Question
Find the angles marked with a question mark shown in the figure.
✓
Answer
In parallelogram $ABCD,$
$CE\ ⊥\ AB$ and $CF\ ⊥\ AD$
$\angle\text{BCE}=40^\circ$
In $\triangle\text{BCE}$,
$\angle\text{BCE}+\angle\text{CEB}+\angle\text{EBC}=180^\circ$ (Sum of angles of a triangle)
$\Rightarrow40^\circ +90^\circ+\angle\text{EBC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{EBC}=180^\circ$
$\Rightarrow\angle\text{EBC}=180^\circ-130^\circ$
$\Rightarrow\angle\text{EBC}=50^\circ$
or $\angle\text{B}=50^\circ$
But $\angle\text{D}=\angle\text{B}$ (Opposite angles)
$\angle\text{D}=50^\circ$ or $\angle\text{ADC}=50^\circ$
Similarly in $\triangle\text{DCF}$,
$\angle\text{DCF}+\angle\text{CFD}+\angle\text{FDC}=180^\circ$
$\Rightarrow\angle\text{DCF}+90^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{DCF}+140^\circ=180^\circ$
$\Rightarrow\angle\text{DCF}=180^\circ-140^\circ$
$\Rightarrow\angle\text{DCF}=40^\circ$
But $\angle\text{C}+\angle\text{B}=180^\circ$ (Sum of adjacent angles)
$\Rightarrow\angle\text{BCE}+\angle\text{ECF}+\angle\text{DCF}+\angle\text{A}=180^\circ$
$\Rightarrow40^\circ+\angle\text{ECF}+40^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{ECF}+130^\circ=180^\circ$
$\Rightarrow\angle\text{ECF}=180^\circ-130^\circ$
$\Rightarrow\angle\text{ECF}=50^\circ$
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