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Question-112076

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Question Number 112076 by I want to learn more last updated on 06/Sep/20
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
Is OD the angle bisector of D?
$$\mathrm{Is}\:\mathrm{OD}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{D}? \\ $$
Commented by som(math1967) last updated on 06/Sep/20
yes you are correct
$$\mathrm{yes}\:\mathrm{you}\:\mathrm{are}\:\mathrm{correct} \\ $$
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
DC is not a chord. I think your solution is incorrect.
$$\mathrm{DC}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{chord}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{your} \\ $$$$\mathrm{solution}\:\mathrm{is}\:\mathrm{incorrect}. \\ $$
Answered by mr W last updated on 06/Sep/20
say radius OC=OB=1 BC=(2/(cos 20°)) ((CD)/(sin 30°))=((BD)/(sin 40°))=((BC)/(sin 70°))=(2/(cos^2 20°)) ⇒CD=(4/(cos^2 20°)) OD^2 =1^2 +(4^2 /(cos^4 20°))−2×(4/(cos^2 20°))×cos 40° =1+((16)/(cos^4 20°))−((8(2 cos^2 20−1))/(cos^2 20°)) =((16)/(cos^4 20°))+(8/(cos^2 20°))−15 =(1/(cos^4 20°))(16+8 cos^2 20°−15 cos^4 20°) ((sin x)/(OC))=((sin 20°)/(OD)) ⇒sin x=((sin 20° cos^2 20°)/( (√(16+8 cos^2 20°−15 cos^4 20°)))) ⇒x≈5.139°
$${say}\:{radius}\:{OC}={OB}=\mathrm{1} \\ $$$${BC}=\frac{\mathrm{2}}{\mathrm{cos}\:\mathrm{20}°} \\ $$$$\frac{{CD}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{BD}}{\mathrm{sin}\:\mathrm{40}°}=\frac{{BC}}{\mathrm{sin}\:\mathrm{70}°}=\frac{\mathrm{2}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°} \\ $$$$\Rightarrow{CD}=\frac{\mathrm{4}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°} \\ $$$${OD}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{4}} \:\mathrm{20}°}−\mathrm{2}×\frac{\mathrm{4}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°}×\mathrm{cos}\:\mathrm{40}° \\ $$$$=\mathrm{1}+\frac{\mathrm{16}}{\mathrm{cos}^{\mathrm{4}} \:\mathrm{20}°}−\frac{\mathrm{8}\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}−\mathrm{1}\right)}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°} \\ $$$$=\frac{\mathrm{16}}{\mathrm{cos}^{\mathrm{4}} \:\mathrm{20}°}+\frac{\mathrm{8}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°}−\mathrm{15} \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \:\mathrm{20}°}\left(\mathrm{16}+\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°−\mathrm{15}\:\mathrm{cos}^{\mathrm{4}} \:\mathrm{20}°\right) \\ $$$$\frac{\mathrm{sin}\:{x}}{{OC}}=\frac{\mathrm{sin}\:\mathrm{20}°}{{OD}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\frac{\mathrm{sin}\:\mathrm{20}°\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°}{\:\sqrt{\mathrm{16}+\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{20}°−\mathrm{15}\:\mathrm{cos}^{\mathrm{4}} \:\mathrm{20}°}} \\ $$$$\Rightarrow{x}\approx\mathrm{5}.\mathrm{139}° \\ $$
Commented by I want to learn more last updated on 06/Sep/20
Wow, thanks sir, i appreciate. Sir, what condition can i say radii OC = OB = 1??
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{appreciate}. \\ $$$$\mathrm{Sir},\:\mathrm{what}\:\mathrm{condition}\:\mathrm{can}\:\mathrm{i}\:\mathrm{say}\:\:\mathrm{radii}\:\:\:\mathrm{OC}\:\:=\:\:\mathrm{OB}\:\:=\:\:\:\mathrm{1}?? \\ $$
Commented by mr W last updated on 06/Sep/20
you can say OC=OB=r, it doesn′t matter, since we are calculating the angles only.
$${you}\:{can}\:{say}\:{OC}={OB}={r},\:{it}\:{doesn}'{t} \\ $$$${matter},\:{since}\:{we}\:{are}\:{calculating}\:{the} \\ $$$${angles}\:{only}. \\ $$
Commented by I want to learn more last updated on 06/Sep/20
Ohh, i understand sir. Thanks so much.
$$\mathrm{Ohh},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}. \\ $$

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