Question

In the given figure, $AB\parallel CD$, $\Delta AOB$ and $\Delta DOC$ are isosceles triangles

$\angle ABOis{32}^{\circ }$, then $\angle BOD$ is

• A. ${62}^{\circ }$
• B. ${31}^{\circ }$
• C. ${64}^{\circ }$
• D. ${32}^{\circ }$

Answer 1

The correct option is C

${64}^{\circ }$

Given that $\angle ABOis{32}^{\circ }$

$\angle BAO={32}^{\circ }$ (Since $\Delta AOB$ is isosceles)

$\angle AOB$ cannot be ${32}^{\circ }$ because $\angle AOB$ must be an obtuse angle

Then $\angle AOB={180}^{\circ }-({32}^{\circ }+{32}^{\circ })={180}^{\circ }-{64}^{\circ }={116}^{\circ }$

$\angle COD={116}^{\circ }$, (Opposite angles)

$\angle BOC={180}^{\circ }$

$\angle BOD+\angle COD={180}^{\circ }$

$\angle BOD={180}^{\circ }-{116}^{\circ }={64}^{\circ }$ (Opposite angles are equal)