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Question Number 221411 by ajfour last updated on 03/Jun/25
Answered by mr W last updated on 04/Jun/25
r=(θ/(2 sin θ))=(π/6)
$${r}=\frac{\theta}{\mathrm{2}\:\mathrm{sin}\:\theta}=\frac{\pi}{\mathrm{6}} \\ $$
Commented by mr W last updated on 07/Jun/25
i think you can know by yourself that it is not a clear and solvable question but a non−sense.
$${i}\:{think}\:{you}\:{can}\:{know}\:{by}\:{yourself} \\ $$$${that}\:{it}\:{is}\:{not}\:{a}\:{clear}\:{and}\:{solvable} \\ $$$${question}\:{but}\:{a}\:{non}−{sense}. \\ $$
Commented by mr W last updated on 04/Jun/25
Commented by mr W last updated on 04/Jun/25
(θ/(tan θ))=2×1 sin (1/2)((π/2)−θ) (θ/(tan θ))=(√2)(cos (θ/2)−sin (θ/2)) (θ^2 /(tan^2 θ))=2(1−sin θ) θ^2 (1+sin θ)=2 sin^2 θ ⇒θ≈0.731365 (41.9°)
$$\frac{\theta}{\mathrm{tan}\:\theta}=\mathrm{2}×\mathrm{1}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$$\frac{\theta}{\mathrm{tan}\:\theta}=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$$\frac{\theta^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \:\theta}=\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\theta^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:\theta\right)=\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\Rightarrow\theta\approx\mathrm{0}.\mathrm{731365}\:\left(\mathrm{41}.\mathrm{9}°\right) \\ $$
Commented by ajfour last updated on 04/Jun/25
yes! Thank you. If blue circle crosses also point (0,1) find θ.
$${yes}!\:{Thank}\:{you}.\:{If}\:{blue}\:{circle}\:{crosses}\:{also} \\ $$$${point}\:\left(\mathrm{0},\mathrm{1}\right)\:{find}\:\theta. \\ $$
Commented by mr W last updated on 05/Jun/25
ORPQ is cyclic. QR is diameter of the smaller circle. α=(π/2)−θ RP=((QP)/(tan θ))=(θ/(tan θ)) RP=2×OP sin (α/2)=2×1 sin (1/2)((π/2)−θ) =2 sin ((π/4)−(θ/2))=(√2) (cos (θ/2)−sin (θ/2)) ⇒(θ/(tan θ))=(√2) (cos (θ/2)−sin (θ/2)) ⇒((θ^2 cos^2 θ)/(sin^2 θ))=2(cos (θ/2)−sin (θ/2))^2 ⇒((θ^2 (1−sin^2 θ))/(sin^2 θ))=2(1−sin θ) ⇒((θ^2 (1+sin θ))/(sin^2 θ))=2 ⇒θ^2 (1+sin θ)=2 sin^2 θ
$${ORPQ}\:{is}\:{cyclic}. \\ $$$${QR}\:{is}\:{diameter}\:{of}\:{the}\:{smaller}\:{circle}. \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}−\theta \\ $$$${RP}=\frac{{QP}}{\mathrm{tan}\:\theta}=\frac{\theta}{\mathrm{tan}\:\theta} \\ $$$${RP}=\mathrm{2}×{OP}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{2}×\mathrm{1}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)=\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\theta}{\mathrm{tan}\:\theta}=\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\theta^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2}\left(\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\theta^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\frac{\theta^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:\theta\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2} \\ $$$$\Rightarrow\theta^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:\theta\right)=\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$
Commented by mr W last updated on 05/Jun/25
Commented by mr W last updated on 04/Jun/25
with θ≈0.731365 (41.9°) we get this:
$${with}\:\theta\approx\mathrm{0}.\mathrm{731365}\:\left(\mathrm{41}.\mathrm{9}°\right)\:{we}\:{get}\:{this}: \\ $$
Commented by mr W last updated on 04/Jun/25
Commented by ajfour last updated on 04/Jun/25
yeah must be true then but i could not follow your solution sir.
