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Question Number 204729 by Amidip last updated on 26/Feb/24

Answered by A5T last updated on 26/Feb/24

$$\frac{sin\left(2x\right)}{AD}=\frac{sin(140-2x)}{AB};\frac{sin(140-3x)}{AD}=\frac{sin\left(x\right)}{DC=AB}$$$$\Rightarrow \frac{sin(140-3x)}{sin\left(x\right)}=\frac{sin\left(2x\right)}{sin(140-2x)}\Rightarrow x=35°$$

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