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Question Number 193609 by Mingma last updated on 17/Jun/23

Answered by Subhi last updated on 17/Jun/23

$$\frac{x}{sin\left(60\right)}=\frac{z}{sin(120-y)}$$$$\frac{z}{sin\left(60\right)}=\frac{1}{sin(60-y)}\Rrightarrow z=\frac{sin\left(60\right)}{sin(60-y)}$$$$x=\frac{{sin}^2\left(60\right)}{sin(60-y)sin(120-y)}$$$$\frac{x+z}{sin\left(120\right)}=\frac{z}{sin(60-y)}$$$$\frac{\frac{{sin}^2\left(60\right)}{sin(120-y)sin(60-y)}+\frac{sin\left(60\right)}{sin(60-y)}}{sin\left(120\right)}=\frac{sin\left(60\right)}{{sin}^2(60-y)}$$$$\frac{{sin}^2\left(60\right).sin(60-y)}{sin(120-y)}+sin\left(60\right)sin(60-y)=sin\left(60\right)sin\left(120\right)$$$$\frac{3}{4}sin(120-y)=\frac{\sqrt{3}}{2}sin(60-y)sin(120-y)+\frac{3}{4}sin(60-y)$$$$\frac{3\sqrt{3}}{8}{cos}^2\left(y\right)-\frac{\sqrt{3}}{8}{sin}^2\left(y\right)-\frac{3}{4}sin\left(y\right)=0$$$${cos}^2\left(y\right)=1-{sin}^2\left(y\right)$$$$\frac{\sqrt{3}}{2}{sin}^2\left(y\right)+\frac{3}{4}sin\left(y\right)-\frac{3\sqrt{3}}{8}=0$$$$sin\left(y\right)=\frac{\sqrt{15}-\sqrt{3}}{4}\Rrightarrow y=32.36$$$$x=\frac{\left({sin}^2\left(60\right)\right)}{sin(60-y)sin(120-y)}=1.618=\frac{1+\sqrt{5}}{2}\left(thegoldenratio\right)$$$$$$$$$$

Commented by Subhi last updated on 17/Jun/23

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