Mymensingh higher secondary 2013#8

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samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm
Mymensingh higher secondary 2013#8

Unread post by samiul_samin » Mon Feb 18, 2019 12:58 am

$ABCD$ is a trapezium with $AB\parallel CD$ and $\angle{ADC}=90^{\circ}$ . $E$ is a point on $CD$ that $BE\perp CD$. $F$ is a point on the extension of $CB$ that $DF\perp CF$.$DF$ and $EB$ intersects at the point $K$.$\angle {EAB}=47^{\circ}$,then$\angle {KCE}=$?

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: Mymensingh higher secondary 2013#8

Unread post by samiul_samin » Mon Feb 18, 2019 8:53 am

Hint
Draw a figure
Answer
$43^{\circ}$

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