The angles of a quadrilateral are in A.P. whose common difference is $10^o$. Find the angles.

Given:

The angles of a quadrilateral are in A.P. whose common difference is $10^o$.

To do:

We have to find the angles.

Solution:

Let the angle be $a-d, a, a+d, a+2d$ where $d=10^o$ is the common difference.
We know that the sum of the angles in a quadrilateral is $360^o$.

This implies,

$a-d+a+a+d+a+2d=360^o$

$4a+2d=360^o$

$4a+2(10^o)=360^o$

$4a=360^o-20^o$

$4a=340^o$

$a=\frac{340^o}{4}$

$a=85^o$

Therefore,

$a-d=85^o-10^o=75^o$

$a+d=85^o+10^o=95^o$

$a+2d=85^o+2(10^o)=85^o+20^o=105^o$

The angles are $75^o, 85^o, 95^o$ and $105^o$.

Tutorialspoint

Updated on: 10-Oct-2022

71 Views

Kickstart Your Career

Get certified by completing the course

Get Started