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Question
In the following figure, PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O.
Calculate the values of:
- ∠QOP
- ∠QCP
Solution
In the figure, PQ = QR ∠RQP = 68°
PC and QC are tangents to the circle with centre O from C.
In ∠PQR,
PQ = QR ...(Given)
∴ ∠PRQ = ∠RPQ
But ∠PRQ + ∠RPQ + ∠RQP = 180° ...(Sum of angles of a triangle)
`\implies` ∠PRQ + ∠PRQ + 68° = 180°
`\implies` 2∠PRQ = 180° – 68° = 112°
∴ `∠PRQ = 112^circ/2 = 56^circ`
Now QC is tangent and PQ is chord
`\implies` ∠PQC = ∠PRQ = 56°
But ∠PQC = ∠QPC ...(∵ PC = QC tangents from C)
∴ ∠QPC = 56°
In ΔPQC,
∠C + ∠PQC + ∠QPC = 180° ...(Angles of a triangle)
∠C + 56° + 56° = 180°
`\implies` ∠C + 112° = 180°
`\implies` ∠C = 180° – 112° = 68°
But ∠POQ + ∠C = 180°
∴ ∠POQ + 68° = 180°
∴ ∠POQ = 180° – 68° = 112°
Hence ∠QOP = 112° and ∠QCP = 68°
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