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Question Number 89647 by otchereabdullai@gmail.com last updated on 18/Apr/20

Answered by mahdi last updated on 18/Apr/20

$$\frac{\mathrm{sinx}}{\mathrm{AC}}=\frac{\sin (120-\mathrm{x})}{\mathrm{AO}}$$$$\frac{\sin 10}{\mathrm{AO}}=\frac{\sin 150}{\mathrm{AB}}$$$$\Rightarrow \frac{\mathrm{sinx}}{\mathrm{AC}}=\frac{\sin (120-\mathrm{x})}{\mathrm{AB}\frac{\sin 10}{\sin 150}}\stackrel{\mathrm{AC}=\mathrm{AB}}{\Rightarrow }$$$$\frac{\mathrm{sinx}}{\sin (120-\mathrm{x})}=\frac{\sin 150}{\sin 10}\Rightarrow \mathrm{x}=100$$

Commented by jagoll last updated on 18/Apr/20

$$\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{the}\mathrm{AO}\mathrm{line}\mathrm{on}\mathrm{the}$$$$\mathrm{matter}\mathrm{sir}?$$

Commented by mahdi last updated on 18/Apr/20

$$\mathrm{i}\mathrm{sorry}\mathrm{jagoll},{\mathrm{point}}^{″}{\mathrm{O}}^{″}\mathrm{is}\mathrm{center}\mathrm{of}\mathrm{\Delta }.$$$$\mathrm{or}\mathrm{C}\widehat{\mathrm{O}A}=\hat{\mathrm{x}}$$

Commented by jagoll last updated on 18/Apr/20

$$\mathrm{thank}\mathrm{you}$$

Commented by otchereabdullai@gmail.com last updated on 18/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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