Question

Question 14
In the figure, O is the centre of the circle BD = OD and CD $\perp$ AB. Find $\angle CAB$.

Answer 1

Given, in the figure BD = OD, CD $\perp$ AB

In $\Delta$ OBD,
BD = OD
$\therefore$ OD = OB [both are the radius of same circle]
OB = OD = BD

Thus, $\Delta$ ODB is an equilateral triangle.
$\therefore \angle BOD=\angle OBD=\angle ODB={60}^{\circ }$

In $\Delta$ MBC and $\Delta$ MBD
MB = MB [common]
$\angle CMB=\angle BMD={90}^{\circ }$
And CM = MD [in a circle, any perpendicular drawn on a chord also bisect the chord]
$\therefore \Delta MBC\cong \Delta MBD$ [ by SAS congruence]
$\therefore \angle MBC=\angle MBD$ [by CPCT]
$\therefore \angle MBC=\angle OBD={60}^{\circ }$
Since, AB is a diameter of the circle
$\angle ACB={90}^{\circ }$

$In\Delta ACB\angle CAB+\angle CBA+\angle ACB={180}^{\circ }$ [by angle sum property of a triangle]
$\angle CAB+{60}^{\circ }+{90}^{\circ }={180}^{\circ }$
$\angle CAB={180}^{\circ }-({60}^{\circ }+{90}^{\circ })={30}^{\circ }$