If in a triangle $A B C, 2 cos A = sin B c o s e c C$ , then

a = h

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b = c

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c = a

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2 a = b c

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Solution

The correct option is C $c = a$
As we know that

$\frac{sin B}{sin C} = \frac{b}{c}$

$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

Given $2 cos A = sin B cosec C$

$⟹ 2 cos A = \frac{sin B}{sin C}$

$⟹ 2 \times \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{b}{c}$

$⟹ b^{2} + c^{2} - a^{2} = b^{2}$

$⟹ c^{2} = a^{2}$

$⟹ c = a$

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Similar questions

\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}

A B C, \frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{a c} . F i n d ∠ A .

A B C

\frac{2 cos A}{a} + \frac{cos B}{b} + \frac{2 cos C}{c} = \frac{a}{b c} + \frac{b}{c a}

\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}

A B C

\frac{2 cos A}{a} + \frac{cos B}{b} + \frac{2 cos C}{c} = \frac{a}{b c} + \frac{b}{c a}

A

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