Question

In a cyclic quadrilateral ABCD,it is being given that $\angle A=(x+y+10{)}^{\circ },\angle B=(y+20{)}^{\circ },\angle C=(x+y-30{)}^{\circ },and\angle D=(x+y{)}^{\circ },Then\angle B=?\left(a\right){70}^{\circ }\left(b\right){80}^{\circ }\left(c\right){100}^{\circ }\left(d\right){110}^{\circ }$

Answer 1

here
A+C=180
B+D=180
x+y+10+x+y-30=180
2x+2y-20=180
x+y=100
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y+20+x+y=180
x+2y=160------------1
x+y=100-------------2

1-2
y=60

so angle B=y+20=80

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