Question

In figure, $\angle AOF$ and $\angle FOG$ form a linear pair. If $\angle EOB=\angle FOC={90}^{\circ }$ and $\angle DOC=\angle FOG=\angle AOB={30}^{\circ }$
i. Name all the right angles

Answer 1

$\angle DOG,\angle AOD,\angle FOC$ and $EOB$
Given:$\angle EOB=\angle FOC={90}^{\circ }$ and $\angle DOC=\angle FOG=\angle AOB={30}^{\circ }$
As $\angle AOF$ and $\angle FOG$ from a linear pair
$\Rightarrow \angle AOF+\angle FOG={180}^{\circ }$
$\Rightarrow \angle AOB+\angle EOB+\angle FOE+\angle FOG={180}^{\circ }$
$\Rightarrow {30}^{\circ }{90}^{\circ }+\angle FOE+{30}^{\circ }={180}^{\circ }\Rightarrow {150}^{\circ }+\angle FOE={180}^{\circ }$
$\angle FOE={180}^{\circ }-{150}^{\circ }$
$\therefore \angle FOE={30}^{\circ }$
Given that :$\angle FOC={90}^{\circ }$
$\Rightarrow \angle FOE+\angle DOE=\angle COD={90}^{\circ }$
$\Rightarrow {30}^{\circ }+\angle DOE+{30}^{\circ }={90}^{\circ }$
$\Rightarrow \angle DOE={90}^{\circ }-{60}^{\circ }$
$\therefore \angle DOE={30}^{\circ }$
Also $\angle EOB={90}^{\circ }$
$\Rightarrow \angle DOE+\angle DOC+\angle COB={90}^{\circ }$
$\Rightarrow {30}^{\circ }+{30}^{\circ }+\angle COB=90^{\circ }$
$\Rightarrow \angle COB={90}^{\circ }-{60}^{\circ }$
$\therefore \angle COB={30}^{\circ }$
Thus, $\angle FOE=\angle COB=\angle DOE={30}^{\circ }$
Now, $\angle DOE+\angle FDE+\angle FOG={30}^{\circ }+{30}^{\circ }+{30}^{\circ }={90}^{\circ }$
Now, $\angle AOB+\angle COB+\angle DOC={30}^{\circ }+{30}^{\circ }+{30}^{\circ }={90}^{\circ }$
$\Rightarrow \angle AOD={90}^{\circ }$
From above, we get two right angles $\angle DOG$ and $\angle AOD$ and two right angles are already given as $\angle FOC$ and $\angle EOB$.
Hence, $\angle DOG$, $\angle AOD$, $\angle FOC$ and $\angle EOB$ are right angles.