Question

If in a triangle a = $\sqrt{3}+1,b=\sqrt{3}-1,C={60}^{\circ }$ then the value of A is ?

• A. ${75}^{\circ }$
• B. ${90}^{\circ }$
• C. ${105}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is C ${105}^{\circ }$

$A+B+C={180}^{\circ }$
$A+B={120}^{\circ }$ .... (1)
From law of tangent, we know that tan $\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}cot\frac{C}{2}$
$=\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)}cot{30}^{\circ }$
$=\frac{2}{2\sqrt{3}}cot{30}^{\circ }$
$tan\left(\frac{A-B}{2}\right)=1$
$A-B={90}^{\circ }$ ..... (2)
By equation 1 & 2, we get
$A={105}^{\circ }$