Question

In the given figure, ABCD is a parallelogram, $\angle ADE={50}^{o}and\angle ACE=\angle BED={90}^o$The value of $\angle EAC+\angle ABC-2\angle DAC$

• A. ${10}^0$
• B. ${15}^0$
• C. ${18}^0$
• D. ${23.5}^0$

Answer 1

The correct option is A ${10}^0$

Given that,

$\angle BED={90}^0$ , Then $\angle AEC={45}^0$

$\angle EAC+\angle EAB={90}^0$ $\therefore BAEC$ is a rectangle.

Then,

$\angle EAC=EAB={45}^0$ and $ABC={45}^0$

Given that, $\angle ADE={50}^0$

In $\Delta ACD$,

$\angle ACD+\angle ADC+\angle DAC={180}^0$

${90}^0+{50}^0+\angle DAC={180}^0$

$\angle DAC={180}^0-{140}^0$

$\angle DAC={40}^0$

Then,

$\angle EAC+\angle ABC-2\angle DAC$

$={45}^0+{45}^0-2\times {40}^0$

$={90}^0-{80}^0={10}^0$

Hence, this is the answer.