## Trigonometry (11th Edition) Clone

The remaining angles and sides are $$\angle B=37.3^\circ\hspace{.75cm}AC\approx51.0ft\hspace{.75cm}BC\approx38.5ft$$
$$\angle A=27.2^\circ\hspace{.75cm}\angle C=115.5^\circ\hspace{.75cm}AB=76.0ft$$ 1) Analysis: - Angles $\angle A$ and $\angle C$ are known. We can always calculate $\angle B$ as the sum of 3 angles in a triangle equals $180^\circ$. - Side $AB$ and its opposite $\angle C$ are known. $\angle A$ is also known, which can help figure out side $BC$. The same for side $AC$ as $\angle B$ is known. (Law of sines is to be applied) 2) Calculate the unknown angle $\angle B$ We know that the sum of 3 angles in a triangle is $180^\circ$. $$\angle A+\angle B+\angle C=180^\circ$$ $$27.2^\circ+\angle B+115.5^\circ=180^\circ$$ $$\angle B+142.7^\circ=180^\circ$$ $$\angle B=180^\circ-142.7^\circ=37.3^\circ$$ 3) Calculate the unknown sides $AC$ and $BC$ a) For $AC$ We know the opposite angle of $AC$: $\angle B=37.3^\circ$, so $\sin B=\sin37.3^\circ\approx0.606$. We also know side $AB=76.0ft$ and its opposite angle $\angle C=115.5^\circ$, $\sin C\approx0.903$ Therefore, using the law of sines: $$\frac{AC}{\sin B}=\frac{AB}{\sin C}$$ $$AC=\frac{AB\sin B}{\sin C}$$ $$AC=\frac{76.0ft\times0.606}{0.903}$$ $$AC\approx51.0ft$$ b) For $BC$ We know the opposite angle of $BC$: $\angle A=27.2^\circ$, so $\sin A=\sin27.2^\circ\approx0.457$. We also know side $AB=76.0ft$ and its opposite angle $\angle C=115.5^\circ$, $\sin C\approx0.903$ Therefore, using the law of sines: $$\frac{BC}{\sin A}=\frac{AB}{\sin C}$$ $$BC=\frac{AB\sin A}{\sin C}$$ $$BC=\frac{76.0ft\times0.457}{0.903}$$ $$BC\approx38.5ft$$ 4) Conclusion: The remaining angles and sides are $$\angle B=37.3^\circ\hspace{.75cm}AC\approx51.0ft\hspace{.75cm}BC\approx38.5ft$$