Question

If ABCD is a square and DCE is an equilateral triangle in the given figure, then $\angle DAE$ is equal to

• A. ${45}^0$
• B. ${30}^0$
• C. ${15}^0$
• D. $22{\frac{1}{2}}^0$

Answer 1

Answer: (C)

${15}^0$

Given, ABCD is a square. DCE is an equilateral triangle.
ABCD is a square,
$AB=BC=CD=DA$
DCE is an equilateral triangle,
$DC=DC=CE$
Hence, $AB=BC=CD=DA=CE=DC$
Now, In $△ADE$,
$AD=DE$
Thus, $\angle AED=\angle DAE=x$
$\angle ADE=\angle ADC+\angle EDC$
$\angle ADE=90+60$
$\angle ADE={150}^{\circ }$
Sum of angles of triangle ADE = 180
$\angle ADE+\angle AED+\angle DAE=180$
$150+x+x=180$
$x={15}^{\circ }$
Hence, $\angle DAE={15}^{\circ }$