Question

# The angles of a quadrilateral are in AP, whose common difference is ${{10}^{\circ }}$. Find the angles of the quadrilateral.

Hint: An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant. We know that the sum of interior angles of any quadrilateral is ${{360}^{\circ }}$ .

Here, we have a quadrilateral whose angles are in A.P with common difference ${{10}^{\circ }}$ and need to determine all angles of it.
Now, we can define any A.P as a sequence of numbers such that the difference between consecutive terms is constant. Successive difference is termed as common difference and whole sequence is called as an Arithmetic progression.
So, let us suppose angles of quadrilateral are
a-3d, a-d, a+d, a+3d
Where the common difference between terms is 2d i.e., (a-d – a+3d = 2d).
As, it is already given that the common difference in angles is ${{10}^{\circ }}$.
Hence, 2d should be equal to ${{10}^{\circ }}.$
2d = 10 or
$d={{5}^{\circ }}.............\left( i \right)$
Hence, now angles can be given as
a-3(5), a-5, a+5, a+3(5) or
a-15, a-5, a+5, a+15
Now, we can use the property of any quadrilateral that is given as “Sum of all interior angles of any quadrilateral is ${{360}^{\circ }}$”.
Hence,
\begin{align} & a-15+a-5+a+5+a+15={{360}^{\circ }} \\ & 4a={{360}^{\circ }} \\ & a={{90}^{\circ }}...............(ii) \\ \end{align}
So, we can evaluate all angles of quadrilateral as
a-3d, a-d, a+d, a+3d
Putting values of a and d, we get
90-3(5), 90-5, 90+5, 90+5(3)
90-15, 85, 95, 90+15
or
${{75}^{\circ }},{{85}^{\circ }},{{95}^{\circ }},{{105}^{\circ }}$
Hence, angles of quadrilateral is given as ${{75}^{\circ }},{{85}^{\circ }},{{95}^{\circ }},{{105}^{\circ }}.$

Note: One can suppose angles of quadrilateral as a, a+d, a+2d, a+3d which is the general representation of an A.P. And use $d={{10}^{\circ }}$ and apply the same property for quadrilateral used in the solution. We have used another representation of angles for simplicity as if we add all terms of AP taken in solution, we get 4a. So, we can get the value of ‘a’ directly.One can equate summation of all angles to ${{180}^{\circ }}$ by mistake as we use for sum of angles in a triangle. Hence, take care of it as well. Terms used in the solution have the common difference ‘2d’. So, don’t take the value of ‘d’ as ${{10}^{\circ }}$ as given in the problem. ‘2d’ should be equal to ${{10}^{\circ }}.$