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Question Number 194029 by Rupesh123 last updated on 26/Jun/23

Commented by Rupesh123 last updated on 26/Jun/23

Perfect ��

Answered by Subhi last updated on 26/Jun/23

$$AD=\sqrt{2Q^2-2Q^{2}cos\left(120\right)}=\sqrt{3}Q$$$$AB=AC=Z$$$$\frac{Z}{sin\left(150\right)}=\frac{\sqrt{3}Q}{sin\left(20\right)}\Rrightarrow Q=\frac{2sin\left(20\right)Z}{\sqrt{3}}$$$$\frac{2sin\left(20\right)Z}{\sqrt{3}.sin(120-x)}=\frac{z-\frac{2sin\left(120\right)Z}{\sqrt{3}}}{sin\left(x\right)}$$$$2sin\left(20\right)sin\left(x\right)=(\frac{\sqrt{3}cos\left(x\right)}{2}+\frac{sin\left(x\right)}{2})(\sqrt{3}-2sin\left(20\right))$$$$x={tan}^{-1}\left(\frac{\frac{\sqrt{3}}{2}(\sqrt{3}-2sin\left(20\right))}{2sin\left(20\right)-\frac{\sqrt{3}-2sin\left(20\right)}{2}}\right)=80$$

Commented by Subhi last updated on 26/Jun/23

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