Question-187482

Question Number 187482 by ajfour last updated on 17/Feb/23

Commented by ajfour last updated on 17/Feb/23

I f {c o n e}^{'} s \frac{R}{h} = m, g r e e n b a l l

r a d i u s = a . F i n d b t h e r a d i i o f

u p p e r b a l l s (e q u a l), a l l i n c o n t a c t

t h e w a y s h o w n .

Answered by a.lgnaoui last updated on 18/Feb/23

\sin θ = \frac{a}{c} c = {O O}_{1} = \frac{O H}{h}

\tan θ = \frac{R}{h} = m \Rightarrow c = \frac{a m}{\sqrt{1 + m^{2}}}

C o n e \sin β = \frac{R \frac{\sqrt{3}}{2}}{R - r \sin α} = \cos α

\sqrt{1 + m^{2}} = \frac{R \sqrt{3}}{2 R - 2 r \frac{m}{\sqrt{1 + m^{2}}}}

2 R \sqrt{1 + m^{2}} - 2 m r = R \sqrt{3}

r = \frac{R (2 \sqrt{1 + m^{2}} - \sqrt{3})}{2 m} .

Commented by a.lgnaoui last updated on 18/Feb/23

r = b

Commented by a.lgnaoui last updated on 18/Feb/23

Answered by mr W last updated on 18/Feb/23

Commented by mr W last updated on 19/Feb/23

\tan θ = \frac{R}{H} = m

A^{'} B = \frac{2 \sqrt{3} b}{3}

O A = \frac{a}{\sin θ}

{A A}^{'} = \sqrt{{(a + b)}^{2} - {(\frac{2 \sqrt{3} b}{3})}^{2}} = \sqrt{a^{2} + 2 a b - \frac{b^{2}}{3}}

{O A}^{'} = \sqrt{a^{2} + 2 a b - \frac{b^{2}}{3}} + \frac{a}{\sin θ}

O C = \frac{a}{\tan θ}

{C C}^{'} = \sqrt{{(a + b)}^{2} - {(b - a)}^{2}} = 2 \sqrt{a b}

{O C}^{'} = 2 \sqrt{a b} + \frac{a}{\tan θ}

{(\sqrt{a^{2} + 2 a b - \frac{b^{2}}{3}} + \frac{a}{\sin θ})}^{2} + {(\frac{2 \sqrt{3} b}{3})}^{2} = {(2 \sqrt{a b} + \frac{a}{\tan θ})}^{2} + b^{2}

\sqrt{a^{2} + 2 a b - \frac{b^{2}}{3}} = (b - a) \sin θ + 2 \sqrt{a b} \cos θ

b^{2} (\frac{4}{3} - \cos^{2} θ) - a^{2} \cos^{2} θ - 2 a b (2 - {3 \cos}^{2} θ) + 2 (b - a) \sqrt{a b} \sin 2 θ = 0

l e t λ = \frac{b}{a}

\Rightarrow (5 - 3 \cos 2 θ) λ^{2} - 6 (1 - 3 \cos 2 θ) λ + 12 \sin 2 θ (λ - 1) \sqrt{λ} - 3 (1 + \cos 2 θ) = 0

e x a m p l e s :

θ = 30 ° \Rightarrow \frac{b}{a} \approx 0.8974

θ = \sin^{- 1} \frac{1}{\sqrt{3}} \approx 35.264 ° \Rightarrow \frac{b}{a} = 1

θ = 60 ° \Rightarrow \frac{b}{a} \approx 1.6445

Commented by a.lgnaoui last updated on 18/Feb/23

c o u p e t r a n s v e r s a l e

Commented by ajfour last updated on 19/Feb/23

Thanks Sir , correct answers.

Answered by ajfour last updated on 18/Feb/23

Commented by ajfour last updated on 18/Feb/23

S e e s o l u t i o n a l s o i n Q . 187535

Commented by a.lgnaoui last updated on 18/Feb/23

M e r c i p o u r e x p l i c a t i o n p a r

g r a p h e .