Question

29. In the figure, ACB = 4ABC, and CP bisects ZACB. Find
(a) ABC
(b) BPC​

Answer 1

Answer:

Let ∠ABC = x,

∵∠ ACB = 4  × ∠ABC,

⇒ ∠ ACB = 4x,

By the diagram,

∠ CAB = 90°,

Thus, in triangle ABC,

∠ ABC + ∠ ACB + ∠ CAB = 180°

x + 4x + 90 = 180

5x + 90 = 180

5x = 90

⇒ x = 18

Hence,  ∠ABC = 18°

(b) Now, CP bisects angle ACB that is, ∠BCP = ∠ACP

∵ ∠ ACB = ∠BCP + ∠ACP = ∠BCP + ∠BCP = 2∠BCP

⇒ 2∠BCP= 4x = 72°

⇒ ∠ BCP = 36°

Now, in triangle BCP,

∠BCP + ∠ BPC + ∠ CBP = 180°

⇒ 36° + ∠ BPC + 18° = 180°

⇒ ∠ BPC + 54° = 180°

⇒ ∠ BPC = 126°

Answer 2

Answer:

blah blah blah..........

here's ur answer guys