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Question Number 193522 by Mingma last updated on 16/Jun/23

Answered by Subhi last updated on 16/Jun/23

$$\frac{x+z}{sin\left(100\right)}=\frac{x}{sin\left(40\right)}$$$$sin\left(100\right)x=sin\left(40\right)x+sin\left(40\right)z\Rrightarrow x=\frac{sin\left(40\right)z}{sin\left(10\right)-sin\left(40\right)}\left(i\right)$$$$\frac{z}{sin(y-40)}=\frac{x+z}{sin\left(u\right)}$$$$u+y-40=180-140=40\Rrightarrow y=80-u\left(ii\right)$$$$\frac{z}{sin(40-u)}=\frac{x+z}{sin\left(u\right)}$$$$sin(40-u)x+sin(40-u)z=zsin\left(u\right)$$$$sin(40-u).\frac{sin\left(40\right)z}{sin\left(100\right)-sin\left(40\right)}+sin(40-u)z=zsin\left(u\right)$$$$(\frac{{sin}^2\left(40\right)}{sin\left(100\right)-sin\left(40\right)}+sin\left(40\right))cos\left(u\right)=(1+cos\left(40\right)+\frac{cos\left(40\right).sin\left(40\right)}{sin\left(100\right)-sin\left(40\right)})sin\left(u\right)$$$$u={tan}^{-1}\left((\frac{{sin}^2\left(40\right)}{sin\left(100\right)-sin\left(40\right)}+sin\left(40\right))\right)/(1+cos\left(40\right)+\frac{\frac{sin\left(80\right)}{2}}{sin\left(100\right)-sin\left(40\right)})={tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)=30$$

Commented by Subhi last updated on 16/Jun/23

Commented by Mingma last updated on 16/Jun/23

Perfect ��

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