Question

The angles of a quadrilateral are in A. P. with common difference ${20}^{\circ }$. Find its angles.

Answer 1

The angles of quadrilateral are in A.P. with common difference, d = ${20}^{\circ }$

Let the first angle be 'a'

Then,

Second angle = a + d = a + ${20}^{\circ }$

Third angle = a + 2d = a + ${40}^{\circ }$

Fourth angle = a + 3d = a + ${60}^{\circ }$

We know that the sum of all angles of a quadrilateral is ${360}^{\circ }$

$\Rightarrow$ a + a+${20}^{\circ }$ + a+${40}^{\circ }$ + a+${60}^{\circ }$ = ${360}^{\circ }$

$\Rightarrow$ 4a + ${120}^{\circ }$ = ${360}^{\circ }$

$\Rightarrow$ 4a = ${360}^{\circ }$ - ${120}^{\circ }$

$\Rightarrow$ 4a = ${240}^{\circ }$

$\Rightarrow$ a = $\frac{{240}^{\circ }}{4}$

$\Rightarrow$ a = ${60}^{\circ }$

First angle = ${60}^{\circ }$

Second angle = ${60}^{\circ }$ + ${20}^{\circ }$ = ${80}^{\circ }$

Third angle = ${60}^{\circ }$ + ${40}^{\circ }$ = ${100}^{\circ }$

Fourth angle = ${60}^{\circ }$ + ${60}^{\circ }$ = ${120}^{\circ }$

Hence, the angles of given quadrilateral are ${60}^{\circ }$, ${80}^{\circ }$, ${100}^{\circ }$, ${120}^{\circ }$