Question

In the adjoining figure, O is the center of the circle and AB is the diameter. Tangent PQ touches the circle at D. $\angle$BDQ = ${48}^{\circ }$. Find the ratio of $\angle$DBA to $\angle$DCB

• A.
• B.
• C.
• D. Can't be determined

Answer 1

The correct option is B

Given that ∠BDQ = ${48}^{\circ }$

$\angle$ODB = $\angle$ODQ - $\angle$BDQ

$\angle$ODB = ${90}^{\circ }$ – ${48}^{\circ }$

$\angle$ODB = ${42}^{\circ }$

OB = OD

Therefore $\angle$ODB = $\angle$OBD = ${42}^{\circ }$

$\angle$ADB = 90 (Angle subtended by the diameter)

$\angle$ADO = $\angle$ADB - $\angle$ODB

$\angle$ADO = ${90}^{\circ }$ – ${42}^{\circ }$

$\angle$ADO = ${48}^{\circ }$

$\angle$DCB = ${180}^{\circ }$ - ∠OAD (Cyclic quadrilateral)

$\angle$DCB = ${180}^{\circ }$ – ${48}^{\circ }$ = ${132}^{\circ }$

Therefore

$\angle$DBA: $\angle$DCB = 42: 132 = 7:22