Question

In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.​

Answer 1

SOLUTION =>

We know that tangents from an external point are equal in length.

∴ PQ = PR

In ∆PQR,

PQ = PR

∴ ∠PQR=∠PRQ (Angles opposite to equal sides are equal.)

Now in ∆PQR,

∠PQR+∠PRQ+∠RPQ=180°

⇒2∠PQR=180°−30°=150°

⇒∠PQR=75

Also, radius is perpendicular to the tangent at the point of contact.

∴ ∠OQP = ∠ORP = 90°

Now, in □PQOR,

∠RPQ + ∠ORP + ∠OQP + ∠QOR = 360°

⇒ 30° + 90° + 90° + ∠QOR = 360°

⇒ ∠QOR = 360° − 210° = 150°

Since, angle subtended by an arc at any point on the circle is half the angle subtended at the centre by the same arc.

∴∠QSR=150∘2=75∘

Also, ∠QSR = ∠SQT (Alternate interior angles)

∴ ∠SQT = 75°

Now,

∠SQT + ∠RQS + ∠PQR = 180°

⇒ 75° + ∠RQS + 75° = 180°

⇒ ∠RQS = 180° − 150°

= 30°

❤Sweetheart❤

$....$

Answer 2

PQ = PR

In ∆PQR,

PQ = PR

∴ ∠PQR=∠PRQ

Now in ∆PQR,

∠PQR+∠PRQ+∠RPQ=180°

⇒2∠PQR=180°−30°=150°

⇒∠PQR=75

∠RPQ + ∠ORP + ∠OQP + ∠QOR = 360°

⇒ 30° + 90° + 90° + ∠QOR = 360°

⇒ ∠QOR = 360° − 210° = 150°

∴∠QSR=150∘2=75∘

also, ∠QSR = ∠SQT (Alternate interior angles)

∴ ∠SQT = 75°

Now,

∠SQT + ∠RQS + ∠PQR = 180°

⇒ 75° + ∠RQS + 75° = 180°

⇒ ∠RQS = 180° − 150°

30°