Question Number 213519 by RojaTaniya last updated on 07/Nov/24
Answered by mr W last updated on 07/Nov/24
$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{3}}{{x}} \\ $$$$\mathrm{tan}\:\left(\alpha+\alpha+\beta\right)=\frac{\mathrm{6}}{{x}} \\ $$$$\frac{\mathrm{6}}{{x}}=\frac{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{3}}{{x}}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}×\frac{\mathrm{3}}{{x}}}=\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{3}} \\ $$$${x}^{\mathrm{2}} =\mathrm{9}\:\Rightarrow{x}=\mathrm{3} \\ $$
Commented by RojaTaniya last updated on 07/Nov/24
$$\:{Sir},\:{perferct}.\:{Thanks}. \\ $$
Answered by A5T last updated on 07/Nov/24
![[ABC]=((3x(√(x^2 +1 ))sinα)/2)=(((√(x^2 +9))×(√(x^2 +36)) sinα)/2) ⇒9x^2 (x^2 +1)=(x^2 +9)(x^2 +36) ⇒9x^4 +9x^2 =x^4 +45x^2 +324⇒2x^4 −9x^2 −81=0 ⇒(x^2 −9)(2x^2 +9)=0⇒x^2 =9⇒x=3](https://www.tinkutara.com/question/Q213533.png)
$$\left[{ABC}\right]=\frac{\mathrm{3}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}\:}{sin}\alpha}{\mathrm{2}}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}×\sqrt{{x}^{\mathrm{2}} +\mathrm{36}}\:{sin}\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)=\left({x}^{\mathrm{2}} +\mathrm{9}\right)\left({x}^{\mathrm{2}} +\mathrm{36}\right) \\ $$$$\Rightarrow\mathrm{9}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{2}} ={x}^{\mathrm{4}} +\mathrm{45}{x}^{\mathrm{2}} +\mathrm{324}\Rightarrow\mathrm{2}{x}^{\mathrm{4}} −\mathrm{9}{x}^{\mathrm{2}} −\mathrm{81}=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{9}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}\right)=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} =\mathrm{9}\Rightarrow{x}=\mathrm{3} \\ $$