Question

A circle with centre O has tangent PQ touching the circle at B. If ABCD is a cyclic quadrilateral and $\angle$DBQ = ${65}^{\circ }$ , then $\angle$BCD is

• A. 35°
• B. 85°
• C. 115°
• D. 90°

Answer 1

The correct option is C

115°

Join OB and OD

We know that OB is perpendicular to PQ

$\angle$OBD = $\angle$OBQ - $\angle$DBQ

$\angle$OBD = ${90}^{\circ }$ – ${65}^{\circ }$

$\angle$OBD = ${25}^{\circ }$

OB = OD (radius)

$\angle$OBD = $\angle$ODB = ${25}^{\circ }$

In $△$ODB

$\angle$OBD + $\angle$ODB + $\angle$BOD = ${180}^{\circ }$

${25}^{\circ }$ + ${25}^{\circ }$ + $\angle$BOD = ${180}^{\circ }$

$\angle$BOD = ${130}^{\circ }$

$\angle$BAD = $\frac{1}{2}$ $\angle$BOD

(Angle subtended by a chord on the center is double the angle subtended on the circle)

$\angle$BAD = $\frac{1}{2}$ ( ${130}^{\circ }$)

$\angle$BAD = ${65}^{\circ }$

ABCD is a cyclic quadrilateral

$\angle$BCD + $\angle$BAD = ${180}^{\circ }$

$\angle$BCD + ${65}^{\circ }$ = ${180}^{\circ }$

$\angle$BCD = ${115}^{\circ }$