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ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Q8.    ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

                    

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As we know that in a quadrilateral the sum of opposite angles is 180 degree.

So, From Here,

4y+20-4x=180

\Rightarrow 4y-4x=160

\Rightarrow y-x=40............(1)

Also,

3y-5-7x+5=180

\Rightarrow 3y-7x=180........(2)

Multiplying (1) by 3 we get,

\Rightarrow 3y-3x=120........(3)

Now,

Subtracting, (2) from (3) we get,

4x=-60

\Rightarrow x=-15

Substituting this value in (1) we get,

y-(-15)=40

\Rightarrow y=40-15

\Rightarrow y=25

Hence four angles of a quadrilateral are :

\angle A =4y+20=4(25)+20=100+20=120^0

\angle B =3y-5=3(25)-5=75-5=70^0

\angle C =-4x=-4(-15)=60^0

\angle D =-7x+5=-7(-15)+5=105+5=110^0

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