In a $△ A B C$ , the tangent of half difference of two angles is one-third the tangent of half the sum of the two angles. The ratio of the sides opposite the angles is

2 : 3

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1 : 3

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2 : 1

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3 : 4

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Solution

The correct option is C $2 : 1$
$tan (\frac{B - C}{2}) = \frac{1}{3} tan (\frac{B + C}{2}) ⟹ sin (\frac{B - C}{2}) cos (\frac{B + C}{2}) = \frac{1}{3} sin (\frac{B + C}{2}) cos (\frac{B - C}{2})$
$⟹ (sin B - sin C) = \frac{1}{3} (sin B + sin C) ⟹ \frac{sin B}{sin C} = 2 ⟹ b : c = 2 : 1$

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In a △ABC, the tangent of half difference of two angles is one-third the tangent of half the sum of the two angles. The ratio of the sides opposite the angles is

The correct option is C 2:1tan(B−C2)=13tan(B+C2)⟹sin(B−C2)cos(B+C2)=13sin(B+C2)cos(B−C2) ⟹(sinB−sinC)=13(sinB+sinC)⟹sinBsinC=2⟹b:c=2:1

In a $△ A B C$ , the tangent of half difference of two angles is one-third the tangent of half the sum of the two angles. The ratio of the sides opposite the angles is

The correct option is C $2 : 1$
$tan (\frac{B - C}{2}) = \frac{1}{3} tan (\frac{B + C}{2}) ⟹ sin (\frac{B - C}{2}) cos (\frac{B + C}{2}) = \frac{1}{3} sin (\frac{B + C}{2}) cos (\frac{B - C}{2})$
$⟹ (sin B - sin C) = \frac{1}{3} (sin B + sin C) ⟹ \frac{sin B}{sin C} = 2 ⟹ b : c = 2 : 1$