In a triangle ABC, with usual notations a = s | vedclass

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(A) Using the tangent rule: $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$.Substitute the values: $a-b = (\sqrt{3}+1)

Vedclass · Accepted Answer

$105^\circ, 15^\circ$

Vedclass · Answer

(A) Using the tangent rule: $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$.Substitute the values: $a-b = (\sqrt{3}+1) - (\sqrt{3}-1) = 2$ and $a+b = (\sqrt{3}+1) + (\sqrt{3}-1) = 2\sqrt{3}$.$	an\left(\frac{A-B}{2}ight) = \frac{2}{2\sqrt{3}} \cot(30^\circ) = \frac{1}{\sqrt{3}} \cdot \sqrt{3} = 1$.So, $\frac{A-B}{2} = 45^\circ \implies A-B = 90^\circ$.Since $A+B+C = 180^\circ$ and $C = 60^\circ$, $A+B = 120^\circ$.Solving the system: $A-B = 90^\circ$ and $A+B = 120^\circ$ gives $2A = 210^\circ \implies A = 105^\circ$ and $B = 15^\circ$.

In a triangle $ABC$, with usual notations $a = \sqrt{3} + 1$, $b = \sqrt{3} - 1$ and $\angle C = 60^\circ$, then the values of $\angle A$ and $\angle B$ respectively are:

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