$ABCD$ is a square (vertices in order $A$, $B$, $C$, $D$) and $BCE$ is an equilateral triangle drawn on side $BC$, with vertex $E$ lying outside the square. What is the measure of angle $DEC$?

Question

Adhir Mishra has three children: Urmila, Raghu and Sumit. Sumit married Roma, the eldest daughter of Mr. and Mrs. Mohan. The Mohans married their youngest daughter to the eldest son of Mr. and Mrs. Sharma and they had two children Sandeep and Shaifali. The Mohans have two more children, Roshan and Bimla, both elder to Sheila. Sohan and Shivendar are sons of Sumit and Roma. Leela is the daughter of Sandeep.

Answer 1

Here is a second way, using coordinate geometry instead of the angle-chasing argument.

Place the square. Let the square have side 1, with $A=(0,1)$, $B=(1,1)$, $C=(1,0)$, $D=(0,0)$.
Locate E. $BCE$ is equilateral on side $BC$ (the segment from $(1,1)$ to $(1,0)$), built outward (to the right). Its apex is at $E = (1+\tfrac{\sqrt3}{2}, 0.5) \approx (1.866, 0.5)$.
Form the vectors from E. $\vec{ED} = D-E \approx (-1.866,-0.5)$ and $\vec{EC} = C-E \approx (-0.866,-0.5)$.
Apply the dot product formula. $\vec{ED}\cdot\vec{EC} \approx (-1.866)(-0.866)+(-0.5)(-0.5) \approx 1.866$. $|\vec{ED}| \approx 1.932$, $|\vec{EC}| = 1$. So $\cos(\angle DEC) \approx 1.866/1.932 \approx 0.9658$.
Solve for the angle. $\angle DEC = \cos^{-1}(0.9658) \approx 15^\circ$.

The coordinate method gives the same result as the angle-chasing method, confirming option A. \[ \boxed{\angle DEC = 15^\circ} \]

Answer 2

Here is a second way, using coordinate geometry instead of the angle-chasing argument.

Place the square. Let the square have side 1, with $A=(0,1)$, $B=(1,1)$, $C=(1,0)$, $D=(0,0)$.
Locate E. $BCE$ is equilateral on side $BC$ (the segment from $(1,1)$ to $(1,0)$), built outward (to the right). Its apex is at $E = (1+\tfrac{\sqrt3}{2}, 0.5) \approx (1.866, 0.5)$.
Form the vectors from E. $\vec{ED} = D-E \approx (-1.866,-0.5)$ and $\vec{EC} = C-E \approx (-0.866,-0.5)$.
Apply the dot product formula. $\vec{ED}\cdot\vec{EC} \approx (-1.866)(-0.866)+(-0.5)(-0.5) \approx 1.866$. $|\vec{ED}| \approx 1.932$, $|\vec{EC}| = 1$. So $\cos(\angle DEC) \approx 1.866/1.932 \approx 0.9658$.
Solve for the angle. $\angle DEC = \cos^{-1}(0.9658) \approx 15^\circ$.

The coordinate method gives the same result as the angle-chasing method, confirming option A. \[ \boxed{\angle DEC = 15^\circ} \]

\(ABCD\) is a square (vertices in order \(A\), \(B\), \(C\), \(D\)) and \(BCE\) is an equilateral triangle drawn on side \(BC\), with vertex \(E\) lying outside the square. What is the measure of angle \(DEC\)?

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The Correct Option is A

Solution and Explanation

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