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Question-219293

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Question Number 219293 by Nicholas666 last updated on 22/Apr/25
Commented by Nicholas666 last updated on 22/Apr/25
find the angle x?
$${find}\:{the}\:{angle}\:{x}? \\ $$$$ \\ $$
Commented by Ghisom last updated on 23/Apr/25
87°
$$\mathrm{87}° \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
how do you find sir? I think this problem is quite difficult to solve
$${how}\:{do}\:{you}\:{find}\:{sir}? \\ $$$${I}\:{think}\:{this}\:{problem}\:{is}\:{quite}\:{difficult}\:{to}\:\:{solve} \\ $$$$ \\ $$
Commented by Frix last updated on 23/Apr/25
I assume 1. these are circles not ellipses and 2. the angle x has not the same value as the x in 7x and 8x. Otherwise we cannot solve it Further the angle x cannot be measured to the green circle line as shown in the picture but to the horizontal axis. My method: Because only the angle is asked for we can let R=1 ⇒ equation of the black circle is C: x^2 +y^2 =1 Small circle: c: (x−m)^2 +(y−r)^2 =r^2 C∩c has exactly 1 solution Further 7x+8x=15x=180° ⇒ x=12°, 8x=96° ⇒ C∩c=P= (((cos 96°)),((sin 84°)) ) Solving the system { ((x^2 +y^2 =1)),(((x−m)^2 +(y−r)^2 =r^2 )) :} 1. Substracting and solving for y=ax+b 2. Inserting in one of the equations and Solving Δ=0 for r, 0<r<(1/2) x^2 +px+q=0 ⇒ Δ=(p^2 /4)−q 3. Insert in y from above to get P= (((cos 96°)),((acos 96° +b)) ) 4. Solve acos 96° +b=sin 84° to get m, −1<m<0 5. Calculate the angle of the line PQ where Q= ((m),(0) ) I also get x=87°
$$\mathrm{I}\:\mathrm{assume}\:\mathrm{1}.\:\mathrm{these}\:\mathrm{are}\:\mathrm{circles}\:\mathrm{not}\:\mathrm{ellipses}\:\mathrm{and} \\ $$$$\mathrm{2}.\:\mathrm{the}\:\mathrm{angle}\:{x}\:\mathrm{has}\:\mathrm{not}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}\:\mathrm{as}\:\mathrm{the} \\ $$$${x}\:\mathrm{in}\:\mathrm{7}{x}\:\mathrm{and}\:\mathrm{8}{x}.\:\mathrm{Otherwise}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{Further}\:\mathrm{the}\:\mathrm{angle}\:{x}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{measured}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{green}\:\mathrm{circle}\:\mathrm{line}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{picture} \\ $$$$\mathrm{but}\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{axis}. \\ $$$$ \\ $$$$\mathrm{My}\:\mathrm{method}: \\ $$$$\mathrm{Because}\:\mathrm{only}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{is}\:\mathrm{asked}\:\mathrm{for}\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{let}\:{R}=\mathrm{1}\:\Rightarrow\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{black}\:\mathrm{circle}\:\mathrm{is} \\ $$$${C}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{Small}\:\mathrm{circle}: \\ $$$${c}:\:\left({x}−{m}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${C}\cap{c}\:\mathrm{has}\:\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution} \\ $$$$\mathrm{Further} \\ $$$$\mathrm{7}{x}+\mathrm{8}{x}=\mathrm{15}{x}=\mathrm{180}°\:\Rightarrow\:{x}=\mathrm{12}°,\:\mathrm{8}{x}=\mathrm{96}° \\ $$$$\Rightarrow\:{C}\cap{c}={P}=\begin{pmatrix}{\mathrm{cos}\:\mathrm{96}°}\\{\mathrm{sin}\:\mathrm{84}°}\end{pmatrix} \\ $$$$\mathrm{Solving}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}}\\{\left({x}−{m}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{1}.