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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 $$40^{\circ}$$.

Mathematics
RS Agarwal
Standard IX

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