Question

In given fig. $40DC-\Delta OBA,\angle BOC={125}^{\circ }$ and $\angle CDO={70}^{\circ }$. Find $\angle DOC,\angle DCO$ and $\angle OAB$ ?

Answer 1

$\angle OAB=\angle OCD$ as they are interior opposite angles

$\angle OAB={55}^o$

$\angle ODC=\angle OBA={70}^o$

$\angle BOC={125}^o$

Finding $\angle DOC,\angle DCO$ and $\angle OAB$

$\angle BOC+\angle COD={180}^o$ (as BD is straight line)

$\angle COD=\angle DOC=180-125={55}^o$

triangle ODC sum of angles = ${180}^o$

$70+55+\angle DCO=180$

$\angle DCO=180-125$

$={55}^o$.