Mathematics

In the given figure, AB || DC, $\angle BAD = 90^0$, $\angle CBD= 28^0$ and $\angle BCE=65^0$. Then $\angle ABD=?$

$37^0$

SOLUTION
$\angle BDC+\angle DBC={ 65 }^{ \circ }$

$(\because Sum\quad of\quad two\quad opposit\quad interior\quad angles\quad of\quad a\quad triangle=exterior\quad angle)$

$\implies\quad \angle BDC+{ 28 }^{ \circ }={ 65 }^{ \circ }$

$\implies\quad \angle BDC={ 37 }^{ \circ }$

$\angle BAD+\angle ADC={ 180 }^{ \circ }\quad (sum\quad of\quad co-interior\quad angle={ 180 }^{ \circ })$

$\implies\quad \angle ADC={ 90 }^{ \circ }$

$\therefore \angle ADB=\angle ADC-\angle BDC={ 90 }^{ \circ }-{ 37 }^{ \circ }={ 53 }^{ \circ }$

$In\triangle ABD:$

$\angle ADB+\angle ABD+\angle DAB={ 180 }^{ \circ }$

$\implies\quad { 53 }^{ \circ }+\angle ABD+{ 90 }^{ \circ }={ 180 }^{ \circ }$

$\implies\quad \angle ABD={ 180 }^{ \circ }-{ 143 }^{ \circ }={ 37 }^{ \circ }$

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Single Correct Medium Published on 09th 09, 2020
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