Question

In the figure, angle POQ and angle ROQ form a linear pair. Find the measure of x.​

Answer 1

• The value of x = 216.

Given :

$\bullet \: \: \: \angle$POQ and $\angle$ROQ form a linear pair.

$\\ \bf \bullet \: \: \: \angle{POQ} = \dfrac{x}{2} \\ \\ \bf \bullet \: \: \: \angle{ROQ} = \dfrac{x}{3}$

To Find :

• The value of x.

Solution :

Given,

$\angle$POQ and $\angle$ROQ form a linear pair.

We know that,

Linear pair = 180°

In linear pair,

Sum of two angles = 180°.

We have,

$\bf \bullet \: \: \: \angle{POQ} = \dfrac{x}{2} \\ \\ \bf \bullet \: \: \: \angle{ROQ} = \dfrac{x}{3}$

So,

$\bf \implies \angle{POQ} + \angle{ROQ} = {180}^{\circ} \\ \\ \\ \bf \implies \dfrac{x}{2} + \dfrac{x}{3} = {180}^{\circ} \\ \\ \\ \bf \implies \dfrac{(x \times 3 )+(x \times 2) }{6} = {180}^{ \circ} \\ \\ \\ \bf \implies \dfrac{3x + 2x}{6} = {180}^{ \circ} \\ \\ \\ \bf \implies 3x + 2x = {180}^{ \circ} \times 6 \\ \\ \\ \bf \implies 5x = {1080}^{ \circ} \\ \\ \\ \bf \implies x = \dfrac{ {1080}^{ \circ} }{5} \\ \\ \\ \bf \implies x = {216}^{ \circ}$

Hence,

The value of x is 216°.

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Verification :

Method I :-

By linear pair,

Sum of two angles is 180°.

$\bf \implies \angle{POQ} + \angle{ROQ} = {180}^{\circ} \\ \\ \\ \bf \implies \dfrac{x}{2} + \dfrac{x}{3} = {180}^{\circ}$

Now, substitute the value of x.

$\bf \implies \bigg(\dfrac{ \not216}{ \not2}\bigg)^{ \circ} + \bigg(\dfrac{ \not216}{ \not3} \bigg)^{ \circ} = {180}^{ \circ} \\ \\ \\ \bf \implies {108}^{ \circ} + {72}^{ \circ} = {180}^{ \circ} \\ \\ \\ \bf \implies {180}^{ \circ} = {180}^{ \circ}$

Hence Verified !

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Method II :-

By linear pair,

Sum of two angles is 180°.

$\bf \implies \angle{POQ} + \angle{ROQ} = {180}^{\circ} \\ \\ \\ \bf \implies \dfrac{x}{2} + \dfrac{x}{3} = {180}^{\circ}$

Now substitute the value of x and after substitute the value of x, we can take L.C.M. (least common multiple) of the numerator and solve it.

$\bf \implies \bigg(\dfrac{216}{2}\bigg)^{ \circ} + \bigg(\dfrac{216}{3} \bigg)^{ \circ} = {180}^{ \circ} \\ \\ \\ \bf \implies {\bigg( \dfrac{(216 \times 3)(216 \times 2)}{6}\bigg)}^{ \circ } = {180}^{ \circ} \\ \\ \\ \bf \implies { \bigg(\dfrac{648 + 432}{6}\bigg)}^{ \circ} = {180}^{ \circ} \\ \\ \\ \bf \implies{\bigg(\dfrac{1080}{6}\bigg)}^{ \circ} = {180}^{ \circ} \\ \\ \\ \bf \implies {180}^{ \circ} = {180}^{ \circ}$

Hence Verified !