## Trigonometry (11th Edition) Clone

The answer to this question is $$\angle A=49^\circ40'\hspace{.75cm}b\approx16.158cm\hspace{.75cm}c\approx25.853cm$$
$$\angle B=38^\circ40'\hspace{.75cm}\angle C=91^\circ40'\hspace{.75cm}a=19.7cm$$ 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 $a$ 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 $\angle A$ We know that the sum of 3 angles in a triangle is $180^\circ$. $$\angle A+\angle B+\angle C=180^\circ$$ $$\angle A+38^\circ40'+91^\circ40'=180^\circ$$ $$\angle A+130^\circ20'=180^\circ$$ $$\angle A=180^\circ-130^\circ20'=49^\circ40'$$ Now it is better to change the angles to the decimal form as they are easier to calculate the sines. $$\angle A=49^\circ40'\approx49.667^\circ$$ $$\angle B=38^\circ40'\approx38.667^\circ$$ $$\angle C=91^\circ40'\approx91.667^\circ$$ 3) Calculate the unknown sides $b$ and $c$ a) For $b$ We know the opposite angle of $b$: $\angle B=38.667^\circ$, so $\sin B=\sin38.667^\circ\approx0.625$. We also know side $a=19.7cm$ and its opposite angle $\angle A=49.667^\circ$, $\sin A\approx0.762$. Therefore, using the law of sines: $$\frac{b}{\sin B}=\frac{a}{\sin A}$$ $$b=\frac{a\sin B}{\sin A}$$ $$b=\frac{19.7cm\times0.625}{0.762}$$ $$b\approx16.158cm$$ b) For $c$ We know the opposite angle of $c$: $\angle C=91.667^\circ$, so $\sin C=\sin91.667^\circ\approx1$. We also know side $a=19.7cm$ and its opposite angle $\angle A=49.667^\circ$, $\sin A\approx0.762$. Therefore, using the law of sines: $$\frac{c}{\sin C}=\frac{a}{\sin A}$$ $$c=\frac{a\sin C}{\sin A}$$ $$c=\frac{19.7cm\times1}{0.762}$$ $$c\approx25.853cm$$ 4) Conclusion: The remaining angles and sides are $$\angle A=49^\circ40'\hspace{.75cm}b\approx16.158cm\hspace{.75cm}c\approx25.853cm$$