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Question -

In the figure, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
(a) 15°
(b) 25°
(c) 70°
(d) 35°



Answer -

In the figure,
 
PQ || RS, ∠AEF = 95°
∠BHS = 110°, ∠ABC = x
∵ PQ || RS,
∴ ∠AEF = ∠1 = 95° (Corresponding anlges)
But ∠1 + ∠2 = 180° (Linear pair)
⇒ ∠2 = 180° – ∠1 = 180° – 95° = 85°
In ∆AGH,
Ext. ∠BHS = ∠2 +x
⇒ 110° = 85° + x
⇒ x= 110°-85° = 25° (b)

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