Question

In figure, $m||n$ and $p||q$. If $\angle 1={75}^o$, prove that $\angle 2=\angle 1+\frac{1}{3}$ of a right angle.

Answer 1

Given: $m\parallel n$ and $p\parallel q$

$\Rightarrow \angle 3=\angle 1$ (corresponding angles)

$\Rightarrow \angle 3={75}^{\circ }$

Since $m\parallel n$,

Now , $\angle 2+\angle 3={180}^{\circ }$ (cointerior angles)

$\Rightarrow \angle 2+{75}^{\circ }={180}^{\circ }$

$\Rightarrow \angle 2={105}^{\circ }$

RHS $=\angle 1+\frac{1}{3}\times {90}^{\circ }$

$={75}^{\circ }+{30}^{\circ }$

$={105}^{\circ }$

$=\angle 2$

$=LHS$

Hence proved