## Trigonometry (11th Edition) Clone

The remaining angles and sides are $$\angle A=37.2^\circ\hspace{.75cm}AB\approx244.015m\hspace{.75cm}BC\approx178.296m$$
$$\angle B=18.7^\circ\hspace{.75cm}\angle C=124.1^\circ\hspace{.75cm}AC=94.6m$$ 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 $$\angle A=37.2^\circ\hspace{.75cm}AB\approx244.015m\hspace{.75cm}BC\approx178.296m$$