The angles of a △ A BC satisfy. c o s A + 2 c o s b c o s A + 2 c o s C = s i n C s i n B th e n △ A BC

Cross-multiplying and de-factorising, we get sin C cos A + sin 2 C = sin B cos A + sin 2 B ⇔ ( sin C − sin B ) cos A = sin 2 B − sin 2 C = 2 cos ( B + C ) sin ( B − C ) = − 2 cos A sin ( B − C ) ⇔ cos A = 0 or I sin C − sin B = − 2 sin ( B − C ) . B = C

Mathematics - Properties of Triangle Question with Solution | TestHub

MathematicsProperties of TriangleGeneral (basic formula based)Medium2 minQB

MathematicsMediummultiple choice

The angles of a $△ A BC$ satisfy. $\frac{c o s A + 2 c o s C}{c o s A + 2 c o s b} = \frac{s i n B}{s i n C} th e n △ A BC$

Options:(select one or more)

Answer:

A, C

Solution:

Cross-multiplying and de-factorising, we get

$sin C cos A + sin 2 C = sin B cos A + sin 2 B \Leftrightarrow$

$(sin C - sin B) cos A = sin 2 B - sin 2 C = 2 cos (B + C) sin (B - C) = - 2$

$cos A sin (B - C)$

$\Leftrightarrow cos A = 0 or \frac{sin C - sin B = - 2 sin ( B - C ) .}{I} B = C$

Stream:JEESubject:MathematicsTopic:Properties of TriangleSubtopic:General (basic formula based)

⏱ 2mℹ️ Source: QB

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