Question

The angles of a quadrilateral are in A.P. and the greatest angle is 120 degrees. Express the angles in radians.

Answer 1

Hint: We already know that a quadrilateral is a polygon with four edges and four vertices or corners. First of all we will suppose the angles are in A.P. with some variable as ‘a’ first term and common difference ‘d’ and then we will use the properties that the sum of internal angles of a quadrilateral is equal to 360 degrees.

Complete step-by-step answer:
Let us suppose the first term and the common difference to be ‘a’ and ‘d’ of the angles of a quadrilateral which are in A.P.
So the angles are a, a+d, a+2d and a+3d.
We know that ${{T}_{n}}$ (nth term) $=a+(n-1)d$ for an A.P.
Now we have been given that the greatest angle is ${{120}^{\circ }}$.
$\Rightarrow a+3d={{120}^{\circ }}.....(1)$
We know that the sum of all interior angles of a quadrilateral is equal to ${{360}^{\circ }}$.
On using the equation (1), we get as follows:
$\Rightarrow a+a+d+a+2d+{{120}^{\circ }}={{360}^{\circ }}$
On taking ${{120}^{\circ }}$ to the right hand side of equality we get as follows:
$\Rightarrow a+a+d+a+2d={{360}^{\circ }}-{{120}^{\circ }}$
On adding the similar terms to the left hand side of the equation, we get as follows:
$\Rightarrow 3a+3d={{240}^{\circ }}$
On taking 3 as common from the equation we get as follows:
$\Rightarrow 3\left( a+d \right)=3({{80}^{\circ }})$
On dividing the equation by 3 to both sides, we get as follows:

& \Rightarrow \dfrac{3}{3}\left( a+d \right)=\dfrac{3}{3}({{80}^{\circ }}) \\\ & \Rightarrow a+d={{80}^{\circ }}....(2) \\\ \end{aligned}$$ On subtracting equation (2) from (1) we get as follows: $$\begin{aligned} & \Rightarrow \left( a+3d \right)-\left( a+d \right)={{120}^{\circ }}-{{80}^{\circ }} \\\ & \Rightarrow a+3d-a-d={{40}^{\circ }} \\\ & \Rightarrow 2d={{40}^{\circ }} \\\ & \Rightarrow d={{20}^{\circ }} \\\ \end{aligned}$$ Hence the common difference is $${{20}^{\circ }}$$. Now we will substitute the value of d = $${{20}^{\circ }}$$ in equation (2) and then we get as follows: $$\begin{aligned} & \Rightarrow a+{{20}^{\circ }}={{80}^{\circ }} \\\ & \Rightarrow a={{80}^{\circ }}-{{20}^{\circ }} \\\ & \Rightarrow a={{60}^{\circ }} \\\ \end{aligned}$$ Hence the first term is $${{60}^{\circ }}$$. $$\Rightarrow a={{60}^{\circ }},\left( a+d \right)={{60}^{\circ }}+{{20}^{\circ }}={{80}^{\circ }},\left( a+2d \right)={{60}^{\circ }}+2\times {{20}^{\circ }}={{100}^{\circ }},\left( a+3d \right)={{120}^{\circ }}$$ Now we know that 1 degree $$=\dfrac{\pi }{180}$$ radians $$\begin{aligned} & \Rightarrow {{60}^{\circ }}=\dfrac{\pi }{180}\times {{60}^{\circ }}=\dfrac{\pi }{3}\text{radians} \\\ & \Rightarrow {{80}^{\circ }}=\dfrac{\pi }{180}\times {{80}^{\circ }}=\dfrac{4\pi }{9}\text{radians} \\\ & \Rightarrow {{100}^{\circ }}=\dfrac{\pi }{180}\times {{100}^{\circ }}=\dfrac{5\pi }{9}\text{radians} \\\ & \Rightarrow {{120}^{\circ }}=\dfrac{\pi }{180}\times {{120}^{\circ }}=\dfrac{2\pi }{3}\text{radians} \\\ \end{aligned}$$ Therefore the angles of quadrilateral in radians are $$\dfrac{\pi }{3}\text{radians},\dfrac{4\pi }{9}\text{radians},\dfrac{5\pi }{9}\text{radians},\dfrac{2\pi }{3}\text{radians}$$. Note: Be careful while converting degrees into radians and use the formula correctly. Don’t use 1 degree $$=\dfrac{180}{\pi }$$ radians in a hurry. Also, be careful of the sign while solving the equation. Also, remember that the sum of internal angles of any four sided polygon is equal to $${{360}^{\circ }}$$. We can assume angles as a-3d, a-d, a+d and a+3d where 2d will be the common difference.