# The angles of a quadrilateral are in A.P.

Question:

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

Solution:

Let the angles be $(A)^{\circ},(A+d)^{\circ},(A+2 d)^{\circ},(A+3 d)^{\circ}$.

Here, d  = 10

So, $(A)^{\circ},(A+10)^{\circ},(A+20)^{\circ},(A+30)^{\circ}$ are the angles of a quadrilateral whose sum is $360^{\circ}$.

$\therefore(A)^{\circ},(A+10)^{\circ},(A+20)^{\circ},(A+30)^{\circ}=360^{\circ}$

$\Rightarrow 4 A=360-60$

$\Rightarrow A=\frac{300}{4}=75^{\circ}$

The angles are as follows:

$75^{\circ},(75+10)^{\circ},(75+20)^{\circ},(75+30)^{\circ}$, i. e. $75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}$