Question

In the given figure$\text{AB ll DC. If}x=\frac{4}{3}y\text{and}y=\frac{3}{8}z$ , find the values of x,y and z.

Answer 1

$AB\parallel DC;$ $x=\frac{4}{3}y$ and $y=\frac{3}{8}z$

$AB\parallel DC,$

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

$\therefore$ $\angle ABD=\angle BDC=x$

In $△BDC,$

$x+y+z={180}^o$

$\Rightarrow$ $\frac{4}{3}y+\frac{3}{8}z+z={180}^o$

First we will solve the $\frac{4}{3}y.$

Since, $y=\frac{3}{8}z$

$\therefore$ $\frac{4}{3}\times \frac{3}{8}z$

$\Rightarrow$ $\frac{12}{24}z$

$\Rightarrow$ $\frac{1}{2}z$

Now, $\frac{1}{2}z+\frac{3}{8}z+z={180}^o$

$\Rightarrow$ $\frac{4z+3z+8z}{8}={180}^o.$

$\Rightarrow$ $\frac{15z}{8}={180}^o.$

$\Rightarrow$ $15z={180}^o\times 8$

$\Rightarrow$ $z=\frac{1440}{15}$

$\Rightarrow$ $z={96}^o.$

Now, $y=\frac{3}{8}z$

$\Rightarrow$ $y=\frac{3\times {96}^o}{8}$

$\Rightarrow$ $y=\frac{288}{8}$

$\Rightarrow$ $y={36}^o.$

Now, $x=\frac{4}{3}y$

$\Rightarrow$ $x=\frac{4\times {36}^o}{3}$

$\Rightarrow$ $x=\frac{144}{3}$

$\Rightarrow$ $x={48}^o.$

$\therefore$ $x={48}^o,y={36}^o$ and $z={96}^o.$