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

Answer 1

(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°