two tangent pq and pr are drawn toa circle with center o from n external point p . prove that angle qpr=2angle oqr
Answers
Answered by
2
[FIGURE IS IN THE ATTACHMENT]
SOLUTION:
Given that: PQ and PR are two tangents drawn to a circle with centre O from an external point P
To Prove: ∠QPR=2∠OQR
Construction: Join QR, OQ and OR
Proof: we know that lengths of a tangent drawn from an external point to a circle are equal.
PQ= PR
∆PQR is an isosceles triangle.
∠PQR= ∠PRQ
In ∆ PQR
∠PQR+ ∠PRQ+∠QPR= 180°
∠PQR+ ∠PQR+∠QPR= 180°
2∠PQR= 180°-∠QPR
∠PQR=1/2 (180°-∠QPR)
∠PQR= 90°-1/2∠QPR
1/2∠QPR=90°-∠PQR………. (1)
Since, OQ Perpendicular PQ
∠OQP= 90°
∠OQR+ ∠PQR=90°
∠OQR =90°- ∠PQR……..(2)
∠OQR =1/2∠QPR
2∠OQR =∠QPR
∠QPR=2∠OQR
hope this helps you
SOLUTION:
Given that: PQ and PR are two tangents drawn to a circle with centre O from an external point P
To Prove: ∠QPR=2∠OQR
Construction: Join QR, OQ and OR
Proof: we know that lengths of a tangent drawn from an external point to a circle are equal.
PQ= PR
∆PQR is an isosceles triangle.
∠PQR= ∠PRQ
In ∆ PQR
∠PQR+ ∠PRQ+∠QPR= 180°
∠PQR+ ∠PQR+∠QPR= 180°
2∠PQR= 180°-∠QPR
∠PQR=1/2 (180°-∠QPR)
∠PQR= 90°-1/2∠QPR
1/2∠QPR=90°-∠PQR………. (1)
Since, OQ Perpendicular PQ
∠OQP= 90°
∠OQR+ ∠PQR=90°
∠OQR =90°- ∠PQR……..(2)
∠OQR =1/2∠QPR
2∠OQR =∠QPR
∠QPR=2∠OQR
hope this helps you
Attachments:
Similar questions