What are the other angles in triangle and what is $\tan(\frac{\beta-\gamma}{2})$, if $\alpha = 60$ and $\frac{b}{c}=2+\sqrt3$? and I tried to solve it using law of sines and law of cosines but I couldn't find answer.
What are the other angles in triangle?
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trigonometry
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0Can you show us your attempt, and exactly where you get stuck? – 2017-01-21
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0Also specify that a,b,c are opposite alph, beta gamma. It's implied. But it's not a given. – 2017-01-21
2 Answers
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we have $$\frac{b}{c}=\frac{\sin(\beta)}{\sin(\gamma)}=2+\sqrt{3}$$ since we have $$\gamma=\frac{2}{3}\pi-\beta$$ we get $$\frac{\sin\left(\frac{2}{3}\pi-\beta\right)}{\sin(\beta)}=\frac{1}{2+\sqrt{3}}$$ can you proceed from here?
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Let $\dfrac{\beta-\alpha}2=x$ and we have $\beta+\alpha=120^\circ$
As $0<\beta,\alpha<\pi,-\dfrac\pi2 $\beta=60^\circ+x,\alpha=60^\circ-x$ $$\dfrac{2+\sqrt3}1=\dfrac{\sin(60^\circ+x)}{\sin(60^\circ-x)}$$ Using Componendo and Dividendo, $$\dfrac{\sin(60^\circ+x)+\sin(60^\circ-x)}{\sin(60^\circ+x)-\sin(60^\circ-x)}=\dfrac{2+\sqrt3+1}{2+\sqrt3-1}$$ $$\iff\tan60^\circ\cot x=\sqrt3=\tan60^\circ$$ Can you take it from here?