In a cyclic quadrilateral ABCD, it is being given that =(x+y+10)^ , =(y+20)^ , =(x+y-30)^ and =(x+y). then, =?

In a cyclic quadrilateral ABCD, it is being given that =(x+y+10)^…

MCQ

In a cyclic quadrilateral $\text{ABCD,}$ it is being given that $\angle\text{A}=(\text{x+y}+10)^\circ,$
$\angle\text{B}=(\text{y}+20)^\circ,$
$ \angle\text{C}=(\text{x+y}-30)^\circ$ and $\angle\text{D}=(\text{x+y}).$ then, $\angle\text{B}=?$

A
$70^\circ$
✓
$80^\circ$
C
$100^\circ$
D
$110^\circ$

✓

Answer

Correct option: B.

$80^\circ$

Given that in cydic quadrilateral $\text{ABCD,}$
$\angle\text{A}=(\text{x}+\text{y}+10)^\circ, \angle\text{B}=(\text{y+20})^\circ,$
$\angle\text{C}=(\text{x}+\text{y}+30)^\circ$ and $\angle\text{D}=(\text{x}+\text{y})^\circ$
We know that,
Opposite angles of a quadrilateral sum upto $180^\circ .$
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow(\text{y}+20)^\circ+(\text{x+y})^\circ=180^\circ$
$\Rightarrow\text{x}+2\text{y}=160\ .....(\text{i})$
Similarly, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow(\text{x+y}+10)^\circ+(\text{x+y}-30)^\circ=180^\circ$
$\Rightarrow2\text{x}+2\text{y}=200$
$\Rightarrow\text{x+y}=100\ ...(\text{ii})$
Subtracting $(ii)$ from $(i),$ we get
$\text{y}=60$
$\angle\text{B}=(60+20)^\circ=80^\circ$