Question
In the figure, given below, $A B C D$ is a square, and $\triangle B E C$ is an equilateral triangle. Find, the case: $\angle B A E$
✓
Answer
We know that the sides of a square are equal and each angle is of $90^{\circ}$
Three sides of an equilateral triangle are equal and each angle is of 60 .
In fig.,
$\therefore A B C D$ is a square and $\triangle B E C$ is an equilateral triangle,
$ \begin{aligned} & \text { (i) } \angle \mathrm{ABE}=\angle \mathrm{ABC}-\angle \mathrm{CBE} \\ & =90^{\circ}-60^{\circ}=30^{\circ} \end{aligned} $
(ii) In $\triangle \mathrm{ABE}$,
$ \begin{aligned} & \angle \mathrm{ABE}+\angle \mathrm{AEB}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots \ldots . . .(\text { Angles of a triangle }) \\ & \Rightarrow 30^{\circ}+\angle \mathrm{BAE}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots . .(\because \mathrm{AB}=\mathrm{BE}) \\ & \Rightarrow 30^{\circ}+2 \angle \mathrm{BAE}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{BAE}=180^{\circ}-30^{\circ}=150^{\circ} \\ & \Rightarrow \angle \mathrm{BAE}=\frac{150^{\circ}}{2}=75^{\circ} \end{aligned} $
Three sides of an equilateral triangle are equal and each angle is of 60 .
In fig.,
$\therefore A B C D$ is a square and $\triangle B E C$ is an equilateral triangle,
$ \begin{aligned} & \text { (i) } \angle \mathrm{ABE}=\angle \mathrm{ABC}-\angle \mathrm{CBE} \\ & =90^{\circ}-60^{\circ}=30^{\circ} \end{aligned} $
(ii) In $\triangle \mathrm{ABE}$,
$ \begin{aligned} & \angle \mathrm{ABE}+\angle \mathrm{AEB}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots \ldots . . .(\text { Angles of a triangle }) \\ & \Rightarrow 30^{\circ}+\angle \mathrm{BAE}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots . .(\because \mathrm{AB}=\mathrm{BE}) \\ & \Rightarrow 30^{\circ}+2 \angle \mathrm{BAE}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{BAE}=180^{\circ}-30^{\circ}=150^{\circ} \\ & \Rightarrow \angle \mathrm{BAE}=\frac{150^{\circ}}{2}=75^{\circ} \end{aligned} $
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