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 angle QRS and angle PQR
Answers
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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difference between the measures of ∠QRS and ∠PQR.
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Answer:
30 degrees
Step-by-step explanation:
PQ is Diameter
=> 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 ∠QPR. = 60° - 30° = 30°