$${yeah}\:{must}\:{be}\:{true}\:{then}\:{but}\:{i}\:{could}\:{not} \\ $$$${follow}\:{your}\:{solution}\:{sir}. \\ $$
Commented by ajfour last updated on 05/Jun/25
thanks sir. Got it.
$${thanks}\:{sir}.\:{Got}\:{it}.\: \\ $$
Commented by Tawa11 last updated on 05/Jun/25
Weldone sirs. Please help when chanced sirs. Q221444
$$\mathrm{Weldone}\:\mathrm{sirs}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{when}\:\mathrm{chanced}\:\mathrm{sirs}.\:\:\mathrm{Q221444} \\ $$
Commented by mr W last updated on 05/Jun/25
do you understand that question? i don′t. so i wont waste my timefor it.
$${do}\:{you}\:{understand}\:{that}\:{question}? \\ $$$${i}\:{don}'{t}.\:{so}\:{i}\:{wont}\:{waste}\:{my}\:{timefor} \\ $$$${it}. \\ $$
Commented by Tawa11 last updated on 05/Jun/25
Sir, I don′t understand too. Please based on your experience, is the diagram drawn correctly. What do you think is not set well in the question. What correction do you think can be mbade?
$$\mathrm{Sir},\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{too}. \\ $$$$\mathrm{Please}\:\mathrm{based}\:\mathrm{on}\:\mathrm{your}\:\mathrm{experience}, \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{diagram}\:\mathrm{drawn}\:\mathrm{correctly}. \\ $$$$\mathrm{What}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{is}\:\mathrm{not}\:\mathrm{set}\:\mathrm{well}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{What}\:\mathrm{correction}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{can}\:\mathrm{be}\:\mathrm{mbade}? \\ $$
Commented by mr W last updated on 05/Jun/25
that question is a non−sense to me. i don′t understand it and wont waste time for it.
$${that}\:{question}\:{is}\:{a}\:{non}−{sense}\:{to}\:{me}. \\ $$$${i}\:{don}'{t}\:{understand}\:{it}\:{and}\:{wont}\:{waste} \\ $$$${time}\:{for}\:{it}.\: \\ $$
Commented by Tawa11 last updated on 05/Jun/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 06/Jun/25
Sir, please check the new modification after I complained. If it is solvable. Please help sir. Q221444
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{new}\:\mathrm{modification} \\ $$$$\mathrm{after}\:\mathrm{I}\:\mathrm{complained}.\:\mathrm{If}\:\mathrm{it}\:\mathrm{is}\:\mathrm{solvable}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{sir}.\:\:\mathrm{Q221444} \\ $$
Commented by Tawa11 last updated on 06/Jun/25
I opened new tab. Q221495
$$\mathrm{I}\:\mathrm{opened}\:\mathrm{new}\:\mathrm{tab}.\:\mathrm{Q221495} \\ $$
Commented by Tawa11 last updated on 07/Jun/25
I am asking incase they have error again in the question.
$$\mathrm{I}\:\mathrm{am}\:\mathrm{asking}\:\mathrm{incase}\:\mathrm{they}\:\mathrm{have}\:\mathrm{error}\:\mathrm{again} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$
Commented by Tawa11 last updated on 07/Jun/25
Sir, your experience is more than my own. That is why I asked you sir. My geometry skill is 1/1000 of your own sir. Hahahahahahaha!!!
$$\mathrm{Sir},\:\mathrm{your}\:\mathrm{experience}\:\mathrm{is}\:\mathrm{more}\:\mathrm{than}\:\mathrm{my}\:\mathrm{own}. \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{why}\:\mathrm{I}\:\mathrm{asked}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{My}\:\mathrm{geometry}\:\mathrm{skill}\:\mathrm{is}\:\:\mathrm{1}/\mathrm{1000}\:\mathrm{of}\:\mathrm{your}\:\mathrm{own}\:\mathrm{sir}. \\ $$$$\mathrm{Hahahahahahaha}!!! \\ $$

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