Question Number 188720 by Rupesh123 last updated on 06/Mar/23
Commented by mr W last updated on 06/Mar/23
$${i}\:{think}\:{it}'{s}\:{impossible}. \\ $$
Commented by ajfour last updated on 06/Mar/23
Answered by a.lgnaoui last updated on 07/Mar/23
![△MOA tan α=(a/h) △MOB tan (2α)=((a+6)/h)=((2tan α)/(1−tanα^2 )) =(((2a)/h)/(1−((a/h))^2 ))=((2ah)/(h^2 −a^2 )) ⇒a^3 +6a^2 +ah^2 −6h^2 =0 ⇒(a=5) ⇒tan α=(5/(5(√(11))))=((√(11))/(11)). •tan 3α=tan (2α+α)=((11+c)/(5(√(11))))=((4(√(11)))/(11)) ⇒ c=9 △MOC ∡(((MOC)/2))=∡MOH=(α+(α/2)) MH⊕ON ⇒∡MOH=(π/2)−(α+(α/2)) θ=(π/2)−[(π/2)−(α+(α/2))]=((3α)/2)](https://www.tinkutara.com/question/Q188840.png)
$$\bigtriangleup{MOA}\:\:\:\mathrm{tan}\:\alpha=\frac{{a}}{{h}} \\ $$$$\bigtriangleup{MOB}\:\:\:\mathrm{tan}\:\left(\mathrm{2}\alpha\right)=\frac{{a}+\mathrm{6}}{{h}}=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}\alpha\:^{\mathrm{2}} } \\ $$$$=\frac{\frac{\mathrm{2}{a}}{{h}}}{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{ah}}{{h}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\boldsymbol{{a}}^{\mathrm{3}} +\mathrm{6}\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{a}}{h}^{\mathrm{2}} −\mathrm{6}{h}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\left(\boldsymbol{{a}}=\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{11}}}=\frac{\sqrt{\mathrm{11}}}{\mathrm{11}}. \\ $$$$\bullet\mathrm{tan}\:\mathrm{3}\alpha=\mathrm{tan}\:\left(\mathrm{2}\alpha+\alpha\right)=\frac{\mathrm{11}+{c}}{\mathrm{5}\sqrt{\mathrm{11}}}=\frac{\mathrm{4}\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{c}}=\mathrm{9} \\ $$$$\bigtriangleup{MOC}\:\:\: \\ $$$$\measuredangle\left(\frac{{MOC}}{\mathrm{2}}\right)=\measuredangle{MOH}=\left(\alpha+\frac{\alpha}{\mathrm{2}}\right) \\ $$$${MH}\oplus{ON}\:\:\Rightarrow\measuredangle{MOH}=\frac{\pi}{\mathrm{2}}−\left(\alpha+\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\theta=\frac{\pi}{\mathrm{2}}−\left[\frac{\pi}{\mathrm{2}}−\left(\alpha+\frac{\alpha}{\mathrm{2}}\right)\right]=\frac{\mathrm{3}\alpha}{\mathrm{2}} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 07/Mar/23
Commented by a.lgnaoui last updated on 07/Mar/23
$$\left({suite}\right) \\ $$$$\bigtriangleup{AOC} \\ $$$$\mathrm{tan}\:\mathrm{3}\alpha=\frac{\mathrm{5}+\mathrm{6}+\mathrm{9}}{\mathrm{5}\sqrt{\mathrm{11}}}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{11}}} \\ $$$$\Rightarrow\mathrm{3}\alpha={Arctg}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{11}}}\right)=\mathrm{85},\mathrm{5631}.. \\ $$$${soit}:\:\:\:\:\boldsymbol{\theta}=\mathrm{42},\mathrm{84} \\ $$$$ \\ $$
Commented by mr W last updated on 08/Mar/23
$${non}−{sense}! \\ $$$${this}\:{is}\:{impossible}: \\ $$
Commented by mr W last updated on 08/Mar/23
