Answer
The remaining angles and sides are
$$\angle A=99^\circ\hspace{.75cm}AC\approx34.3cm\hspace{.75cm}AB\approx21.108cm$$
Work Step by Step
$$\angle B=52^\circ\hspace{.75cm}\angle C=29^\circ\hspace{.75cm}BC=43cm$$
1) Analysis:
- 2 angles are given. We can always find out the third one using the law of the sum of 3 angles in a triangle.
- Side $BC$ and its opposite $\angle A$ are known. $\angle B$ is also known, which can help figure out side $AC$. 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 $180^\circ$.
$$\angle A+\angle B+\angle C=180^\circ$$
$$\angle A+52^\circ+29^\circ=180^\circ$$
$$\angle A+81^\circ=180^\circ$$
$$\angle A=180^\circ-81^\circ=99^\circ$$
3) Calculate the unknown sides $AC$ and $AB$
a) For $AC$
We know the opposite angle of $AC$: $\angle B=52^\circ$, so $\sin B=\sin52^\circ\approx0.788$.
We also know side $BC=43cm$ and its opposite angle $\angle A=99^\circ$, $\sin A\approx0.988$
Therefore, using the law of sines:
$$\frac{AC}{\sin B}=\frac{BC}{\sin A}$$
$$AC=\frac{BC\sin B}{\sin A}$$
$$AC=\frac{43cm\times0.788}{0.988}$$
$$AC\approx34.3cm$$
b) For $AB$
We know the opposite angle of $AB$: $\angle C=29^\circ$, so $\sin C=\sin29^\circ\approx0.485$.
We also know side $BC=43cm$ and its opposite angle $\angle A=99^\circ$, $\sin A\approx0.988$
Therefore, using the law of sines:
$$\frac{AB}{\sin C}=\frac{BC}{\sin A}$$
$$AB=\frac{BC\sin C}{\sin A}$$
$$AB=\frac{43cm\times0.485}{0.988}$$
$$AB\approx21.108cm$$
4) Conclusion:
The remaining angles and sides are
$$\angle A=99^\circ\hspace{.75cm}AC\approx34.3cm\hspace{.75cm}AB\approx21.108cm$$
