Question

In the figure, O is the centre of the circle. Find the size of each lettered angle:

Answer 1

In the figure,
AC is the diameter of the circle, with centre as O,
OD || CB and $\angle CAB={40}^o$
In $\Delta ABC$,
$\angle B={90}^o$
(Angle in a semicircle)
By angle sum property of triangle, we get,
$\angle BCA+\angle ABC+\angle BAC={180}^o$
$\angle BCA+\angle BAC+{90}^o={180}^o$
$\angle BCA+\angle BAC={180}^o-{90}^o$
$\angle BCA+\angle BAC={90}^o$
$x+{40}^o={90}^o$
$x={90}^o-{40}^o$
We get,
$x={50}^o$
$\because$ OD || CB
Hence,
$\angle AOD=\angle BCA$ (corresponding angles)
$\angle AOD={50}^o$
But $\angle AOD+\angle DOC={180}^o$ (Linear pair)
${50}^o+y={180}^o$
$y={180}^o-{50}^o$
We get,
$y={130}^o$
Therefore, $x={50}^o$ and $y={130}^o$