Question

HELP ME ASAP PLEASE ITS THE SAME CONCEPT

Answer 1

Answer :

• m∠Q = 133°

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Step-by-step explanation :

We know that,

• Sum of all angles of a triangle = 180°.

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Therefore,

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${ \longrightarrow \sf \qquad \: m \angle \: P +m \angle Q + m\angle R = 180 {}^{ \circ} }$

${ \longrightarrow \sf \qquad \: {(x + 13)}^{ \circ} +{(10x + 13)}^{ \circ} + {(2x - 2)}^{ \circ} = 180 {}^{ \circ} }$

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Adding like terms we get :

${ \longrightarrow \sf \qquad \: {(x + 10x + 2x)} +(13^{ \circ} + 13^{ \circ} - 2^{ \circ} ) = 180 {}^{ \circ} }$

${ \longrightarrow \sf \qquad \: 13x +24^{ \circ} = 180 {}^{ \circ} }$

${ \longrightarrow \sf \qquad \: 13x = 180 {}^{ \circ} - 24^{ \circ} }$

${ \longrightarrow \sf \qquad \: 13x = 156 {}^{ \circ} }$

${ \longrightarrow \sf \qquad \: x = \dfrac{156 {}^{ \circ} }{13} }$

${ \longrightarrow \qquad \: { \pmb{ x = 12 {}^{ \circ} } }}$

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Therefore,

• The value of x = 12°

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Now, Substituting the value of x in m∠Q :

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${ \longrightarrow \qquad \: { \sf{ m \angle Q =( 10x + 13) {}^{ \circ} } }}$

${ \longrightarrow \qquad \: { \sf{ m \angle Q =10(12) {}^{ \circ} + 13{}^{ \circ} } }}$

${ \longrightarrow \qquad \: { \sf{ m \angle Q =120 {}^{ \circ} + 13{}^{ \circ} } }}$

${ \longrightarrow \qquad \: { \pmb{ m \angle Q =133 {}^{ \circ} } }}$

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