Question
In the figure given alongside, $AB || CD, EF || BC$, $\angle\text{BAC}=60^\circ$ and $\angle\text{DHE}=50^\circ.$ Find $\angle\text{GCH}$ and $\angle\text{AGH}.$ 
✓
Answer
$AB || CD$ and $AC$ is the transversal.
$\Rightarrow\angle\text{BAC}=\angle\text{ACD}=60^\circ$ (alternate angles) i. e.
$\angle\text{BAC}=\angle\text{GCH}=60^\circ$
Now, $\angle\text{DHF}=\angle\text{CHG}=50^\circ$ (vertically opposite angles) In $\triangle\text{GCH},$
by angle sum property, $\angle\text{GCH}+\angle\text{CHG}+\angle\text{CGH}=180^\circ$
$\Rightarrow60^\circ+50^\circ\angle\text{CGH}=180^\circ$
$\Rightarrow\angle\text{CGH}=70^\circ$
Now, $\angle\text{CGH}+\angle\text{AGH}=180^\circ$ (linear pair)
$\Rightarrow70^\circ+\angle\text{AGH}=180^\circ$
$\Rightarrow\angle\text{BAC}=\angle\text{ACD}=60^\circ$ (alternate angles) i. e.
$\angle\text{BAC}=\angle\text{GCH}=60^\circ$
Now, $\angle\text{DHF}=\angle\text{CHG}=50^\circ$ (vertically opposite angles) In $\triangle\text{GCH},$
by angle sum property, $\angle\text{GCH}+\angle\text{CHG}+\angle\text{CGH}=180^\circ$
$\Rightarrow60^\circ+50^\circ\angle\text{CGH}=180^\circ$
$\Rightarrow\angle\text{CGH}=70^\circ$
Now, $\angle\text{CGH}+\angle\text{AGH}=180^\circ$ (linear pair)
$\Rightarrow70^\circ+\angle\text{AGH}=180^\circ$
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