#### Answer

The remaining angles and sides are
$$\angle A=56^\circ\hspace{.75cm}BC\approx307.382ft\hspace{.75cm}AB\approx361.146ft$$

#### Work Step by Step

$$\angle B=20^\circ50'\approx20.833^\circ\hspace{.75cm}\angle C=103^\circ10'\approx103.167^\circ\hspace{.75cm}AC=132ft$$
Here $AC$ is the side $b$, so $b=132ft$
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 $b$ 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 $\angle A$
We know that the sum of 3 angles in a triangle is $180^\circ$.
$$\angle A+\angle B+\angle C=180^\circ$$
$$\angle A+20^\circ50'+103^\circ10'=180^\circ$$
$$\angle A+124^\circ=180^\circ$$
$$\angle A=180^\circ-124^\circ=56^\circ$$
3) Calculate the unknown sides $a$ and $c$
a) For $a$
We know the opposite angle of $a$: $\angle A=56^\circ$, so $\sin A=\sin56^\circ\approx0.829$.
We also know side $b=132ft$ and its opposite angle $\angle B=20.833^\circ$, $\sin B\approx0.356$.
Therefore, using the law of sines:
$$\frac{a}{\sin A}=\frac{b}{\sin B}$$
$$a=\frac{b\sin A}{\sin B}$$
$$a=\frac{132ft\times0.829}{0.356}$$
$$a\approx307.382ft$$
So, $BC\approx307.382ft$
b) For $c$
We know the opposite angle of $c$: $\angle C=103.167^\circ$, so $\sin C=\sin103.167^\circ\approx0.974$.
We also know side $b=132ft$ and its opposite angle $\angle B=20.833^\circ$, $\sin B\approx0.356$.
Therefore, using the law of sines:
$$\frac{c}{\sin C}=\frac{b}{\sin B}$$
$$c=\frac{b\sin C}{\sin B}$$
$$c=\frac{132ft\times0.974}{0.356}$$
$$c\approx361.146ft$$
So, $AB\approx361.146ft$
4) Conclusion:
The remaining angles and sides are
$$\angle A=56^\circ\hspace{.75cm}BC\approx307.382ft\hspace{.75cm}AB\approx361.146ft$$