Question

ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

Answer 1

Use the basic property of cyclic quadrilateral which is given as “sum of opposite angles of any cyclic quadrilateral is $180{}^\circ$. Solve equations to get ‘x’ and ‘y’.

Complete step-by-step answer:
Cyclic quadrilateral ABCD is given as:-

We know the property of a cyclic quadrilateral that the sum of opposite angles is $180{}^\circ$. It means the sum of $\angle B$ and $\angle D$ is $180{}^\circ$, similarly, the sum of $\angle A$ and $\angle C$ is $180{}^\circ$. Hence from the diagram, we can write two equations as

$\angle A+\angle C=180{}^\circ$

$4y+20{}^\circ +(-4x)=180{}^\circ$

Or

$-4x+4y=180-20$

$\Rightarrow -4x+4y=160$

On dividing the whole equation by 4, we get

$-x+y=40\ldots \ldots (1)$

Now, another equation can be written as

$\angle B+\angle D=180{}^\circ$

$\Rightarrow (3y-5)+(-7x+5)=180{}^\circ$

$\Rightarrow -7x+3y-5+5=180{}^\circ$

$\Rightarrow -7x+3y=180{}^\circ \ldots \ldots (2)$

Now, we can use the substitution method to solve equations (1) and (2).

We have $-x+y=40$ from equation (1).

So, value of y can be written in terms of x as,

$y=x+40\ldots \ldots (3)$

We can put value of ‘y’ from equation (3) in equation (2) i.e. $-7x+3y=180$;

Hence, we get

$-7x+3\left( x+40 \right)=180$

$\Rightarrow -7x+3x+120=180$

$\Rightarrow -4x=60$

$\Rightarrow x=-\dfrac{60}{4}=-15$

So, we get $x=-15$

Now, we can calculate value of ‘y’ by substituting $x=-15$ in equation (3), we get value of y as

$y=-15+40=25$

Hence we get,

$x=-15,y=25$

Now, angles A, B, C, D can be given as

$\angle A=4y+20=4\left( 25 \right)+20=120{}^\circ$

$\angle B=3y-5=3\left( 25 \right)-5=70{}^\circ$

$\angle C=-4x=-4\left( -15 \right)=60{}^\circ$

$\angle D=-7x+5=-7\left( -15 \right)+5=110{}^\circ$

Hence, we get angles of given cyclic quadrilateral be $120{}^\circ ,70{}^\circ ,60{}^\circ ,110{}^\circ$.

Note: One can prove the given property of cyclic quadrilateral i.e. sum of opposite angles in cyclic quadrilateral is $180{}^\circ$ in following way:-

As we know that angles in the same segment are equal.
Hence, $\angle 1=\angle 6,\angle 5=\angle 8,\angle 2=\angle 4,\angle 7=\angle 3$.
Now, we know that the sum of interior angles of a quadrilateral is $360{}^\circ$.
Use $\angle A+\angle B+\angle C+\angle D=360{}^\circ$ and we have equations to prove $\angle A+\angle C=180{}^\circ$ and $\angle B+\angle D=180{}^\circ$.
One cannot get angles by using the property of quadrilateral that the sum of all interior angles is $360{}^\circ$.