Given: ∠AOC=90° 1Now COD is a straight line ∠AOC+∠AOD=180° Linear Pair⇒90°+∠AOD=180° using 1⇒∠AOD=90° 2Now ,∠AOD=∠AOR+∠RODSince RS is the bisector of ∠AOD⇒∠AOD=∠AOR+∠AOR⇒∠AOD=2∠AOR⇒∠AOR=∠AOD2=90°2 Using 2=45°⇒∠AOR=∠ROD=45°and ∠AOR=∠BOS Vertically opposite angles⇒∠BOS=45°and ∠ROD=∠COS Vertically opposite angles⇒∠COS=45°and AOB is a straight line ⇒∠AOD+∠DOB=180° Linear Pair⇒∠DOB=90°and PQ is the bisector of ∠BOD⇒∠BOD=2∠DOQ⇒∠DOQ=45°and ∠DOQ=∠POC=45° Vertically opposite anglesand ∠QOB=45°⇒∠QOB=∠AOP= 45°Vertically opposite anglesand COD is a straight line⇒∠DOB+∠BOC=180° Linear Pair⇒∠BOC=90°

please answer this

Given : ∠ AOC = 90 ° . . . (1) Now COD is a straight line ∠ AOC + ∠ AOD = 180 ° [Linear Pair] \Rightarrow 90 ° + ∠ AOD = 180 ° [using (1)] \Rightarrow ∠ AOD = 90 ° . . . (2) Now, ∠ AOD = ∠ AOR + ∠ ROD Since RS is the bisector of ∠ AOD \Rightarrow ∠ AOD = ∠ AOR + ∠ AOR \Rightarrow ∠ AOD = 2 ∠ AOR \Rightarrow ∠ AOR = \frac{∠ AOD}{2} = \frac{90 °}{2} [Using (2)] = 45 ° \Rightarrow ∠ AOR = ∠ ROD = 45 ° and ∠ AOR = ∠ BOS (Vertically opposite angles) \Rightarrow ∠ BOS = 45 ° and ∠ ROD = ∠ COS (Vertically opposite angles) \Rightarrow ∠ COS = 45 ° and AOB is a straight line \Rightarrow ∠ AOD + ∠ DOB = 180 ° [Linear Pair] \Rightarrow ∠ DOB = 90 ° and PQ is the bisector of ∠ BOD \Rightarrow ∠ BOD = 2 ∠ DOQ \Rightarrow ∠ DOQ = 45 ° and ∠ DOQ = ∠ POC = 45 ° (Vertically opposite angles) and ∠ QOB = 45 ° \Rightarrow ∠ QOB = ∠ AOP = 45 ° (Vertically opposite angles) and COD is a straight line \Rightarrow ∠ DOB + ∠ BOC = 180 ° [Linear Pair] \Rightarrow ∠ BOC = 90 °

View Full Answer