Question

In the given figure, O is the centre of the circle PS = SO = PQ/2 and PM = MO, then find the difference between the measures of ∠QRS and ∠PQR.​

Answer 1

Given : O is the centre of the circle PS = SO = PQ/2 and PM = MO

To Find : difference between the measures of ∠QRS and ∠PQR.

Solution:

PQ is Diameter as it passes through center O

=> PO  = Radius = Diameter/2  =  PQ/2

PS = SO = PQ/2   Given

=>  PS = SO  = PO

=> ∠POS = 60°

∠POS +  ∠QOS = 180°  Linear Pair

=> 60° +  ∠QOS = 180°

=>   ∠QOS = 120°

∠QRS  = (1/2)  ∠QOS

=> ∠QRS  = (1/2) 120°

=> ∠QRS  =60°

in ΔPMS  & ΔOMS

PM = OM   given

MS = MS   common

PS = SO   given

=> ΔPMS  ≅ ΔOMS

=> ∠PMS  = ∠OMS

∠PMS  + ∠OMS = 180°

=> ∠PMS  = ∠OMS = 90°

∠PMS = ∠RMQ  ( vertically opposite angles)

=>  ∠RMQ = 90°

in   Δ QMR

∠RMQ  + ∠QRM  + ∠MQR = 180°​

∠QRM = ∠QRS  =60°

∠MQR  = ∠PQR

=> 90°  + 60° +  ∠PQR = 180°​

=> ∠PQR = 30°​

difference between the measures of ∠QRS and ∠PQR.​   = 60° - 30°​ = 30°​

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