Question

Given angle $A={60}^{\circ },c=\sqrt{3}-1,b=\sqrt{3}+1$. Solve the triangle

• A. $a=\sqrt{6},B={15}^{\circ },C={100}^{\circ }$
• B. $a=\sqrt{6},B={15}^{\circ },C={105}^{\circ }$
• C. $a=\sqrt{12},B={15}^{\circ },C={105}^{\circ }$
• D. $a=\sqrt{6},B={105}^{\circ },C={15}^{\circ }$

Answer 1

The correct option is D

$a=\sqrt{6},B={105}^{\circ },C={15}^{\circ }$

$A={60}^{\circ }\Rightarrow B+C={120}^{\circ }Weknow,tan\frac{B-C}{2}=\frac{b-c}{b+c}cot\frac{A}{2}\Rightarrow tan\frac{B-C}{2}=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)}cot\left(\frac{60}{2}\right)=\frac{2}{2\sqrt{3}}\left(\sqrt{3}\right)=1\therefore \frac{B-C}{2}={45}^{\circ }\Rightarrow B-C={90}^{\circ }B+C={120}^{\circ }\therefore B={105}^{\circ }andC={15}^{\circ },Fromsinerule,a=\frac{bsinA}{sinB}=\sqrt{3}+1\left(\frac{sin60}{sin105}\right)sin105=sin({60}^{\circ }+{45}^{\circ })=\frac{\sqrt{3}}{2}\left(\frac{1}{\sqrt{2}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}a=\frac{(\sqrt{3}+1)\left(\sqrt{3}\right)\left(2\right)\left(\sqrt{2}\right)}{(\sqrt{3}+1)2}=\sqrt{6}\therefore a=\sqrt{6},B={105}^{\circ },C={15}^{\circ }$
Option d is correct.