Question

9.In the given figure p is a transversal of the lines m
and n and <1 = 60° and <2 = 2/3 of a right angle.Prove that m || n.
I request you please help me ​

Answer 1

$\bf \Huge \implies \: Given :$

$\bf \large \longrightarrow \angle1 = 60 \degree$

$\bf \large \rightarrow \angle \: 2 = \frac{2}{3} \: \: of \: \: a \: \: right \: \: angle$

$\bf \large \leadsto \: = \frac{2}{3} \times 90 \degree \: = 60 \degree$

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$\bf \Huge \implies \: To \: \: Prove:$

Prove that m || n.

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$\bf \Huge \implies \: Solution :$

$\bf \small \leadsto \angle \: 3 = \angle \: 2 \:  (vertically \: \: opposite \: \: angles)$

$\bf \large \iff \angle \: 3 = \: 60 \degree$

$\bf \small \iff \: Since, ∠1=∠3 \: corresponding \: \: angles$

So, corresponding angles are equal .

$\:$

Then m ∥ n.

$\: \:$

Hence , Proved

$\: \:$

I am Ayan

Also from Delhi