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Question
In the given figure, show that: ∠a = ∠b + ∠c

(i) If ∠b = 60° and ∠c = 50° ; find ∠a.
(ii) If ∠a = 100° and ∠b = 55° : find ∠c.
(iii) If ∠a = 108° and ∠c = 48° ; find ∠b.
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Solution
∵ AB || CD
∴ b = c and ∠A = ∠C ........(Alternate angles)
Now in Δ PCD,
Ext. ∠APC = ∠C + ∠D
⇒ a = b + c
(i) If b = 60°, c = 50°, then
a = b + c = 60° + 50° = 110°
(ii) If a = 100° and b = 55°,
then a = b + c
⇒ 100° = 55°+ c
⇒ c = 100°− 55° = 45°
(iii) If a = 108° and c = 48°
then a = b + c
⇒ 108° = b + 48°
⇒ b = 108°− 48° = 60°
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