Question

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Answer 1

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(vii) In line AD,

∠ACB + ∠ACD = 180°       (Linear Pair)

⇒ y + 120° = 180°

⇒ y = 180° - 120°

⇒ y = 60°

Since ΔABC is an isosceles triangle, x = y.

⇒ x = y = 60°

(viii) In line PS,

∠QRP + ∠QRS = 180°       (Linear Pair)

⇒ y + 110° = 180°

⇒ y = 180° - 110°

⇒ y = 70°

Since ΔPQR is an isosceles triangle,

∠RPQ = ∠RQP = x

In ΔPQR,

∠PQR + ∠RPQ + ∠RQP = 180°

⇒ x + x + y = 180°

⇒ 2x + 70° = 180°

⇒ 2x = 180° - 70°

⇒ 2x = 110°

⇒ x = 110° ÷ 2

⇒ x = 55°