Question-71428

Question Number 71428 by TawaTawa last updated on 15/Oct/19

Answered by ajfour last updated on 15/Oct/19

Commented by ajfour last updated on 15/Oct/19

P [r \cos (θ - ϕ), r \sin (θ - ϕ)]

\tan θ = \frac{r \sin (θ - ϕ)}{r - a}

\Rightarrow ϕ = θ - \sin^{- 1} [\frac{(r - a) \tan θ}{r}]

\Rightarrow \sin ϕ = \sin θ \sqrt{1 - {[\frac{(r - a) \tan θ}{r}]}^{2}} - \frac{(r - a) \sin θ}{r}

Q [- r \cos (ϕ + α), r \sin (ϕ + α)]

\tan α = \frac{r \sin (ϕ + α)}{a + r \cos (ϕ + α)}

\Rightarrow a \sin α + r \cos (ϕ + α) \sin α

= r \sin (ϕ + α) \cos α

\Rightarrow r \sin ϕ = a \sin α

\Rightarrow r {\sin θ \sqrt{1 - {[\frac{(r - a) \tan θ}{r}]}^{2}} - \frac{(r - a) \sin θ}{r}}

= a \sin α

\dots . .

Commented by TawaTawa last updated on 15/Oct/19

God bless you sir

Commented by ajfour last updated on 15/Oct/19

w h e r e t h e p u r p l e a n d b l u e a n g l e s

m e e t, i s t h a t m i d p o i n t o f r a d i u s ?

Commented by TawaTawa last updated on 15/Oct/19

Yes sir

Commented by ajfour last updated on 15/Oct/19

i w i s h s o m e o n e e l s e h e l p s . .

Commented by mind is power last updated on 15/Oct/19

let β = OAB

we have in OAQ

\frac{R}{\sin (α)} = \frac{a}{\sin (\emptyset)} . . 1

in OAB

\frac{a}{\sin (\emptyset)} = \frac{R}{\sin (β)} \dots 2

1 & 2 \Rightarrow \frac{a}{\sin (\emptyset)} = \frac{R}{\sin (β)} = \frac{R}{\sin (α)} \Rightarrow \sin (α) = \sin (β) \Rightarrow {\begin{cases} α = β \\ β = 180 - α \end{cases}

α = β \Rightarrow P = Q not possible \Rightarrow β = 180 - α

θ = PAB = 180 - β = 180 - (180 - α) = α

\Rightarrow α = θ

Commented by TawaTawa last updated on 15/Oct/19

God bless you sir

Commented by mind is power last updated on 16/Oct/19

y^{'} re welcom