Question

In the given figure, $PQ$ is a tangent to a circle at $A,DB$ is the diameter and $O$ is the centre of the circle. If $\angle ADB={30}^{o}and\angle CBD={60}^o$, calculate:
i) $\angle QAB$
ii) $\angle PAD$
iii) $\angle CDB$

Answer 1

In $△ADB$

$\angle DAB={90}^o$

$\angle ADB+\angle DAB+\angle ABD={180}^o$

$30+90+\angle ABD={180}^o$

$ABD={60}^o$

similarly

$\angle CDB+\angle CBD+\angle DCB=180$

$\angle CDB+60+90=180$

$\angle CDB={30}^o$

In $△ADBAO=OB$

$\angle OAB=\angle OBA={60}^o$

$60+\angle BAQ={90}^o$

$\angle BAQ={30}^o$

In $△AODOO=OA$

$\angle ODA=\angle OAD={30}^o$

$\angle OAD+\angle OAP={90}^o$

$30+\angle DAP=90$

$\angle DAP={60}^o$