Question

In the given figure, O is the center of the circle BD = OD and $CD\perp AB$. Find $\angle CAB$ (in degrees)

Answer 1

$BD=OD$ [ Given ]

$OD=OB$ [ Radius of a circle. ]

$\therefore$ $AD=OB=BD$

$\therefore$ $△OBD$ is an equilateral triangle.

$\Rightarrow$ $\angle AOD+\angle DOB={180}^o$ [ Linear pair ]

$\Rightarrow$ $\angle AOD+{60}^o={180}^o$ [ Angle of an equilateral triangle ]

$\Rightarrow$ $\angle AOD={120}^o$

$\Rightarrow$ $\angle ACD=\frac{1}{2}\angle AOD$ [ Angle at the center is twice the angle at the circumference ]

$\Rightarrow$ $\angle ACD=\frac{1}{2}\times {120}^o$

$\therefore$ $\angle ACD={60}^o$ ---- ( 1 )

$\Rightarrow$ $\angle CEA={90}^o$ [ Since $CD\perp AB$ ] --- ( 2 )

In $△CAE,$

$\Rightarrow$ $\angle CAE+\angle CEA+\angle ACE={180}^o$

$\Rightarrow$ $\angle CAE+{90}^o+{60}^o={180}^o$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\angle CAE+{150}^o={180}^o$

$\Rightarrow$ $\angle CAE={180}^o-{150}^o$

$\Rightarrow$ $\angle CAE={30}^o$

$\therefore$ $\angle CAB={30}^o.$