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Question Number 165270 by cortano1 last updated on 28/Jan/22

Answered by TheSupreme last updated on 28/Jan/22

$$B+C=120$$$$\frac{sin\left(C\right)}{sin\left(B\right)}=2+\sqrt{3}$$$$sin\left(C\right)=(2+\sqrt{3})sin(120-C)$$$$sin\left(C\right)=(2+\sqrt{3})(\frac{cos\left(C\right)}{2}-\frac{\sqrt{3}}{2}sin\left(C\right))$$$$sin\left(C\right)(1+(2+\sqrt{3})\frac{\sqrt{3}}{2})=(2+\sqrt{3})\frac{cos\left(C\right)}{2}$$$$sin\left(C\right)(\frac{5}{2}+\sqrt{3})=(1+\frac{\sqrt{3}}{2})cos\left(C\right)$$$$tan\left(C\right)=\frac{1+\frac{\sqrt{3}}{2}}{\frac{5}{2}+\sqrt{3}}=\frac{2+\sqrt{3}}{5+2\sqrt{3}}=\frac{(2+\sqrt{3})(5-2\sqrt{3})}{25-12}=\frac{4+\sqrt{3}}{13}$$$$C=arctan\left(\frac{4+\sqrt{3}}{13}\right)$$$$B=120-C$$$$$$

Answered by mr W last updated on 28/Jan/22

$$a^2=b^2+c^2-2bc\cos 60°$$$${\left(\frac{a}{c}\right)}^2={(2+\sqrt{3})}^2+1-2(2+\sqrt{3})\times \frac{1}{2}$$$$\frac{a}{c}=\sqrt{3(2+\sqrt{3})}$$$$\frac{a}{c}=\frac{\sin A}{\sin C}=\frac{\sqrt{3}}{2\sin C}=\sqrt{3(2+\sqrt{3})}$$$$\sin C=\frac{\sqrt{3}}{2\sqrt{3(2+\sqrt{3})}}=\frac{\sqrt{2-\sqrt{3}}}{2}$$$$\Rightarrow C={\sin }^{-1}\frac{\sqrt{2-\sqrt{3}}}{2}=15°or165°$$$$\frac{a}{b}=\frac{a}{c}\times \frac{c}{b}=\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2\sin B}=\frac{1}{2+\sqrt{3}}\times \sqrt{3(2+\sqrt{3})}$$$$\sin B=\frac{\sqrt{2+\sqrt{3}}}{2}$$$$\Rightarrow B={\sin }^{-1}\frac{\sqrt{2+\sqrt{3}}}{2}=75°or105°$$$$sinceB+C=180°-60°=120°,$$$$\Rightarrow B=105°andC=15°isthesolution.$$

Commented by mr W last updated on 29/Jan/22

$$anotherway:$$$$B+C=180-60=120°$$$$\frac{\sin B}{\sin C}=\frac{b}{c}=2+\sqrt{3}$$$$\frac{\sin (120-C)}{\sin C}=2+\sqrt{3}$$$$\frac{\sqrt{3}\cos C+\sin C}{2\sin C}=2+\sqrt{3}$$$$\frac{\sqrt{3}}{\tan C}=3+2\sqrt{3}$$$$\tan C=\frac{\sqrt{3}}{3+2\sqrt{3}}=2-\sqrt{3}$$$$\Rightarrow C=15°$$$$\Rightarrow B=120-15=105°$$

Commented by cortano1 last updated on 29/Jan/22

$$nice$$

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