Answer
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Hint: 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:-
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 $.
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 $.
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