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Question Number 90514 by I want to learn more last updated on 24/Apr/20

Commented by $@ty@m123 last updated on 24/Apr/20

tan 20^o =((BC)/(4+x)) BC=0.364(4+x)....(1) tan 80^o =((BC)/x) BC=5.671x ...(2) From (1) &(2) 5.671x=0.364(4+x) x=0.2743

$$\mathrm{tan}\:\mathrm{20}^{\mathrm{o}} =\frac{{BC}}{\mathrm{4}+{x}} \\ $$$${BC}=\mathrm{0}.\mathrm{364}\left(\mathrm{4}+{x}\right)....\left(\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\mathrm{80}^{\mathrm{o}} =\frac{\mathrm{B}{C}}{{x}} \\ $$$${BC}=\mathrm{5}.\mathrm{671}{x}\:...\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right) \\ $$$$\mathrm{5}.\mathrm{671}{x}=\mathrm{0}.\mathrm{364}\left(\mathrm{4}+{x}\right) \\ $$$${x}=\mathrm{0}.\mathrm{2743} \\ $$

Commented by me2love2math last updated on 24/Apr/20

x=3.....dont know to use the app...

$${x}=\mathrm{3}.....{dont}\:{know}\:{to}\:{use}\:{the}\:{app}... \\ $$

Commented by $@ty@m123 last updated on 24/Apr/20

If x=3 then cos 20^o =(7/(10))=0.7 But Actual value of cos 20^o =0.94

$${If}\:{x}=\mathrm{3}\: \\ $$$${then}\:\mathrm{cos}\:\mathrm{20}^{\mathrm{o}} =\frac{\mathrm{7}}{\mathrm{10}}=\mathrm{0}.\mathrm{7} \\ $$$$\mathrm{B}{ut}\:{Actual}\:{value}\:{of}\:\mathrm{cos}\:\mathrm{20}^{\mathrm{o}} =\mathrm{0}.\mathrm{94} \\ $$

Commented by me2love2math last updated on 24/Apr/20

the triangle is not complete....extend the diagram and apply triangle theorem

$${the}\:{triangle}\:{is}\:{not}\:{complete}....{extend}\: \\ $$$${the}\:{diagram}\:{and}\:{apply}\:{triangle}\:{theorem} \\ $$

Commented by I want to learn more last updated on 24/Apr/20

Thanks sirs, but which answer is correct please.

$$\mathrm{Thanks}\:\mathrm{sirs},\:\mathrm{but}\:\mathrm{which}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{please}. \\ $$

Commented by me2love2math last updated on 24/Apr/20

my English bad but my maths is good I want to learn how to use the app

$${my}\:{English}\:{bad}\:{but}\:{my}\:{maths}\:{is}\:{good} \\ $$$${I}\:{want}\:{to}\:{learn}\:{how}\:{to}\:{use}\:{the}\:{app} \\ $$

Answered by mahdi last updated on 24/Apr/20

cos20=((4+x)/(10))⇒2cos^2 10−1=((4+x)/(10))⇒ cos^2 10=((14+x)/(20)) ; tan^2 10=(1/(cos^2 10))−1=((6−x)/(12+x)) ⇒tan10=((DC)/(BC))⇒tan^2 10=((DC^2 )/(BC^2 ))=((DC^2 )/(AB^2 −AC^2 )) ⇒((6−x)/(12+x))=(x/(100−(4+x)^2 ))⇒x≈4.092

$$\mathrm{cos20}=\frac{\mathrm{4}+\mathrm{x}}{\mathrm{10}}\Rightarrow\mathrm{2cos}^{\mathrm{2}} \mathrm{10}−\mathrm{1}=\frac{\mathrm{4}+\mathrm{x}}{\mathrm{10}}\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{10}=\frac{\mathrm{14}+\mathrm{x}}{\mathrm{20}}\:;\:\mathrm{tan}^{\mathrm{2}} \mathrm{10}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{10}}−\mathrm{1}=\frac{\mathrm{6}−\mathrm{x}}{\mathrm{12}+\mathrm{x}} \\ $$$$\Rightarrow\mathrm{tan10}=\frac{\mathrm{DC}}{\mathrm{BC}}\Rightarrow\mathrm{tan}^{\mathrm{2}} \mathrm{10}=\frac{\mathrm{DC}^{\mathrm{2}} }{\mathrm{BC}^{\mathrm{2}} }=\frac{\mathrm{DC}^{\mathrm{2}} }{\mathrm{AB}^{\mathrm{2}} −\mathrm{AC}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{6}−\mathrm{x}}{\mathrm{12}+\mathrm{x}}=\frac{\mathrm{x}}{\mathrm{100}−\left(\mathrm{4}+\mathrm{x}\right)^{\mathrm{2}} }\Rightarrow\mathrm{x}\approx\mathrm{4}.\mathrm{092} \\ $$

Commented by I want to learn more last updated on 24/Apr/20

Thanks sir, we are having different amswers sir. which one is correct please

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{we}\:\mathrm{are}\:\mathrm{having}\:\mathrm{different}\:\mathrm{amswers}\:\mathrm{sir}. \\ $$$$\mathrm{which}\:\mathrm{one}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{please} \\ $$

Commented by mr W last updated on 24/Apr/20

your question is wrong! so you can′t get one correct answer!

$${your}\:{question}\:{is}\:{wrong}!\:{so}\:{you}\:{can}'{t} \\ $$$${get}\:{one}\:{correct}\:{answer}! \\ $$

Answered by mr W last updated on 24/Apr/20

question is wrong! AD=4 is wrong or ∠BDC=80° is wrong! say AD=4 is correct, then AC=10×cos 20° BC=10×sin 20° x=10×cos 20°−4 tan ∠BDC=((10×sin 20°)/(10×cos 20°−4)) ∠BDC=tan^(−1) (((10×sin 20°)/(10×cos 20°−4)))=32.363°≠80°

$${question}\:{is}\:{wrong}! \\ $$$${AD}=\mathrm{4}\:{is}\:{wrong}\:{or}\:\angle{BDC}=\mathrm{80}°\:{is}\:{wrong}! \\ $$$$ \\ $$$${say}\:{AD}=\mathrm{4}\:{is}\:{correct},\:{then} \\ $$$${AC}=\mathrm{10}×\mathrm{cos}\:\mathrm{20}° \\ $$$${BC}=\mathrm{10}×\mathrm{sin}\:\mathrm{20}° \\ $$$${x}=\mathrm{10}×\mathrm{cos}\:\mathrm{20}°−\mathrm{4} \\ $$$$\mathrm{tan}\:\angle{BDC}=\frac{\mathrm{10}×\mathrm{sin}\:\mathrm{20}°}{\mathrm{10}×\mathrm{cos}\:\mathrm{20}°−\mathrm{4}} \\ $$$$\angle{BDC}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{10}×\mathrm{sin}\:\mathrm{20}°}{\mathrm{10}×\mathrm{cos}\:\mathrm{20}°−\mathrm{4}}\right)=\mathrm{32}.\mathrm{363}°\neq\mathrm{80}° \\ $$

Commented by I want to learn more last updated on 25/Apr/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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