In a ABC, and C = 60°. Then the ordered pair is equal to

JEE MainMathematicsModerate

In a $\Delta$ ABC, $\frac{a}{b} = 2 + \sqrt{3}$ and $\angle$ C =
60°. Then the ordered pair $(\angle A,\angle B)$ is equal to

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\frac{a}{b} = \frac{2 + \sqrt{3}}{1}

\therefore\frac{a + b}{a - b} = \frac{3 + \sqrt{3}}{\sqrt{3} + 1} = \sqrt{3}

tan\frac{A - B}{2} = \frac{a - b}{a + b}cot\frac{C}{2} = \frac{1}{\sqrt{3}}cot30^{\circ} = 1

\frac{A - B}{2}

= 45

\Rightarrow

A + B = 120 A = 105°, B = 15°

Source: Grass-G / M-24 Mathematics Paper 31 Dec.docx

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