In a triangle {ABC},\;a/b=2+{3} and C=60^{}. Then the measure of A is

Correct Answer is : 105∘105^{}105∘

Correct Answer is : 105∘105^{}105∘

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AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 26 of 160

Marks: +1, -0

In a triangle

\text{ABC},\;\frac{a}{b}=2+\sqrt{3}

\angle C=60^{\circ}

\angle A

Solution: 👈: Video Solution

Given that

A B C

is a triangle

So,

A+B+C=180^{\circ}

Given,

C=60^{\circ}

\Rightarrow \;\; A+B=120^{\circ} \text{.......} (i)

We have by sine formula

\;\; \frac{\sin A}{a} = \frac{\sin B}{b}

\Rightarrow \;\; \frac{\sin A}{\sin B} = \frac{a}{b} = 2+\sqrt{3} \; \text{(given)} \;

Applying componendo and dividendo, we get

\;\; \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2+\sqrt{3}+1}{2+\sqrt{3}-1} = \frac{3+\sqrt{3}}{\sqrt{3}+1}

\Rightarrow \; \frac{2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}

= \;\; \frac{(3+\sqrt{3})(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}

\Rightarrow \; \frac{2 \left(\frac{\sqrt{3}}{2}\right)}{2 \left(\frac{1}{2}\right)} \cot \left(\frac{A-B}{2}\right) = \sqrt{3}

\Rightarrow \sqrt{3} \cot \left(\frac{A-B}{2}\right) = \sqrt{3} \Rightarrow \left(\frac{A-B}{2}\right) = 45^{\circ}

\Rightarrow \;\; A-B=90 \text{........} (ii)

Solving Eqs. (i) and (ii), we get

\angle A=105^{\circ}

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