Question

In the figure(not drawn to scale), ABCD is a square, ADE is an equilateral triangle and BFE is a straight line, find $y$.

• A. ${90}^o$
• B. ${45}^o$
• C. ${75}^o$
• D. ${15}^o$

Answer 1

Answer: (C)

${75}^o$

Given that $ABCD$ is a square

$AB=BC=CD=DA...................\left(1\right)$

ADE is an equilateral triangle

$AE=ED=DA............................\left(2\right)$

from $\left(1\right)$ and $\left(2\right)$

$AB=AE$

Now in triangle $\Delta ABE$

$\angle BAE=\angle BAD+\angle EAD$

$\angle BAE={90}^{\circ }+{60}^{\circ }$

$\Rightarrow \angle BAE={150}^{\circ }$

$AB=AE$ proved above

$\Rightarrow \angle ABE=\angle AEB$ angles opposite to equal sides are also equal

the sum of all angles of a triangle $={180}^{\circ }$

$\Rightarrow \angle BAE+\angle ABE+\angle AEB={180}^{\circ }$

$\Rightarrow {150}^{\circ }+2\angle AEB={180}^{\circ }$

$\Rightarrow 2\angle AEB={30}^{\circ }$

$\Rightarrow \angle AEB=\angle AEF={15}^{\circ }$

$\angle y=\angle EAF+\angle EAF$ exterior angle property of a triangle $\Delta AEF$

$\angle y={15}^{\circ }+{60}^{\circ }$

$\angle y={75}^{\circ }$