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Question
In the given figure, PQ = PS. The value of x is ______.
Options
35°
45°
55°
70°
Solution
In the given figure, PQ = PS. The value of x is 45°.
Explanation:
In ΔPQS, ∠2 + ∠3 = 110° ......(i) [Exterior angle property of a triangle]
∠2 + ∠3 + ∠4 = 180° ......[Angle sum property of a triangle]
⇒ ∠4 = 180° = 110° ......[Using (i)]
∴ ∠4 = 70°
Now, PQ = PS ......(Given)
⇒ ∠2 = ∠4 = 70° ......(ii)
(ii) Now, in ∆PRS,
∠2 = x + 25° ......[Exterior angle property of a triangle]
⇒ x = 70° – 25° = 45° ......[Using (ii)]
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