Question

In the adjoining fig. $10.33,ABCD$ is a rhombus.
Find the measure of the following angles, if $\angle ACB={30}^{\circ }$
(a) $\angle BOC$
(b) $\angle CBO$
(c) $\angle OAD$
(d) $\angle ABO$

Answer 1

From the figure:-

$\angle DAO=\angle BCO$ (Alternate interior Angle)

$={30}^o$

(Two parallel lines are crossed by a transversal)

We know that Diagonals in a rhombus intersect at right angle so $\angle BOC=90$ …………..$\left(1\right)$

through vertically opposite Angles $AOD={90}^o$

Now takes $\Delta BOC$

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

${90}^o+{30}^o+\angle CBO={180}^o$

$\angle CBO=180-{120}^o$

$\angle CBO={60}^o$ ……………$\left(2\right)$ Ans

$\angle OCB=\angle OAD$ (Alternate interior Angle)

$={30}^o$ …………..$\left(3\right)$ Ans

All the four triangle should be similar by this in $\Delta AOB$.

$\angle AOB={90}^o$

$\angle OAB={30}^o$

$\angle ABO={60}^o$.