Question

$DA$ and $DB$ are the triangle drwan from D, to the circle with centre $O$, at A and B respectively $\angle ADO={35}^{\circ }$ find $\angle AOB$

Answer 1

We are given,

Circle of centre $O$ is having radius $OA$ and $OB$ (r)

Now, In $\Delta OAD$ & $\Delta OBD$ is equal as two sides are equal and one angle is equal $(OA=OB=r)$ and $OD$ also

$m\angle OAD=m\angle OBD={90}^o$

Now, given that $m\angle OBD={90}^o$

Now, given that $m\angle ADO={35}^o$

then $m\angle AOD=?$

$\Delta AOD,m\angle AOD+m\angle ADO+m\angle OAD={180}^o$

$m\angle AOD=180-m\angle OAD-m\angle ADO$

$m\angle AOD=180-90-35={55}^o$

So, $m\angle AOB=m\angle AOD+m\angle DOB$

$={55}^o+{55}^o$

$m\angle AOB={110}^o$