Question

In Fig., AB || DC. If x = $\frac{4}{3}$ y and y = $\frac{3}{8}$ z, find $\angle$BCD, $\angle$ABC and $\angle$BAD?
[3 MARKS]

Answer 1

Angles : 1 Mark each

Since AB || DC and transversal BD intersects them at B and D respectively.

$\angle$ABD = $\angle$BDC [Alternate angles]

And $\angle$CBD = $\angle$ADB

$\angle$BDC = $x^{\circ }$ and y = ${36}^{\circ }$

$\angle$ABD =$x^{\circ }$ and $\angle$ADB = ${36}^{\circ }$ (Given)

But, it is given that,

$x=\frac{4}{3}yandy=\frac{3}{8}z$

$x=\frac{4}{3}\times 36and36=\frac{3}{8}z$

$x=48andz=\frac{36\times 8}{3}=96$

Now, in BAD, we have

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

(\angle\)BAD + ${36}^{\circ }$ + $x^{\circ }$ = ${180}^{\circ }$

$\angle$BAD + ${36}^{\circ }$ + ${48}^{\circ }$ = ${180}^{\circ }$

$\angle$BAD = ${96}^{\circ }$

Thus, $\angle$BCD = $z^{\circ }$ = ${96}^{\circ }$,

$\angle$ABC = $x^{\circ }$ + $y^{\circ }$= ${48}^{\circ }$ + ${36}^{\circ }$ =${84}^{\circ }$ and $\angle$BAD = ${96}^{\circ }$