Question

# In a $$\triangle ABC$$, if $$3\angle A = 4 \angle B = 6\angle C$$, calculate the angles.

Solution

## Consider $$3 \angle A = 4 \angle B = 6 \angle C = x$$So we can write it as$$3 \angle A = x$$$$\angle A = x/3$$$$4 \angle B = x$$$$\angle B = x/4$$$$6 \angle C = x$$$$\angle C = x/6$$We know that the sum of all the angles in a triangle is $$180^{\circ}$$.So we can write it as$$\angle A + \angle B + \angle C = 180^{\circ}$$By substituting the values in the above equation we get$$(x/3) + (x/4) + (x/6) = 180^{\circ}$$$$LCM$$ of $$3, 4$$ and $$6$$ is $$12$$So we get$$(4x + 3x + 2x)/12 = 180^{\circ}$$By addition$$9x/12 = 180^{\circ}$$By cross multiplication$$9x = 180 \times 129$$$$x = 2160$$By division$$x = 2160/9$$$$x = 240$$By substituting the values of $$x$$$$\angle A = x/3 = 240/3 = 80^{\circ}$$$$\angle B = x/4 = 240/4 = 60^{\circ}$$$$\angle C = x/6 = 240/6 = 40^{\circ}$$Therefore, the value of $$\angle A, \angle B$$ and $$\angle C$$ is $$80^{\circ}, 60^{\circ}$$ and $$40^{\circ}$$.MathematicsRS AgarwalStandard IX

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