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Question-197670

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Question Number 197670 by mr W last updated on 26/Sep/23
Commented by mr W last updated on 26/Sep/23
unsolved old question Q#197017
$${unsolved}\:{old}\:{question}\:{Q}#\mathrm{197017} \\ $$
Commented by ajfour last updated on 26/Sep/23
https://youtu.be/Y2LCMHAw3Fw?si=Vt7lXmpu1xLMVuB9
Commented by ajfour last updated on 26/Sep/23
A simple educational video i posted hours back, on dy/dx if y=x and if y= x^2
Commented by mr W last updated on 26/Sep/23
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Answered by mr W last updated on 26/Sep/23
Commented by mr W last updated on 26/Sep/23
s=side length of ΔABC tan α=(s/3)×((√3)/2)×(1/(s−(s/3)×(1/2)))=((√3)/5) a=(s/2)×tan α=(s/2)×((√3)/( 5))=((s(√3))/(10)) b=(s/2) β=90°−60°=30° c=b−2a cos β=(s/2)−2×((s(√3))/(10))×((√3)/2)=(s/5) ((blue area)/(ΔABC))=((c/s))^2 =((1/5))^2 =(1/(25)) ✓
$${s}={side}\:{length}\:{of}\:\Delta{ABC} \\ $$$$\mathrm{tan}\:\alpha=\frac{{s}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{{s}−\frac{{s}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$$${a}=\frac{{s}}{\mathrm{2}}×\mathrm{tan}\:\alpha=\frac{{s}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\:\mathrm{5}}=\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{10}} \\ $$$${b}=\frac{{s}}{\mathrm{2}} \\ $$$$\beta=\mathrm{90}°−\mathrm{60}°=\mathrm{30}° \\ $$$${c}={b}−\mathrm{2}{a}\:\mathrm{cos}\:\beta=\frac{{s}}{\mathrm{2}}−\mathrm{2}×\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{10}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{{s}}{\mathrm{5}} \\ $$$$\frac{{blue}\:{area}}{\Delta{ABC}}=\left(\frac{{c}}{{s}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{25}}\:\checkmark \\ $$
Commented by mr W last updated on 26/Sep/23
you are right sir! i took tan α wrongly as sin α.
$${you}\:{are}\:{right}\:{sir}! \\ $$$${i}\:{took}\:\mathrm{tan}\:\alpha\:{wrongly}\:{as}\:\mathrm{sin}\:\alpha. \\ $$
Commented by ajfour last updated on 26/Sep/23
triangle sides 6s and 2c ((6s)/2)×tan 30°=3stan α+(c/(cos 30°)) ..(i) ((sin α)/(2s))=((sin (α+60°))/(6s)) ⇒ 3sin α=(1/2)sin α+((√3)/2)cos α ⇒ tan α=((√3)/5) now from (i) s(√3)=((3(√3))/5)s+((2c)/( (√3))) (c/s)=(3/5) ⇒ ((2c)/(6s))=(1/5) r_(area) =(1/(25))
$${triangle}\:{sides}\:\:\:\mathrm{6}{s}\:\:{and}\:\:\mathrm{2}{c} \\ $$$$\frac{\mathrm{6}{s}}{\mathrm{2}}×\mathrm{tan}\:\mathrm{30}°=\mathrm{3}{s}\mathrm{tan}\:\alpha+\frac{{c}}{\mathrm{cos}\:\mathrm{30}°}\:\:..\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{2}{s}}=\frac{\mathrm{sin}\:\left(\alpha+\mathrm{60}°\right)}{\mathrm{6}{s}} \\ $$$$\Rightarrow\:\:\mathrm{3sin}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$$${now}\:\:\:{from}\:\:\left({i}\right) \\ $$$${s}\sqrt{\mathrm{3}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{5}}{s}+\frac{\mathrm{2}{c}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{c}}{{s}}=\frac{\mathrm{3}}{\mathrm{5}}\:\:\:\Rightarrow\:\:\:\frac{\mathrm{2}{c}}{\mathrm{6}{s}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${r}_{{area}} =\frac{\mathrm{1}}{\mathrm{25}} \\ $$

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