Question

In figure, AB ll DC, if x = $\left(\frac{4}{3}\right)yandy=\left(\frac{3}{8}\right)z.Find\angle BAD.$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\text{Since AB ll DC and transversal BD intersects them at B and D respectively}.$
$Therefore,\angle ABD=\angle BDC\left(Alternateangles\right)$
$And\angle CBD=\angle ADB$
$\angle BDC=x^{0}andy={36}^{\circ }$
$\angle ABD=x^{0}and\angle ADB={36}^{\circ }\left(Given\right)$
$Itisgiventhat,x=\left(\frac{4}{3}\right)yandy=\left(\frac{3}{8}\right)z$

$x=\frac{4}{3}\times {36}^{\circ }and{36}^{\circ }=\left(\frac{3}{8}\right)z$

$x={48}^{\circ }andz={36}^{\circ }\times \left(\frac{8}{3}\right)={96}^{\circ }$
$Now,in\Delta BAD,wehave$
$\angle BAD+\angle ADB+\angle ABD={180}^{\circ }$
$\angle BAD+{36}^{\circ }+x^0={180}^{\circ }$
$\angle BAD={96}^{\circ }$
$Thus,\angle BCD=z^0={96}^{\circ }$
$\angle ABC=x^0+y^0={48}^{\circ }+{36}^{\circ }={84}^{\circ }and\angle BAD={96}^{\circ }$