Question

A =90°, angle B=90°,OB=4.5cm,OA=6CM and AP=4cm.find theQB

Answer 1

In ∆AOP and ∆QOB
∠A = ∠B = 90°
∠AOP = ∠QOB (vertically opposite angle)
OA ≠ OB
∴ ΔAOP and ΔQOB are similar Δs
∴ OA/OB = AP/QB (in similar Δs the ratio of corresponding sides are constants)
QB = AP*OB/OA
QB = 3cm

Answer 2

In ∆AOP and ∆ BOQ,
angle O = angle O .........( Vertically opp. angles)
angle A = angle B ..........( each 90° )
∆ AOP ~ ∆ BOQ ...........(AA test )
AO/ BO = AP/ BQ = OP/ OQ
6/4.5 = 4/BQ
60/ 45 =. 4/BQ
4BQ = 12
BQ = 3 cm.