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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

((sinx)/(AC))=((sin(120−x))/(AO)) ((sin10)/(AO))=((sin150)/(AB)) ⇒((sinx)/(AC))=((sin(120−x))/(AB((sin10)/(sin150)))) ⇒^(AC=AB) ((sinx)/(sin(120−x)))=((sin150)/(sin10))⇒x=100

$$\frac{\mathrm{sinx}}{\mathrm{AC}}=\frac{\mathrm{sin}\left(\mathrm{120}−\mathrm{x}\right)}{\mathrm{AO}} \\ $$$$\frac{\mathrm{sin10}}{\mathrm{AO}}=\frac{\mathrm{sin150}}{\mathrm{AB}} \\ $$$$\Rightarrow\frac{\mathrm{sinx}}{\mathrm{AC}}=\frac{\mathrm{sin}\left(\mathrm{120}−\mathrm{x}\right)}{\mathrm{AB}\frac{\mathrm{sin10}}{\mathrm{sin150}}}\:\:\:\overset{\mathrm{AC}=\mathrm{AB}} {\Rightarrow}\: \\ $$$$\frac{\mathrm{sinx}}{\mathrm{sin}\left(\mathrm{120}−\mathrm{x}\right)}=\frac{\mathrm{sin150}}{\mathrm{sin10}}\Rightarrow\mathrm{x}=\mathrm{100} \\ $$

Commented by jagoll last updated on 18/Apr/20

i don′t see the AO line on the matter sir?

$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{the}\:\mathrm{AO}\:\mathrm{line}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{matter}\:\mathrm{sir}? \\ $$

Commented by mahdi last updated on 18/Apr/20

i sorry jagoll,point′′O′′ is center of Δ. or CO^� A=x^�

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

Commented by jagoll last updated on 18/Apr/20

thank you

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

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

thank you sir

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

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