Mathematics
In adjoining figure, P and Q are the centers of two circles touching externally at R and CD is the common tangent.
If ∠CAR = 38°, then find ∠DBR.
Circles
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Answer
Join PC and QD.
We know that,
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ PC ⊥ CD and QD ⊥ CD
⇒ ∠PCD = 90° and ∠QDC = 90°
The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
⇒ ∠CPR = 2 x ∠CAR = 2 x 38° = 76°
⇒ ∠CPR = ∠CPQ = 76°
CDQP is quadrilateral.
∴ ∠PCD + ∠QDC + ∠DQP + ∠CPQ = 360°
⇒ 90° + 90° + ∠DQP + 76° = 360°
⇒ 256° + ∠DQP = 360°
⇒ ∠DQP = 360° - 256°
⇒ ∠DQP = 104°
From figure,
⇒ ∠DQR = ∠DQP
⇒ ∠DQR = 104°
The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
⇒ ∠DQR = 2 x ∠DBR
⇒ 104° = 2 x ∠DBR
⇒ ∠DBR =
⇒ ∠DBR = 52°.
Hence, ∠DBR = 52°.
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