Question Number 89647 by otchereabdullai@gmail.com last updated on 18/Apr/20
Answered by mahdi last updated on 18/Apr/20
$$\frac{\mathrm{sinx}}{\mathrm{AC}}=\frac{\mathrm{sin}\left(\mathrm{120}−\mathrm{x}\right)}{\mathrm{AO}} \\ $$$$\frac{\mathrm{sin10}}{\mathrm{AO}}=\frac{\mathrm{sin150}}{\mathrm{AB}} \\ $$$$\Rightarrow\frac{\mathrm{sinx}}{\mathrm{AC}}=\frac{\mathrm{sin}\left(\mathrm{120}−\mathrm{x}\right)}{\mathrm{AB}\frac{\mathrm{sin10}}{\mathrm{sin150}}}\:\:\:\overset{\mathrm{AC}=\mathrm{AB}} {\Rightarrow}\: \\ $$$$\frac{\mathrm{sinx}}{\mathrm{sin}\left(\mathrm{120}−\mathrm{x}\right)}=\frac{\mathrm{sin150}}{\mathrm{sin10}}\Rightarrow\mathrm{x}=\mathrm{100} \\ $$
Commented by jagoll last updated on 18/Apr/20
$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{the}\:\mathrm{AO}\:\mathrm{line}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{matter}\:\mathrm{sir}? \\ $$
Commented by mahdi last updated on 18/Apr/20