Q. 214.2( 5 Votes )

# In a cyclic quadrilateral ABCD, it is being given that ∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?A. 70°B. 80°C. 100°D. 110°

It is given in the question that,

In cyclic quadrilateral ABCD, we have:

A = (x + y + 10)o

B = (y + 20)o

C = (x + y – 30)o

D = (x + y)o

As ABCD is a cyclic quadrilateral

A + C = 180o and B + D = 180o

Now, A + C = 180o

(x + y + 10)o + (x + y – 30)o = 180o

2x + 2y – 20o = 180o

x + y = 100o (i)

Also, B + D = 180o

(y + 20)o + (x + y)o = 180o

x + 2y + 20o = 180o

x + 2y = 160o (ii)

On subtracting (i) from (ii), we get

y = (160 – 100)o

y = 60o

Putting the value of y in (i), we get

x + 60o = 100o

x = 100o – 60o

x = 40o

B = (y + 20)o

B = 60o + 20o = 80o

Hence, option B is correct

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