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

MCQ

In a cyclic quadrilateral 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
70º
✓
80º
C
100º
D
110º

✓

Answer

Correct option: B.

80º

Given that in cydic quadrilateral 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°.
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow(\text{y}+20)^\circ+(\text{x+y})^\circ=180^\circ$
$\Rightarrow\text{x}+2\text{y}=160\ .....(\text{i})$
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
$\text{y}=60$
$\angle\text{B}=(60+20)^\circ=80^\circ$

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