\:\mathrm{Substracting}\:\mathrm{and}\:\mathrm{solving}\:\mathrm{for}\:{y}={ax}+{b} \\ $$$$\mathrm{2}.\:\mathrm{Inserting}\:\mathrm{in}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{and} \\ $$$$\:\:\:\:\:\mathrm{Solving}\:\Delta=\mathrm{0}\:\mathrm{for}\:{r},\:\mathrm{0}<{r}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:\Rightarrow\:\Delta=\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q} \\ $$$$\mathrm{3}.\:\mathrm{Insert}\:\mathrm{in}\:{y}\:\mathrm{from}\:\mathrm{above} \\ $$$$\:\:\:\:\:\mathrm{to}\:\mathrm{get}\:{P}=\begin{pmatrix}{\mathrm{cos}\:\mathrm{96}°}\\{{a}\mathrm{cos}\:\mathrm{96}°\:+{b}}\end{pmatrix} \\ $$$$\mathrm{4}.\:\mathrm{Solve}\:{a}\mathrm{cos}\:\mathrm{96}°\:+{b}=\mathrm{sin}\:\mathrm{84}°\:\mathrm{to}\:\mathrm{get}\:{m},\:−\mathrm{1}<{m}<\mathrm{0} \\ $$$$\mathrm{5}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:{PQ}\:\mathrm{where} \\ $$$$\:\:\:\:\:{Q}=\begin{pmatrix}{{m}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{I}\:\mathrm{also}\:\mathrm{get}\:{x}=\mathrm{87}° \\ $$
Commented by Ghisom last updated on 23/Apr/25
let the angle 7x=α, there′s a formula x=((α/2)+45)° in this case α=7x=84° x=(((48)/2)+45)°=87°
$$\mathrm{let}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{7}{x}=\alpha,\: \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{formula} \\ $$$${x}=\left(\frac{\alpha}{\mathrm{2}}+\mathrm{45}\right)° \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\alpha=\mathrm{7}{x}=\mathrm{84}° \\ $$$${x}=\left(\frac{\mathrm{48}}{\mathrm{2}}+\mathrm{45}\right)°=\mathrm{87}° \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
thank you very much sir, you can really see something hidden
$${thank}\:{you}\:{very}\:{much}\:{sir}, \\ $$$${you}\:{can}\:{really}\:{see}\:{something}\:{hidden} \\ $$$$ \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
thank you
$${thank}\:{you} \\ $$
Answered by mr W last updated on 23/Apr/25
Commented by mr W last updated on 23/Apr/25
7x=Rα 8x=Rβ (α/β)=(7/8) α+β=π ⇒α=((7π)/(7+8))=((7π)/(15)) 2((π/2)−x)=(π/2)−α ⇒x=(1/2)((π/2)+α)=(1/2)((π/2)+((7π)/(15)))=((29π)/(60)) ✓ (=87°) ⇒R=((7x)/α)=((15x)/π)=((15)/π)×((29π)/(60))=((29)/4)
$$\mathrm{7}{x}={R}\alpha \\ $$$$\mathrm{8}{x}={R}\beta \\ $$$$\frac{\alpha}{\beta}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\alpha+\beta=\pi \\ $$$$\Rightarrow\alpha=\frac{\mathrm{7}\pi}{\mathrm{7}+\mathrm{8}}=\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+\frac{\mathrm{7}\pi}{\mathrm{15}}\right)=\frac{\mathrm{29}\pi}{\mathrm{60}}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\left(=\mathrm{87}°\right) \\ $$$$\Rightarrow{R}=\frac{\mathrm{7}{x}}{\alpha}=\frac{\mathrm{15}{x}}{\pi}=\frac{\mathrm{15}}{\pi}×\frac{\mathrm{29}\pi}{\mathrm{60}}=\frac{\mathrm{29}}{\mathrm{4}} \\ $$
Commented by Ghisom last updated on 23/Apr/25
yes!
$$\mathrm{yes}! \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
thank you for your solution professor
$${thank}\:{you}\:{for}\:{your}\:{solution}\:{professor} \\ $$

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