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: $\angle A B E$
✓
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 .
Therefore, In fig., $A B C D$ 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}=150^{\circ} \end{aligned} $
(ii) But in $\triangle \mathrm{ABE}$
$ \begin{aligned} & \angle \mathrm{ABE}+\angle \mathrm{BEA}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots \ldots . .(\text { Angles of a triangle }) \\ & \Rightarrow 150^{\circ}+\angle \mathrm{BAE}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots . .(\because \mathrm{AB}=\mathrm{BE}) \\ & \Rightarrow 150^{\circ}+2 \angle \mathrm{BAE}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{BAE}=180^{\circ}-150^{\circ}=30^{\circ} \\ & \therefore \angle \mathrm{BAE}=\frac{30^{\circ}}{2}=15^{\circ} \end{aligned} $
Three sides of an equilateral triangle are equal and each angle is of 60 .
Therefore, In fig., $A B C D$ 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}=150^{\circ} \end{aligned} $
(ii) But in $\triangle \mathrm{ABE}$
$ \begin{aligned} & \angle \mathrm{ABE}+\angle \mathrm{BEA}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots \ldots . .(\text { Angles of a triangle }) \\ & \Rightarrow 150^{\circ}+\angle \mathrm{BAE}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots . .(\because \mathrm{AB}=\mathrm{BE}) \\ & \Rightarrow 150^{\circ}+2 \angle \mathrm{BAE}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{BAE}=180^{\circ}-150^{\circ}=30^{\circ} \\ & \therefore \angle \mathrm{BAE}=\frac{30^{\circ}}{2}=15^{\circ} \end{aligned} $
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