Question

In quadrilateral ABCD $\angle$B$={90}^o$, $\angle$C $-\angle$D$={60}^o$ and $\angle$A $-\angle$C$-\angle$D $={10}^o$. Find $\angle$A, $\angle$C and $\angle$D.

Answer 1

$\begin{array}{c}Thisfigureisaquadrilateral\\ Then,sumofallanglesis{360}^0\\ \angle A+\angle B+\angle C+\angle D={360}^0\to \left(i\right)\\ \angle B={90}^0\left(given\right)\\ \angle C-\angle D={60}^0\to \left(ii\right)\\ \angle A-\angle C-\angle D={10}^0\to \left(iii\right)\\ Find\angle A,\angle C\&\angle D\\ Addingequation\left(i\right)and\left(iii\right)weget\angle A\\ \Rightarrow 2\angle A+\angle {370}^0\\ \Rightarrow 2\angle A={370}^0-{90}^0\\ \angle A=\frac{280}{2}={140}^0\\ Againaddingequation\left(i\right)and\left(ii\right)\\ \angle A+\angle B+2\angle C={420}^0\\ {140}^0+{90}^0+2\angle C={420}^0\\ 2\angle C={190}^0\\ \therefore \angle C={95}^0\\ fromequation\left(ii\right)\\ \angle C={60}^0=\angle D\\ {95}^0-{60}^0=\angle D\\ \therefore \angle D={35}^{0}Ans.\end{array}$