Question
In the given figure, if $\text{AB }||\text{ CD},\text{EF }||\text{ BC},\angle\text{BAC}=65^\circ$ and $\angle\text{DHF}=35^\circ,$ find $\angle\text{AGH}.$
✓
Answer
In the given figure, if $\text{AB }||\text{ CD},\text{EF }||\text{ BC},\angle\text{BAC}=65^\circ$ and $\angle\text{DHF}=35^\circ$ We need to find $\angle\text{AGH}.$
Here, $GF$ and $CD$ are straight lines intersecting at point $H$,
so using the property, "vertically opposite angles are equal",
we get, $\angle\text{DHF}=\angle\text{GHC}$
$\angle\text{GHC}=35^\circ$ Further, as $AB \| CD$ and $AC$ is the transversal Using the property, "alternate interior angles are equal" $\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{BAC}=65^\circ$ Further applying angle sum property of the triangle In $\triangle\text{GHC}$
$\angle\text{DHF}+\angle\text{HCG}+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}+35^\circ+65^\circ=180^\circ$
$100^\circ+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}=180^\circ-100^\circ$
$\angle\text{CGH}=80^\circ$ Hence, applying the property, "angles forming a linear pair are supplementary" As $AGC$ is a straight line $\angle\text{CGH}+\angle\text{AGH}=180^\circ$
$\angle\text{AGH}+80^\circ=180^\circ$
$\angle\text{AGH}=180^\circ-80^\circ$
$\angle\text{AGH}=100^\circ$
Therefore,$\angle\text{AGH}=100^\circ$
Here, $GF$ and $CD$ are straight lines intersecting at point $H$,
so using the property, "vertically opposite angles are equal",
we get, $\angle\text{DHF}=\angle\text{GHC}$
$\angle\text{GHC}=35^\circ$ Further, as $AB \| CD$ and $AC$ is the transversal Using the property, "alternate interior angles are equal" $\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{BAC}=65^\circ$ Further applying angle sum property of the triangle In $\triangle\text{GHC}$
$\angle\text{DHF}+\angle\text{HCG}+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}+35^\circ+65^\circ=180^\circ$
$100^\circ+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}=180^\circ-100^\circ$
$\angle\text{CGH}=80^\circ$ Hence, applying the property, "angles forming a linear pair are supplementary" As $AGC$ is a straight line $\angle\text{CGH}+\angle\text{AGH}=180^\circ$
$\angle\text{AGH}+80^\circ=180^\circ$
$\angle\text{AGH}=180^\circ-80^\circ$
$\angle\text{AGH}=100^\circ$
Therefore,$\angle\text{AGH}=100^\circ$
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