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Question
The adjoining figure shows two straight lines AB and CD intersecting at point P. If ∠BPC = 4x – 5° and ∠APD = 3x + 15°; find:
(i) the value of x.
(ii) ∠APD
(iii) ∠BPD
(iv) ∠BPC
Solution
(i) 4x - 5 = 3x + 15 (∵ ∠CPB = ∠APD; opposite angles)
⇒ 4x - 3x = 15 + 5
⇒ x = 20
(ii) ∠APD = 3x + 15
= (3 × 20) + 15
= 60 + 15 = 75°
(iii) ∠BPD = 180 - ∠BPC
= 180 - (4x - 5)
= 180 - 4x + 5
= 185 - (4 × 20)
= 185 - 80 = 105°
(iv) ∠BPC = 4x - 5
= (4 × 20) - 5
= 80 - 5
= 75°
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