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Question Number 173948 by dragan91 last updated on 21/Jul/22

Commented by dragan91 last updated on 21/Jul/22

Value of Angle ?

$$\mathrm{Value}\:\mathrm{of}\:\mathrm{Angle}\:? \\ $$

Commented by a.lgnaoui last updated on 22/Jul/22

2α+β=(π/2) and x+α=(π/2) cos 2α=(1/2) 2α=(π/3) α=(π/6) x=(π/2)−(π/6) x=(π/3)

$$\mathrm{2}\alpha+\beta=\frac{\pi}{\mathrm{2}}\:\:\:{and}\:\:\:\:\:{x}+\alpha=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\mathrm{2}\alpha=\frac{\pi}{\mathrm{3}}\:\:\alpha=\frac{\pi}{\mathrm{6}} \\ $$$${x}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{\pi}{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 22/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 23/Jul/22

how are you sure that cos 2α=(1/2)? 2α+β=(π/2) is not true! (wrongly marked in the question) you have ignored the dotted line.

$${how}\:{are}\:{you}\:{sure}\:{that}\:\mathrm{cos}\:\mathrm{2}\alpha=\frac{\mathrm{1}}{\mathrm{2}}? \\ $$$$\mathrm{2}\alpha+\beta=\frac{\pi}{\mathrm{2}}\:{is}\:{not}\:{true}! \\ $$$$\left({wrongly}\:{marked}\:{in}\:{the}\:{question}\right) \\ $$$${you}\:{have}\:{ignored}\:{the}\:{dotted}\:{line}. \\ $$

Answered by mr W last updated on 23/Jul/22

Commented by mr W last updated on 23/Jul/22

1^2 +(2−r)^2 =(1+r)^2 ⇒r=(2/3) tan θ=((2−r)/1)=(4/3)=((2 tan (θ/2))/(1−tan^2 (θ/2))) 2 tan^2 (θ/2)+3 tan (θ/2)−2=0 (2 tan (θ/2)−1)(tan (θ/2)+2)=0 ⇒tan (θ/2)=(1/2) ⇒sin (θ/2)=(1/( (√5))) 2α+θ=π ⇒α=(π/2)−(θ/2) 2β+((π/2)−θ)=π ⇒β=(π/4)+(θ/2) γ=π−α−β=π−(π/2)+(θ/2)−(π/4)−(θ/2)=(π/4) ⇒EF=DE=2−OD=2−2×1×cos α =2−2 cos ((π/2)−(θ/2)) =2−2 sin (θ/2) =2(1−(1/( (√5))))=((2((√5)−1))/( (√5))) tan x=((OE)/(EF))=((2×(√5))/(2((√5)−1)))=((5+(√5))/4) ⇒x=tan^(−1) ((5+(√5))/4)≈61.07°

$$\mathrm{1}^{\mathrm{2}} +\left(\mathrm{2}−{r}\right)^{\mathrm{2}} =\left(\mathrm{1}+{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}−{r}}{\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}} \\ $$$$\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}+\mathrm{3}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\mathrm{1}\right)\left(\mathrm{tan}\:\frac{\theta}{\mathrm{2}}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\mathrm{2}\alpha+\theta=\pi \\ $$$$\Rightarrow\alpha=\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{2}\beta+\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\pi \\ $$$$\Rightarrow\beta=\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}} \\ $$$$\gamma=\pi−\alpha−\beta=\pi−\frac{\pi}{\mathrm{2}}+\frac{\theta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{EF}={DE}=\mathrm{2}−{OD}=\mathrm{2}−\mathrm{2}×\mathrm{1}×\mathrm{cos}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}−\mathrm{2}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}−\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{tan}\:{x}=\frac{{OE}}{{EF}}=\frac{\mathrm{2}×\sqrt{\mathrm{5}}}{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{4}}\approx\mathrm{61}.\mathrm{07}° \\ $$

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