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Answer :
(a) As we know the sum of interior angles of a quadrilateral is \(360^{\circ}\).
∴ \(x+ 120^{\circ} +130^{\circ} +50^{\circ}= 360^{\circ}\)
\(\Rightarrow x+300^{\circ} = 360^{\circ} \)
\(\Rightarrow x=360^{\circ} - 300^{\circ}= 60^{\circ} \)
(b) As we know the sum of interior angles of a quadrilateral is \(360^{\circ}\).
∴ \(x+ 70^{\circ} +60^{\circ} +90^{\circ}= 360^{\circ}\)
\(\Rightarrow x+220^{\circ} = 360^{\circ} \)
\(\Rightarrow x=360^{\circ} - 220^{\circ}= 140^{\circ} \)
(c) As we see the given figure with 5 sides, so the sum of interior angles of this polygon is =\((n-2) \times 180^{\circ} = 3 \times 180^{\circ}=540^{\circ}\)
So in the figure we have ,
\(m∠1 + 60^{\circ} =180^{\circ} \quad\)[Angle of straight line is \(180^{\circ}\)]
\(\Rightarrow m∠1=180^{\circ}- 60^{\circ} =120^{\circ} \)
and also, \(m∠2 + 70^{\circ} =180^{\circ} \quad \)[Angle of straight line is \(180^{\circ}\)]
\(\Rightarrow m∠2=180^{\circ}- 70^{\circ} =110^{\circ} \)
Therefore, the sum of internal angles:
\(\qquad m∠1+m∠1+x+30^{\circ} +x= 540^{\circ}\)
\(\Rightarrow 120^{\circ}+110^{\circ}+2x+30^{\circ}=540^{\circ} \)
\(\Rightarrow 2x+260^{\circ}=540^{\circ} \)
\(\Rightarrow 2x=540^{\circ}-260^{\circ}=280^{\circ} \)
\(\Rightarrow x=\frac{280^{\circ}}{2}=140^{\circ} \)
(d) As we see the given figure with 5 sides, so the sum of interior angles of this polygon is =\((n-2) \times 180^{\circ} = 3 \times 180^{\circ}=540^{\circ}\)
Therefore, the sum of internal angles:
\(\qquad x+x+x+x+x= 540^{\circ}\)
\(\Rightarrow 5x= 540^{\circ}\)
\(\Rightarrow x=\frac{280^{\circ}}{5}=108^{\circ} \)