.75cm}\angle C=124.1^\circ\hspace{.75cm}AC=94.6m$$
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 $AC$ and its opposite $\angle B$ are known. $\angle A$ is also known, which can help figure out side $BC$. The same for side $AB$ as $\angle C$ is known. (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 $180^\circ$.
$$\angle A+\angle B+\angle C=180^\circ$$
$$\angle A+18.7^\circ+124.1^\circ=180^\circ$$
$$\angle A+142.8^\circ=180^\circ$$
$$\angle A=180^\circ-142.8^\circ=37.2^\circ$$
3) Calculate the unknown sides $AB$ and $BC$
a) For $AB$
We know the opposite angle of $AB$: $\angle C=124.1^\circ$, so $\sin C=\sin124.1^\circ\approx0.828$.
We also know side $AC=94.6m$ and its opposite angle $\angle B=18.7^\circ$, $\sin B\approx0.321$.
Therefore, using the law of sines:
$$\frac{AB}{\sin C}=\frac{AC}{\sin B}$$
$$AB=\frac{AC\sin C}{\sin B}$$
$$AB=\frac{94.6m\times0.828}{0.321}$$
$$AB\approx244.015m$$
b) For $BC$
We know the opposite angle of $BC$: $\angle A=37.2^\circ$, so $\sin A=\sin37.2^\circ\approx0.605$.
We also know side $AC=94.6m$ and its opposite angle $\angle B=18.7^\circ$, $\sin B\approx0.321$.
Therefore, using the law of sines:
$$\frac{BC}{\sin A}=\frac{AC}{\sin B}$$
$$BC=\frac{AC\sin A}{\sin B}$$
$$BC=\frac{94.6m\times0.605}{0.321}$$
$$BC\approx178.296m$$
4) Conclusion:
The remaining angles and sides are
$$\angle A=37.2^\circ\hspace{.75cm}AB\approx244.015m\hspace{.75cm}BC\approx178.296m$$
