Question

In a rectangle $ABCD$. If $\angle BAC={35}^o$ then find:

(i) $\angle ACB$
(ii) $\angle ABD$
(iii) $\angle COD$
(iv) $\angle BOC$

Answer 1

In $△ABC,$

$\Rightarrow$ $\angle ACB+\angle CBA+\angle BAC={180}^o$

$\Rightarrow$ $\angle ACB+{90}^o+{35}^o={180}^o$

$\Rightarrow$ $\angle ACB+{125}^o={180}^o$

$\Rightarrow$ $\angle ACB={55}^o$

We know that, diagonals of rectangle are equal and bisect each other.

$\therefore$ $OA=OB$

$\Rightarrow$ $\angle BAO=\angle ABO$ [ Base angles of equal sides are equal ]

$\therefore$ $\angle ABO={35}^o$

$\therefore$ $\angle ABD={35}^o$

In $△ABO,$

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

$\Rightarrow$ ${35}^o+\angle BOA+{35}^o={180}^o$

$\Rightarrow$ ${70}^o+\angle BOA={180}^o$

$\therefore$ $\angle BOA={110}^o$

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

$\therefore$ $\angle COD={110}^o$

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

$\Rightarrow$ ${110}^o+\angle BOC={180}^o$

$\therefore$ $\angle BOC={70}^o$

So we get,

$\left(i\right)$ $\angle ACB={55}^o$

$\left(ii\right)$ $\angle ABD={35}^o$

$\left(iii\right)$ $\angle COD={110}^o$

$\left(iv\right)$ $\angle BOC={70}^o$