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

Answers (1)

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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 :

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

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

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

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

Posted by

Pankaj Sanodiya

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