Question
In Fig.$\angle1=60^\circ$ and $\angle6=120^\circ$ Show that the lines m and n are parallel.
✓
Answer
We have, $\angle5+\angle6=180^\circ$ [linear pair Angles]
$\Rightarrow \angle5+120^\circ=180^\circ$
$\Rightarrow \angle5=180^\circ-120^\circ=60^\circ$ Now, $\angle1=\angle5$
$[\text{Each}=60^\circ]$ But, these are corresponding angles.
Therefore, the lines m and n are parallel.
$\Rightarrow \angle5+120^\circ=180^\circ$
$\Rightarrow \angle5=180^\circ-120^\circ=60^\circ$ Now, $\angle1=\angle5$
$[\text{Each}=60^\circ]$ But, these are corresponding angles.
Therefore, the lines m and n are parallel.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?
If both $( x -2)$ and $\left(x-\frac{1}{2}\right)$ are factors of $p x^2+5 x+r$, Show that $p = r$.
→The sides of a triangular plot are in the ratio of $3 : 5 : 7$ and its perimeter is $300 m.$ Find its area.
→How many lead balls, each of radius $1\ cm$, can be made from a sphere of radius $8\ cm$? $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
→Factorize the following expressions: $32a^3 + 108b^3$
→A traffic signal board, indicating $\text{SCHOOLAHEAD}$ is an equilateral triangle with side a Find the area of the signal board, using Heron's formula. If its perimeter is $180 \ cm$, what will be the area of the signal board?
→How many spheres $12\ cm$ in diameter can be made from a metallic cylinder of diameter $8\ cm$ and height $90\ cm?$
→Rationales the denominator and simplify: $\frac{1+\sqrt2}{3-2\sqrt2}$
→Find the volume and surface area of a sphere whose radius is: $\big(\text{Take}\ \pi=\frac{22}{7}\big). 3.5\ cm$
→In the given figure, two parallel line $l$ and $m$ are intersected by two parallel lines $p$ and $q.$
Show that $\triangle\text{ABC}\cong\triangle\text{CDA}.$
→Show that $\triangle\text{ABC}\cong\triangle\text{CDA}.$