Question

In cyclic $\square ABCD,m\angle A+m\angle B=150$ and $m\angle B+m\angle C={190}^o$. Find $m\angle D$

Answer 1

$A+B=150⟶\left(1\right)$

$\Rightarrow C+D=360-150=210⟶\left(2\right)$

$B+C=190⟶\left(3\right)$

$\Rightarrow A+D=360-190=170⟶\left(4\right)$

$A+C=180⟶\left(5\right)$

From $\left(1\right)\&\left(2\right)$

$\Rightarrow A-C=-40$

$C-A=40⟶\left(6\right)$

From $\left(5\right)\&\left(6\right)$

$\Rightarrow C=110$

$A=70$

$D=100$

$B=80$

Hence, solved.