In figure, and from a linear pair. If = =90^ and = = =30^ . i. Fi…

Question

In figure, $\angle\text{AOF}$ and $\angle\text{FOG}$ from a linear pair. If $\angle\text{EOB}=\angle\text{FOC}=90^\circ$ and $\angle\text{DOC}=\angle\text{FOG}=\angle\text{AOB}=30^\circ.$

$i.$ Find the measure of $\angle\text{FOE},\angle\text{COB}$ and $\angle\text{DOE}.$
$ii.$ Name all the right angles.
$iii.$ Name three pairs of adjacent complementary angles.
$iv.$ Name three pairs of adjacent supplementary angles.
$v.$ Name three pairs of adjacent angles

✓

Answer

$i. \angle\text{FOE}=\text{x},\angle\text{DOE}=\text{y}$ and $\angle\text{BOC}=\text{z}$
Since $\angle\text{AOF},\angle\text{FOG}$ is a linear pair
$\angle\text{AOF}+30=180$
$\angle\text{AOF}=180-30$
$\angle\text{AOF}=150$
$\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}+\angle\text{DOE}+\angle\text{EOF}=150$
$30 + z + 30 + y + x = 150$
$x + y + z =150 - 30 - 30$
$x + y + z = 90 ...(1)$
$\angle\text{FOC}=90^\circ$
$\angle\text{FOE}+\angle\text{EOD}+\angle\text{DOC}=90^\circ$
$x + y + 30 = 90$
$x + y = 90 - 30$
$x + y = 60 ...(2)$
Substituting $(2)$ in $(1)$
$x + y + z = 90$
$60 + z = 90$
$z = 90 - 60 = 30$
Given $BOE = 90$
$\angle\text{BOC}+\angle\text{COD}+\angle\text{DOE}=90^\circ$
$30 + 30 + DOE = 90$
$DOE = 90 - 60 = 30$
$DOE = x = 30$
We also know that,
$x + y = 60$
$y = 60 - x$
$y = 60 - 30$
$y = 30$
Thus we have $\angle\text{FOE}=30,\angle\text{COB}=30$ and $\angle\text{DOE}=30$
$ii.$ Right angles are:
$\angle\text{DOG},\angle\text{COF},\angle\text{BOF},\angle\text{AOD}.$
$iii.$ Adjacent complementary angles are:
$(\angle\text{AOB},\angle\text{BOD});(\angle\text{AOC},\angle\text{COD});(\angle\text{BOC},\angle\text{COE}).$
$iv.$ Adjacent supplementary angles are:
$(\angle\text{AOB},\angle\text{BOG});(\angle\text{AOC},\angle\text{COG});(\angle\text{AOD},\angle\text{DOG}).$
$v.$ Adjacent angles are:
$(\angle\text{BOC},\angle\text{COD});(\angle\text{COD},\angle\text{DOE});(\angle\text{DOE},\angle\text{EOF}).$