In triangles ABC and DEF,AB = DE,AC = EF and | vedclass

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Question

(C) Given that the area of $\Delta ABC = 	ext{Area of } \Delta DEF$. Using the area formula $\frac{1}{2}bc \sin A$,we have: $\frac{1}{2}(AB)(AC) \sin

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(C) Given that the area of $\Delta ABC = 	ext{Area of } \Delta DEF$. Using the area formula $\frac{1}{2}bc \sin A$,we have: $\frac{1}{2}(AB)(AC) \sin A = \frac{1}{2}(DE)(EF) \sin E$. Since $AB = DE$ and $AC = EF$,this simplifies to: $\sin A = \sin E$. Given $\angle A = 2\angle E$,we substitute $A = 2E$: $\sin(2E) = \sin E$. $2 \sin E \cos E = \sin E$. Since $\sin E 
eq 0$ (as it is an angle of a triangle),we divide by $\sin E$: $2 \cos E = 1 \Rightarrow \cos E = \frac{1}{2}$. Thus,$E = \frac{\pi}{3}$. Therefore,$A = 2E = 2 	imes \frac{\pi}{3} = \frac{2\pi}{3}$.

In triangles $ABC$ and $DEF$,$AB = DE$,$AC = EF$ and $\angle A = 2\angle E$. The two triangles will have the same area if angle $A$ is equal to:

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