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Question Number 204145 by cortano12 last updated on 07/Feb/24

Answered by deleteduser1 last updated on 07/Feb/24

Commented by cortano12 last updated on 08/Feb/24

$$\mathrm{i}\mathrm{got}\mathrm{x}=40°$$

Commented by deleteduser1 last updated on 07/Feb/24

$$\frac{sin\left(20°\right)}{DC}=\frac{sin\left(110°\right)}{BC}\Rightarrow BF=DCcot\left(20°\right)-CF$$$$\frac{sin(110-x)}{CF}=\frac{sin(20+x)}{DC}\Rightarrow CF=\frac{DCsin(110-x)}{sin(20+x)}$$$$\Rightarrow BF=DC(cot20-\frac{sin(110-x)}{sin(20+x)})$$$$\frac{sin\left(x\right)}{BF}=\frac{sin\left(20\right)}{DF};\frac{sin\left(50\right)}{DF}=\frac{sin(20+x)}{DC}$$$$\Rightarrow DF=\frac{DCsin\left(50\right)}{sin(20+x)}\Rightarrow \frac{sin\left(x\right)}{BF}=\frac{sin\left(20\right)sin(20+x)}{DCsin\left(50\right)}$$$$\Rightarrow \frac{sin\left(x\right)sin(20+x)}{cot20sin(20+x)-sin(110-x)}=\frac{sin\left(20\right)sin(20+x)}{sin50}$$$$\Rightarrow x=20°$$

Commented by deleteduser1 last updated on 08/Feb/24

$$Yoursolution?$$

Commented by A5T last updated on 27/Aug/24

$$\frac{sin\left(20°\right)}{DC}=\frac{sin\left(110°\right)}{BC}\Rightarrow BF=DCcot\left(20°\right)-CF$$$$\frac{sin(110-x)}{CF}=\frac{sin(20+x)}{DC}\Rightarrow CF=\frac{DCsin(110-x)}{sin(20+x)}$$$$\Rightarrow BF=DC(cot20-\frac{sin(110-x)}{sin(20+x)})...\left(i\right)$$$$\frac{sin\left(x\right)}{BF}=\frac{sin\left(20\right)}{DF};\frac{sin\left(50\right)}{DF}=\frac{sin(20+x)}{DC}$$$$\Rightarrow DF=\frac{DCsin\left(50\right)}{sin(20+x)}\Rightarrow \frac{sin\left(x\right)}{BF}=\frac{sin\left(20\right)sin(20+x)}{DCsin\left(50\right)}$$$$\Rightarrow BF=\frac{DCsin50sinx}{sin20sin(20+x)}...\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right)$$$$\Rightarrow sin50sinx=sin(20+x)cos20-sin(110-x)sin20$$$$\Rightarrow x=40°$$

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