Question

Given $AD||AC$

$\angle ECF=3\times \angle ACE$ and $\angle ADB={72}^o$.

Find the value of $x$.

Answer 1

$AD\parallel EC$

So,

$\angle ACE=\angle DAC$ (Alternate angles)

$\angle ACE=x$ ............... (1)

$\angle ECF=3\angle ACE$

$\angle ECF=3x$ .............. (2)

$\angle ACD={180}^{\circ }-\angle ACE-\angle ECF$

$\angle ACD={180}^{\circ }-x-3x$

$\angle ACD={180}^{\circ }-4x$ ............ (3)

For triangle $ADC$

$\angle ADB=\angle DAC+\angle ACD$ (Exterior angle is equal to the sum of the opposite two interior angles)

$72=x+180-4x$

$72=180-3x$

$3x=180-72$

$3x=108$

$x={36}^{\circ }$