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 [rcos (θ−φ), rsin (θ−φ)] tan θ=((rsin (θ−φ))/(r−a)) ⇒ φ=θ−sin^(−1) [(((r−a)tan θ)/r)] ⇒ sin φ=sin θ(√(1−[(((r−a)tan θ)/r)]^2 ))−(((r−a)sin θ)/r) Q[−rcos (φ+α), rsin (φ+α)] tan α=((rsin (φ+α))/(a+rcos (φ+α))) ⇒ asin α+rcos (φ+α)sin α =rsin (φ+α)cos α ⇒ rsin φ=asin α ⇒ r{sin θ(√(1−[(((r−a)tan θ)/r)]^2 ))−(((r−a)sin θ)/r)} = asin α .....

$${P}\:\left[{r}\mathrm{cos}\:\left(\theta−\phi\right),\:{r}\mathrm{sin}\:\left(\theta−\phi\right)\right] \\ $$$$\mathrm{tan}\:\theta=\frac{{r}\mathrm{sin}\:\left(\theta−\phi\right)}{{r}−{a}} \\ $$$$\Rightarrow\:\:\phi=\theta−\mathrm{sin}^{−\mathrm{1}} \left[\frac{\left({r}−{a}\right)\mathrm{tan}\:\theta}{{r}}\right] \\ $$$$\Rightarrow\:\mathrm{sin}\:\phi=\mathrm{sin}\:\theta\sqrt{\mathrm{1}−\left[\frac{\left({r}−{a}\right)\mathrm{tan}\:\theta}{{r}}\right]^{\mathrm{2}} }−\frac{\left({r}−{a}\right)\mathrm{sin}\:\theta}{{r}} \\ $$$${Q}\left[−{r}\mathrm{cos}\:\left(\phi+\alpha\right),\:{r}\mathrm{sin}\:\left(\phi+\alpha\right)\right] \\ $$$$\mathrm{tan}\:\alpha=\frac{{r}\mathrm{sin}\:\left(\phi+\alpha\right)}{{a}+{r}\mathrm{cos}\:\left(\phi+\alpha\right)} \\ $$$$\Rightarrow\:{a}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\left(\phi+\alpha\right)\mathrm{sin}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:={r}\mathrm{sin}\:\left(\phi+\alpha\right)\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:\:{r}\mathrm{sin}\:\phi={a}\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\:{r}\left\{\mathrm{sin}\:\theta\sqrt{\mathrm{1}−\left[\frac{\left({r}−{a}\right)\mathrm{tan}\:\theta}{{r}}\right]^{\mathrm{2}} }−\frac{\left({r}−{a}\right)\mathrm{sin}\:\theta}{{r}}\right\} \\ $$$$\:\:\:\:=\:{a}\mathrm{sin}\:\alpha \\ $$$$….. \\ $$

Commented by TawaTawa last updated on 15/Oct/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by ajfour last updated on 15/Oct/19

$${where}\:{the}\:{purple}\:{and}\:{blue}\:{angles} \\ $$$${meet},\:{is}\:{that}\:{midpoint}\:{of}\:{radius}? \\ $$

Commented by TawaTawa last updated on 15/Oct/19

$$\mathrm{Yes}\:\mathrm{sir} \\ $$

Commented by ajfour last updated on 15/Oct/19

$${i}\:{wish}\:{someone}\:{else}\:{helps}.. \\ $$

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

let β=OAB we have in OAQ (R/(sin(α)))=(a/(sin(∅)))..1 in OAB (a/(sin(∅)))=(R/(sin(β)))...2 1&2⇒(a/(sin(∅)))=(R/(sin(β)))=(R/(sin(α)))⇒sin(α)=sin(β)⇒ { ((α=β)),((β=180−α)) :} α=β⇒P=Q not possible⇒β=180−α θ=PAB=180−β=180−(180−α)=α ⇒α=θ

$$\mathrm{let}\:\beta=\mathrm{OAB} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{in}\:\mathrm{OAQ} \\ $$$$\frac{\mathrm{R}}{\mathrm{sin}\left(\alpha\right)}=\frac{\mathrm{a}}{\mathrm{sin}\left(\emptyset\right)}..\mathrm{1} \\ $$$$\mathrm{in}\:\mathrm{OAB} \\ $$$$\frac{\mathrm{a}}{\mathrm{sin}\left(\emptyset\right)}=\frac{\mathrm{R}}{\mathrm{sin}\left(\beta\right)}…\mathrm{2} \\ $$$$\mathrm{1\&2}\Rightarrow\frac{\mathrm{a}}{\mathrm{sin}\left(\emptyset\right)}=\frac{\mathrm{R}}{\mathrm{sin}\left(\beta\right)}=\frac{\mathrm{R}}{\mathrm{sin}\left(\alpha\right)}\Rightarrow\mathrm{sin}\left(\alpha\right)=\mathrm{sin}\left(\beta\right)\Rightarrow\begin{cases}{\alpha=\beta}\\{\beta=\mathrm{180}−\alpha}\end{cases} \\ $$$$\alpha=\beta\Rightarrow\mathrm{P}=\mathrm{Q}\:\mathrm{not}\:\mathrm{possible}\Rightarrow\beta=\mathrm{180}−\alpha \\ $$$$\theta=\mathrm{PAB}=\mathrm{180}−\beta=\mathrm{180}−\left(\mathrm{180}−\alpha\right)=\alpha \\ $$$$\Rightarrow\alpha=\theta \\ $$$$ \\ $$

Commented by TawaTawa last updated on 15/Oct/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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

$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$$$ \\ $$

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