Question

ABCD is a cyclic quadrilateral whose side AD is a diameter of the circle through A, B, C and D. If $\angle ABC={110}^{\circ }$, find $\angle BDAC$.

• A. 30º
• B. 50º
• C. 60º
• D. 40º

Answer 1

Answer: (B), (C)

The correct options are
B

50º

C

60º

Given: ABCD is a cyclic quadrilateral whose side AD is the diameter of the circle and
$\angle ABC={110}^{\circ }$.

To find $\angle BAC,$

$\angle D+\angle B={180}^{\circ }$
$\text{(Opposite angles of a cyclic quadilateral )}$

${110}^{\circ }+\angle D={180}^{\circ }$

$\angle D={180}^{\circ }-{110}^{\circ }={70}^{\circ }$

$\angle ACD={90}^{\circ }$
$\text{(Angle in a semicircle)}$

$\text{In}\Delta ADC,$

$\angle BAC+{70}^{\circ }+{90}^{\circ }={180}^{\circ }$
$\text{(Since sum of angles of a triangle is}{180}^{\circ })$

$\angle BAC={180}^{\circ }-{90}^{\circ }-{70}^{\circ }={20}^{\circ }$