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Question Number 191676 by ajfour last updated on 28/Apr/23

Commented by ajfour last updated on 28/Apr/23

$I f f i r s t t i m e d i p p e d a r e a o f$ $b i s c u i t i s A_{1} .$ $F i n d m a x . o f \frac{A_{2}}{A_{1}} .$ $A_{2} t h e a r e a d i p p e d t h e s e c o n d$ $t i m e, a f t e r A_{1} i s r e m o v e d .$
Commented by mr W last updated on 29/Apr/23

	Answered by mr W last updated on 29/Apr/23

	Commented by mr W last updated on 29/Apr/23
	$sin φ=(a/r) a=r sin φ h=r cos φ A_1 =φr^2 −((r^2 sin 2φ)/2)=r^2 (φ−((sin 2φ)/2)) OA=(h/(cos θ))=((r cos φ)/(cos θ)) cos ϕ=((r^2 +((r^2 cos^2 φ)/(cos^2 θ))−a^2 )/(2r×((r cos φ)/(cos θ)))) ⇒ϕ=cos^(−1) [(((1−4 sin^2 φ) cos^2 θ+cos^2 φ)/(2 cos φ cos θ))] ((sin γ)/r)=((sin ϕ)/(2a)) ⇒γ=sin^(−1) (((sin ϕ)/(2 sin φ))) A_2 =(r^2 /2)(ϕ+θ−φ)+((rh [sin (φ−θ)−sin ϕ])/(2 cos θ)) A_2 =(r^2 /2)(ϕ+θ−φ)+((r^2 cos φ [sin (φ−θ)−sin ϕ])/(2 cos θ)) λ=(A_2 /A_1 )=((ϕ+θ−φ+((cos φ [sin (φ−θ)−sin ϕ])/(cos θ)))/(2φ−sin 2φ)) for λ_(max) : (dλ/dθ)=0 or (d/dθ){ϕ+θ−φ+((cos φ [sin (φ−θ)−sin ϕ])/(cos θ))}=0 this can only be nummerically solved. example: r=b=5, a=3 ⇒θ≈0.2261 or 12.955°$
	$\sin ϕ = \frac{a}{r}$ $a = r \sin ϕ$ $h = r \cos ϕ$ $A_{1} = ϕ r^{2} - \frac{r^{2} \sin 2 ϕ}{2} = r^{2} (ϕ - \frac{\sin 2 ϕ}{2})$ $O A = \frac{h}{\cos θ} = \frac{r \cos ϕ}{\cos θ}$ $\cos φ = \frac{r^{2} + \frac{r^{2} \cos^{2} ϕ}{\cos^{2} θ} - a^{2}}{2 r \times \frac{r \cos ϕ}{\cos θ}}$ $\Rightarrow φ = \cos^{- 1} [\frac{(1 - 4 \sin^{2} ϕ) \cos^{2} θ + \cos^{2} ϕ}{2 \cos ϕ \cos θ}]$ $\frac{\sin γ}{r} = \frac{\sin φ}{2 a}$ $\Rightarrow γ = \sin^{- 1} (\frac{\sin φ}{2 \sin ϕ})$ $A_{2} = \frac{r^{2}}{2} (φ + θ - ϕ) + \frac{r h [\sin (ϕ - θ) - \sin φ]}{2 \cos θ}$ $A_{2} = \frac{r^{2}}{2} (φ + θ - ϕ) + \frac{r^{2} \cos ϕ [\sin (ϕ - θ) - \sin φ]}{2 \cos θ}$ $λ = \frac{A_{2}}{A_{1}} = \frac{φ + θ - ϕ + \frac{\cos ϕ [\sin (ϕ - θ) - \sin φ]}{\cos θ}}{2 ϕ - \sin 2 ϕ}$ $f o r λ_{m a x} : \frac{d λ}{d θ} = 0 o r$ $\frac{d}{d θ} {φ + θ - ϕ + \frac{\cos ϕ [\sin (ϕ - θ) - \sin φ]}{\cos θ}} = 0$ $t h i s c a n o n l y b e n u m m e r i c a l l y s o l v e d .$ $e x a m p l e :$ $r = b = 5, a = 3 \Rightarrow θ \approx 0.2261 o r 12.955 °$
	Commented by mr W last updated on 29/Apr/23

