A) \[{{60}^{{}^\circ }}\]
B) \[{{50}^{{}^\circ }}\]
C) \[{{90}^{{}^\circ }}\]
D) \[{{70}^{{}^\circ }}\]
Correct Answer: D
Solution :
(d): Given \[5x=4y=a\Rightarrow x=\frac{\alpha }{5},y=\frac{\alpha }{4}\] \[\therefore \]\[AB\parallel CD\] \[\therefore \]\[a=x\] \[\therefore \]\[a+y={{180}^{{}^\circ }}\] \[\therefore \]\[\frac{\alpha }{5}+\frac{\alpha }{4}={{180}^{{}^\circ }}\] \[\therefore \]\[9\alpha =180\times 20\Rightarrow \alpha =400\] \[\therefore \] \[y=\frac{400}{4}={{100}^{{}^\circ }}\] \[\therefore \]\[z=y+10={{100}^{{}^\circ }}+{{10}^{{}^\circ }}={{110}^{{}^\circ }}\] and \[w+z={{180}^{{}^\circ }}\] (Linear pair) \[w+{{110}^{{}^\circ }}={{180}^{{}^\circ }}\] \[w={{70}^{{}^\circ }}\]
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