tangents PQ and PR are drawn from an external point P to a circle with centre
o such that angle RPQ = 30°. A chord RS is drawn parllel to the tangent PQ find angle RQS
Answers
Given:
✰ Tangents PQ and PR are drawn from an external point P to a circle with centre
O.
✰ RPQ = 30°
✰ A chord RS is drawn parallel to the tangent PQ.
To find:
✠ ∠RQS
Solution:
We know that, tangents PQ and PR are drawn from an external point P. Tangents drawn to a circle are equal in length.
∴ PQ = PR
In ∆PQR, which is an isosceles triangle ( Two sides of an isosceles triangle are equal. )
➛ ∠PQR = ∠PRQ ( Opposite angles to equal sides are equal )
According to angle sum property of a triangle,
In ∆PQR
➛ ∠PQR + ∠PRQ + ∠RPQ = 180°
➛ ∠PQR + ∠PQR + 30° = 180°
➛ 2∠PQR = 180° - 30°
➛ 2∠PQR = 150°
➛ ∠PQR = 150/2
➛ ∠PQR = 75°
Radius is parallel to tangent drawn at the point of contact between two,
∴ ∠OQP = ∠ORP = 90°
In □ PQOR,
Sum of interior angles of a rectangle.
⤳ ∠RPQ + ∠OQR + ∠ORP + ∠QOR = 360°
⤳ 30° + 90° + 90° + ∠QOR = 360°
⤳ 210° + ∠QOR = 360°
⤳ ∠QOR = 360° - 210°
⤳ ∠QOR = 150°
We know that the angle subtended by an arc at any point on the given circle is half the angle subtended at the centre of the circle by the same arc of circle.
➛ ∠QSR = 150°/2
➛ ∠QSR = 75°
RS || PQ [ RS is parallel to PQ ]
➛ ∠QSR = ∠SQT = 75° [ Alternate angles ]
∴ ∠SQT = 75°
Now,
➤ ∠SQT + ∠RQS + ∠PQR = 180°
➤ 75° + ∠RQS + 75° = 180°
➤ ∠RQS + 150° = 180°
➤ ∠RQS = 180° - 150°
➤ ∠RQS = 30°
_______________________________
