Question

In the given figure, arc APB : arc BQC
2: 3 and angleAOC = 150°. Find :
O
(i) angleAOB
(ii) angleBOC
(iii) angleOBA
(iv) angleOCB
(v) angleABC​

Answer 1

Here , it is given that:-

arc APB  :  arc BQC = 2:3

∠ AOC = 150°

∵ arc APB  :  arc BQC

∠ AOB : ∠ BOC = 2x : 3x

We observe that ,

∠ AOB + ∠ BOC = ∠ AOC = 150°

2x + 3x =  150°

5x = 150°

So, x = 150/5 = 30°

(i) angleAOB

∠ AOB =  2x = 2 × 30 = 60°

(ii) angleBOC

∠ BOC = 3x = 3 × 30 = 90°

(iii) angleOBA

OA  = OB [ Radii of same circle]

∠ OBA = ∠ OAB

Here OAB , forms a triangle .

So, ∠ OBA + ∠ OAB + ∠AOB = 180°

∵ ∠ OBA = ∠ OAB

2  ∠ OBA +  ∠AOB =  180°

2  ∠ OBA + 60 = 180°

2  ∠ OBA = 120°

∠ OBA = 60°

(iv) angleOCB

OB = OC [Radii]

∠ OBC = ∠ OCB

Here also a triangle OBC is formed.

So, ∠ OCB + ∠ BOC + ∠ OCB = 180°

∵ ∠ OBC = ∠ OCB

2 ∠ OCB  + 90° = 180°

2 ∠ OCB  = 180° -  90°

2 ∠ OCB =   90°

∠ OCB  = 90/2 = 45°

(v) angleABC

Here,

∠ ABC = ∠ OCB + ∠ AOB

∠ ABC = 45° + 60° = 105°