Answer
The remaining angles and sides are
$$\angle C=57.36^\circ\hspace{.75cm}b\approx11.119ft\hspace{.75cm}c\approx11.544ft$$
Work Step by Step
$$\angle A=68.41^\circ\hspace{.75cm}\angle B=54.23^\circ\hspace{.75cm}a=12.75ft$$
1) Analysis:
- Angles $\angle A$ and $\angle B$ are known. We can always calculate $\angle C$ as the sum of 3 angles in a triangle equals $180^\circ$.
- Side $a$ and its opposite $\angle A$ are known. $\angle B$ and $\angle C$ are also known, so they are helpful to finding out sides $b$ and $c$. (Law of sines is to be applied)
2) Calculate the unknown angle $\angle C$
We know that the sum of 3 angles in a triangle is $180^\circ$.
$$\angle A+\angle B+\angle C=180^\circ$$
$$68.41^\circ+54.23^\circ+\angle C=180^\circ$$
$$122.64^\circ+\angle C=180^\circ$$
$$\angle C=180^\circ-122.64^\circ=57.36^\circ$$
3) Calculate the unknown sides $b$ and $c$
a) For $b$
We know the opposite angle of $b$: $\angle B=54.23^\circ$, so $\sin B=\sin54.23^\circ\approx0.811$.
We also know side $a=12.75ft$ and its opposite angle $\angle A=68.41^\circ$, $\sin A\approx0.93$.
Therefore, using the law of sines:
$$\frac{b}{\sin B}=\frac{a}{\sin A}$$
$$b=\frac{a\sin B}{\sin A}$$
$$b=\frac{12.75ft\times0.811}{0.93}$$
$$b\approx11.119ft$$
b) For $c$
We know the opposite angle of $c$: $\angle C=57.36^\circ$, so $\sin C=\sin57.36^\circ\approx0.842$.
We also know side $a=12.75ft$ and its opposite angle $\angle A=68.41^\circ$, $\sin A\approx0.93$.
Therefore, using the law of sines:
$$\frac{c}{\sin C}=\frac{a}{\sin A}$$
$$c=\frac{a\sin C}{\sin A}$$
$$c=\frac{12.75ft\times0.842}{0.93}$$
$$c\approx11.544ft$$
4) Conclusion:
The remaining angles and sides are
$$\angle C=57.36^\circ\hspace{.75cm}b\approx11.119ft\hspace{.75cm}c\approx11.544ft$$
