Question

In the given figure, ABCD is a quadrilateral inscribed in a circle. Diagonals AC and BD are joined. If $\angle$CAD = 60${}^o$ and $\angle$BDC = 25${}^o$. Then $\angle$BCD is

• A. 85${}^o$
• B. 120${}^o$
• C. 115${}^o$
• D. 95${}^o$

Answer 1

The correct option is D 95${}^o$

Given- O is the centre of a circle. $\overline{AOC}\&\overline{BOD}$ are two diameters. $\angle CAD={60}^o$.
To find out $\angle BCD=?$
Solution- We join AC & BD & BC. Now OA, OC & OD, OB are radii of the given circle since they belong to the diameters AC & BD respectively.
$\therefore OA=OC=OD=OB$.
Again, BC subtends $\angle CDB$ & $\angle CAB$ to the circumference of the given circle. $\therefore \angle CAB=\angle CDB={25}^o$ since angles, subtended by a chord of a circle to its circumference, are equal. Now ABCD is a cyclic quadrilateral.
$\therefore \angle DAB+\angle DCB={180}^0$
$\therefore \angle BCD={95}^0$
Ans- Option D.