In a triangle ABC, angle A is greater than B, if the measures of angle A and B satisfy the equation
$3 \sin x - 4 \sin^{3} x - k = 0, 0 < K < 1$ Then the measure of angle C is

Question

In a triangle ABC, angle A is greater than B, if the measures of angle A and B satisfy the $$equation3sinx$$−$$4sin3x$$−$$k=0$$, $$0$$

Infinity Learn · Accepted Answer

Given A>B and, A and B satisfy the equation           $$3sinx$$−$$4sin3x$$−$$k=0$$⇒$$sin3x-k=0$$           (or)  $$sin3A$$−$$k=0$$  and  $$sin3B$$−$$K=0$$                         ∴$$sin3A=sin3B=$$$$sin$$π$$-3B$$ ⇒                          $$3A=$$π−$$3B$$    ∵A>B⇒                           $$A+B=$$π$$3$$ Then $$C=$$π−π$$3=2$$$$π$$$$3$$

In a triangle ABC, angle A is greater than B, if the measures of angle A and B satisfy the equation
$3 \sin x - 4 \sin^{3} x - k = 0, 0 < K < 1$ Then the measure of angle C is

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In a triangle ABC, angle A is greater than B, if the measures of angle A and B satisfy the equation3sinx−4sin3x−k=0, 0<K<1Then the measure of angle C is

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free study material

In a triangle ABC, angle A is greater than B, if the measures of angle A and B satisfy the equation
$3 \sin x - 4 \sin^{3} x - k = 0, 0 < K < 1$ Then the measure of angle C is