Question: If ABCD be a cyclic quadrilateral in which $ \angle A=4x{}^\circ , $ $ \angle B=7x{}^\circ , $ $ \angle C=5y{}^\circ , $ $ \angle D=y{}^\circ , $ then $ x:y $ is
Options:
A) 3 : 4
B) 4 : 3
C) 5 : 4
D) 4 : 5
Show Answer
Answer:
Correct Answer: B
Solution:
- In a cyclic quadrilateral.
Sum of opposite angle $ =180{}^\circ $
So, $ 4x+5y=7x+y=180{}^\circ $
$ \Rightarrow $ $ 4x+5y=7x+y $
$ \Rightarrow $ $ 5y-y=7x-4x $
$ \Rightarrow $ $ 4y=3x $
$ \Rightarrow $ $ \frac{4}{3}=\frac{x}{y} $
$ \therefore $ $ x:y=4:3 $