Question

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

A) \[{45^ \circ }\]

B) \[{30^ \circ }\]

C) \[{15^ \circ }\]

D) \[22{\dfrac{1}{2}^ \circ }\]

Answer 1

Hint: Angle sum property: The sum of all the angles of any triangle is \[{180^ \circ }.\]

Each angle of equilateral triangle is \[{60^ \circ }.\]

Each angle of square is \[{90^ \circ }.\]

The angles of the opposite equal side of the isosceles triangle are equal.

In the following problem we are going to use the above properties to find\[\angle DEA\] which is further divided by two to find the required angle.

Complete step by step answer:
Here, \[ABCD\] is a square and \[DCE\] is an equilateral triangle.

We know that, each angle of a square is always \[{90^ \circ }\] and each angle of an equilateral triangle is always \[{60^ \circ }\] since the sides of the square are equal.
So, let us consider the following angles to solve the problem\[\angle ADC = {90^ \circ }\] and \[\angle EDC = {60^ \circ }\]
Therefore,
From the given figure we can find the following equality,
\[\angle EDA = \angle ADC + \angle EDC\]
Here let us substitute the values of \[\angle ADC = {90^ \circ }\] and \[\angle EDC = {60^ \circ }\] in the above expression therefore we get,
\[\angle EDA = {90^ \circ } + {60^ \circ } = {150^ \circ }\]
The angle of EDA is found to be \[{150^ \circ }\]
This fact is further used to find the angle of ADE
It is a well-known fact that, the sum of the angles of triangle is \[{180^ \circ }.\]
From the triangle \[ADE\] we get,
\[\angle DEA = \angle DAC + \angle DEC = {180^ \circ } - {150^ \circ } = {30^ \circ }\]
Here,
\[DC = DE = AD\]
So, triangle\[ADE\] is an isosceles triangle. So, \[\angle DAE = \angle DEA\]
Therefore, the angle is found by dividing it into half,
\[\angle DAE = \dfrac{{{{30}^ \circ }}}{2} = {15^ \circ }\]

Hence, the correct option is (C) \[{15^ \circ }\].

Note:
Triangle \[ADE\] is an isosceles triangle it will lead to a fact that \[\angle DAE = \angle DEA\] as \[\angle DAE = {15^ \circ }\] we will lead to the fact that \[\angle DAE = \angle DEA = {15^ \circ }\]. Similarly, here we can find the values of the angles of the triangle \[BCE\] by the same method. The properties of triangles play a major role in finding the solution.