	Commented by ajfour last updated on 29/Apr/23

	$T h a n k y o u s i r, i s h a l l f o l l o w .$
	Answered by a.lgnaoui last updated on 29/Apr/23
	$Area A_1 (∅)=b^2 (2φ−sin ∅cos ∅) sin ∅=(a/(2b)) ∅=sin^(−1) ((a/(2b))) soit θ=2γ−φ A_2 (γ)=b^2 (2γ−sin γcos γ) (A_2 /A_1 )=((2𝛄−sin γcos γ)/(2∅−sin ∅cos ∅)) =(((θ+φ)−sin (((θ+∅)/2))cos( ((θ+∅)/2)))/(2∅−sin ∅cos ∅)) =((θ+∅−(1/2)sin (θ+∅))/(2∅−(a/(2b))×(√(1−(a^2 /(4b^2 )))))) =((θ+sin^(−1) ((a/(2b)))−(1/2)[(1/(2b))sin θ(√(4b^2 −a^2 )) +((bcos θ)/(2a))])/(2sin^(−1) ((a/(2b)))−(1/(4a))(√(4b^2 −a^2 ))))= (((θ+∅)−(1/4)((b/a)cos θ)+(((sin θ)/b))(√(4b^2 −a^2 )) ])/(2sin^(−1) ((a/(2b)))−(1/(4a))(√(4b^2 −a^2 )))) ((A2)/A_1 )=(1/2)[(((θ+φ)−(1/4)(((bcos θ)/a)+((sin θ(√(4b^2 −a^2 )))/b)))/(∅−((√(4b^2 −a^2 ))/(8a))))] (A_2 /A_1 ) max⇒∅=sin^(−1) ((a/(2b)))→(1/(8a))(√(4b^2 −a^2 )) ? { ((γ=((θ+∅)/2))),((∅: given with (a, b))) :} ....................?$
	$Area A_{1} (\emptyset) = b^{2} (2 ϕ - \sin \emptyset \cos \emptyset)$ $\sin \emptyset = \frac{a}{2 b} \emptyset = \sin^{- 1} (\frac{a}{2 b})$ $soit θ = 2 γ - ϕ$ $A_{2} (γ) = b^{2} (2 γ - \sin γ \cos γ)$ $\frac{A_{2}}{A_{1}} = \frac{2 γ - \sin γ \cos γ}{2 \emptyset - \sin \emptyset \cos \emptyset}$ $= \frac{(θ + ϕ) - \sin (\frac{θ + \emptyset}{2}) \cos (\frac{θ + \emptyset}{2})}{2 \emptyset - \sin \emptyset \cos \emptyset}$ $= \frac{θ + \emptyset - \frac{1}{2} \sin (θ + \emptyset)}{2 \emptyset - \frac{a}{2 b} \times \sqrt{1 - \frac{a^{2}}{{4 b}^{2}}}}$ $= \frac{θ + \sin^{- 1} (\frac{a}{2 b}) - \frac{1}{2} [\frac{1}{2 b} \sin θ \sqrt{{4 b}^{2} - a^{2}} + \frac{bcos θ}{2 a}]}{{2 \sin}^{- 1} (\frac{a}{2 b}) - \frac{1}{4 a} \sqrt{{4 b}^{2} - a^{2}}} =$ $\frac{(θ + \emptyset) - \frac{1}{4} (\frac{b}{a} \cos θ) + (\frac{\sin θ}{b}) \sqrt{{4 b}^{2} - a^{2}}]}{{2 \sin}^{- 1} (\frac{a}{2 b}) - \frac{1}{4 a} \sqrt{{4 b}^{2} - a^{2}}}$ $\frac{A 2}{A_{1}} = \frac{1}{2} [\frac{(θ + ϕ) - \frac{1}{4} (\frac{bcos θ}{a} + \frac{\sin θ \sqrt{{4 b}^{2} - a^{2}}}{b})}{\emptyset - \frac{\sqrt{{4 b}^{2} - a^{2}}}{8 a}}]$ $\frac{A_{2}}{A_{1}} \max \Rightarrow \emptyset = \sin^{- 1} (\frac{a}{2 b}) \to \frac{1}{8 a} \sqrt{{4 b}^{2} - a^{2}} ?$ ${\begin{cases} γ = \frac{θ + \emptyset}{2} \\ \emptyset : given with (a, b) \end{cases}$ $. . . . . . . . . . . . . . . . . . . . ?$
	Commented by mr W last updated on 29/Apr/23

	$c a l c u l a t i o n o f A_{2} i s t o t a l l y w r o n g!$ $s h a p e o f A_{2} i s n o t a s e g m e n t o f c i r c l e!$
	Commented by mr W last updated on 29/Apr/23

	Commented by ajfour last updated on 29/Apr/23

	$T h a n k s s i r, i t h i n k i s h o u l d n o t$ $a t t e m p t .$
	Commented by ajfour last updated on 29/Apr/23

	Commented by ajfour last updated on 29/Apr/23
	$sin α=(a/b) A_1 =b^2 sin^(−1) (a/b)−a(√(b^2 −a^2 )) let C≡(−p, −(√(b^2 −a^2 )))≡(−p, −q) E≡(−p+acos θ, −q+asin θ) −p+acos θ=bcos φ −q+asin θ=bsin φ 2β=(π/2)+φ−α A_2 =(((p+a)(−q+asin θ))/2) +b^2 β−b^2 sin βcos β now asin θ=bsin φ+q ⇒ asin θ=q+bsin (2β+α−(π/2)) p+a=a(1+cos θ)−bcos φ =a(1+cos θ)−bcos (2β+α−(π/2)) =a+(√(a^2 −{q−bcos (2β+α)}^2 )) −bsin (2β+α) hence A_2 (β).$
	$\sin α = \frac{a}{b}$ $A_{1} = b^{2} \sin^{- 1} \frac{a}{b} - a \sqrt{b^{2} - a^{2}}$ $l e t C \equiv (- p, - \sqrt{b^{2} - a^{2}}) \equiv (- p, - q)$ $E \equiv (- p + a \cos θ, - q + a \sin θ)$ $- p + a \cos θ = b \cos ϕ$ $- q + a \sin θ = b \sin ϕ$ $2 β = \frac{π}{2} + ϕ - α$ $A_{2} = \frac{(p + a) (- q + a \sin θ)}{2}$ $+ b^{2} β - b^{2} \sin β \cos β$ $n o w a \sin θ = b \sin ϕ + q$ $\Rightarrow a \sin θ = q + b \sin (2 β + α - \frac{π}{2})$ $p + a = a (1 + \cos θ) - b \cos ϕ$ $= a (1 + \cos θ) - b \cos (2 β + α - \frac{π}{2})$ $= a + \sqrt{a^{2} - {q - b \cos (2 β + α)}^{2}}$ $- b \sin (2 β + α)$ $h e n c e A_{2} (β) .$
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