Question

In the adjoining figure, $ABCD$ is a parallelogram in which $\angle CAD={40}^{\circ },\angle BAC={35}^{\circ }$ and $\angle COD={65}^{\circ }$. Calculate (i) $\angle ABD$ (ii) $\angle BDC$ (iii) $\angle CBD$

Answer 1

$\left(i\right)$

$\angle COD={65}^o$ [ Given ]

$\Rightarrow$ $\angle COD=\angle AOB$ [ Vertically opposite angles ]

$\therefore$ $\angle AOB={65}^o.$

In $△AOB,$

$\Rightarrow$ $\angle BAC+\angle AOB+\angle ABO={180}^o$ [ Sum of angles of a triangle is ${180}^o.$ ]

$\Rightarrow$ ${35}^o+{65}^o+\angle ABO={180}^o.$

$\Rightarrow$ ${100}^o+\angle ABO={180}^o$

$\Rightarrow$ $\angle ABO={80}^o$

$\therefore$ $\angle ABD={80}^o$

$\left(ii\right)$

$AB\parallel CD$ abd $BD$ is transversal.

$\therefore$ $\angle ABD=\angle BDC$ [ Alternate angles ]

$\therefore$ $\angle BDC={80}^o$

$\left(iii\right)$

$\Rightarrow$ $\angle AOB+\angle BOC={180}^o$ [ Linear pair ]

$\Rightarrow$ ${65}^o+\angle BOC={180}^o$

$\therefore$ $\angle BOC={115}^o$

$\Rightarrow$ $\angle DAO=\angle OCB$ [ Alternate angles ]

$\therefore$ $\angle OCB={40}^o$

In $△OCB$

$\Rightarrow$ $\angle BOC+\angle OCB+\angle CBO={180}^o.$

$\Rightarrow$ ${115}^o+{40}^o+\angle CBO={180}^o$

$\Rightarrow$ ${155}^o+\angle CBD={180}^o$

$\Rightarrow$ $\angle CBD={25}^o.$

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