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Question Number 193906 by cherokeesay last updated on 22/Jun/23

Answered by Subhi last updated on 22/Jun/23

$$$$$$\frac{1}{sin\left(18\right)}=\frac{1+y}{sin(144-x)}$$$$\frac{1}{sin\left(18\right)}=\frac{y}{sin\left(x\right)}\Rrightarrow y=\frac{sin\left(x\right)}{sin\left(18\right)}$$$$\frac{1}{sin\left(18\right)}=\frac{1+\frac{sin\left(x\right)}{sin\left(18\right)}}{sin(144-x)}$$$$sin(144-x)=sin\left(18\right)+sin\left(x\right)$$$$sin\left(144\right)cos\left(x\right)-cos\left(144\right)sin\left(x\right)=sin\left(18\right)+sin\left(x\right)$$$$sin\left(144\right)cos\left(x\right)+(-cos\left(144\right)-1)sin\left(x\right)=sin\left(18\right)$$$$putsin\left(x\right)=u$$$$sin\left(144\right).\sqrt{1-u^2}={(1+cos\left(144\right))}^{2}u+sin\left(18\right)$$$$(-{sin}^2\left(144\right)-{(1+cos\left(144\right))}^2)u^2-2sin\left(18\right)(1+cos\left(144\right))u+({sin}^2\left(144\right)-{sin}^2\left(18\right))=0$$$$u=0.66913$$$$x={sin}^{-1}(0.66913)=42$$$$\frac{z}{sin(162-x)}=\frac{1}{sin\left(18\right)}$$$$z=\frac{sin(162-x)}{sin\left(18\right)}=2.8$$

Commented by Subhi last updated on 22/Jun/23

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