Question

In quadrilateral ABCD, $\angle B={90}^0,\angle C-\angle D={60}^0$ and $\angle A-\angle C-\angle D={10}^0$, Find $\angle A,\angle C,\angle D$.

• A. $140°,85°$ and $35°$
• B. $140°,95°$ and $45°$
• C. $130°,95°$ and $35°$
• D. $140°,95°$ and $35°$

Answer 1

The correct option is D $140°,95°$ and $35°$



$\angle B={90}^{\circ }$
$\angle C-\angle D={60}^{\circ }$
$\angle A-\angle C-\angle D={10}^{\circ }$.........(1)
$\angle A+\angle C+\angle D={360}^{\circ }-{90}^{\circ }={270}^{\circ }$ (from Angle sum property of quadrilateral)

On adding,
$\angle A+\angle C+\angle D={270}^{\circ }$
$\angle A-\angle C-\angle D={10}^{\circ }$
___________________
$2\angle A={280}^{\circ }$
$\angle A=140°$
$\Rightarrow$ from (1)
$\angle A={140}^{\circ },\angle A-(\angle C+\angle D)={10}^{\circ }$
${140}^{\circ }-(\angle C+\angle D)={10}^{\circ }$
$\angle C+\angle D={130}^{\circ }$ ................(2)
If $\angle C-\angle D={60}^{\circ }$ & $\angle C+\angle D={130}^{\circ }$
$2\angle C={190}^{\circ },\angle C={95}^{\circ }$
$\angle D=130°-95°$ - from (2)
$\angle D={35}^{\circ }$
So, the correct answer is option (d).