In a cyclic quadrilateral ABCD, it is being given that $\angle$A = (x + y + 10)°, $\angle$B = (y + 20)°, $\angle$C = (x + y – 30)° and $\angle$D = (x + y)°. Then, $\angle$B = ?

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$80^\circ$ Explanation: In a cyclic quadrilateral, opposite angles are supplementary and sum to $180^\circ$. So, A + C = $180^\circ$: (x + y + 10) + (x + y – 30) = $180^\circ$ $\Rightarrow$ 2(x + y) – 20 = $180^\circ$ $\Rightarrow$ x + y = $100^\circ$ Thus, D = x + y = $100^\circ$ and B + D = $180^\circ$ $\Rightarrow$ B = $180^\circ$ – $100^\circ$ = $80^\circ$

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