Question

3tan^2theta - 4√3 tan theta + 3 = 0 find theta acute angle​

Answer 1

AnsWer :

$\underline{\dag\:\textsf{According \: to \: given \: in question:}}$

$\normalsize\ : \implies\sf\ 3tan^2\theta - 4\sqrt{3} + 3 = 0$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{ Using \: middle \: term \: factorization:}) }$

$\normalsize\ : \implies\sf\ 3tan^2\theta - 3\sqrt{3} - \sqrt{3} + 3 = 0 \\ \\ \normalsize\ : \implies\sf\ 3tan\theta(tan\theta - \sqrt{3}) - \sqrt{3}(tan\theta - \sqrt{3}) = 0$

$\normalsize\ : \implies\sf\underbrace{(tan\theta - \sqrt{3})}_{\orange{case \: \: 1} } - \underbrace{(3tan\theta - \sqrt{3})}_{\green{case \: \: 2} } = 0$

$\rule{100}2$

$\underline{\dag\:\textsf{Solving \: with \: case \: 1}}$

$\normalsize\ : \implies\sf\ tan\theta - \sqrt{3} = 0 \\ \\ \normalsize\ : \implies\sf\ tan\theta = \sqrt{3} \\ \\ \normalsize\ : \implies\sf\ tan\theta = tan60^{\circ}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{tan60^{\circ} = \sqrt{3}}) }$

$\normalsize\ : \implies\sf\cancel{tan} \theta = \cancel{tan} 60^{\circ} \\ \\ \normalsize\ : \implies{\boxed{\sf{\theta = 60^{\circ} }}}$

$\underline{\dag\:\textsf{ Solving \: with \: case \: 2}}$

$\normalsize\ : \implies\sf\ 3tan\theta - \sqrt{3} = 0$

$\normalsize\ : \implies\sf\ 3tan\theta = \sqrt{3}$

$\normalsize\ : \implies\sf\ tan\theta = \frac{\cancel{\sqrt{3}}}{\cancel{3}}$

$\normalsize\ : \implies\sf\ tan\theta = \frac{1}{\sqrt{3}} \\ \\ \normalsize\ : \implies\sf\ tan\theta = tan30^{\circ}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{tan30^{\circ} = \frac{1}{\sqrt{3}}}) }$

$\normalsize\ : \implies\sf\cancel{tan} \theta = \cancel{tan} 60^{\circ} \\ \\ \normalsize\ : \implies{\boxed{\sf{ \theta = 60^{\circ} }}}$

$\rule{100}2$

$\bullet$ The Value of $\bf\theta$ is either $\bf\ 60{^\circ}$ or $\bf\ 30^{\circ}$.

Answer 2

Given :

$\sf3tan^2\theta - 4\sqrt{3}tan\theta + 3 = 0$

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To find :

The value of theta

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Solution :

$\begin{lgathered}\normalsize\ \implies\sf\ 3tan^2\theta - 3\sqrt{3} - \sqrt{3} + 3 = 0 \\ \\ \normalsize\ \implies\sf\ 3tan\theta(tan\theta - \sqrt{3}) - \sqrt{3}(\tan\theta - \sqrt{3}) = 0\end{lgathered}$

$\normalsize\implies \sf\ (tan\theta-\sqrt{3})(3tan\theta-\sqrt{3})= 0$

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Now we have two cases

$\mathbb\pink{Case\:1}$

$\begin{lgathered}\normalsize\ \implies\sf\ tan\theta - \sqrt{3} = 0 \\ \\ \normalsize\ \implies\sf\ tan\theta = \sqrt{3} \\ \\ \normalsize\ \implies\sf\ tan\theta = tan60^{\circ}\end{lgathered}$

$\begin{lgathered}\normalsize\ \implies\sf{tan} \theta = {tan} 60^{\circ} \\ \\ \normalsize\ \implies{\boxed{\bf{\blue{\theta = 60^{\circ} }}}}\end{lgathered}$

$\mathbb\pink{Case\:2}$

$\normalsize\ \implies\sf\ 3tan\theta - \sqrt{3} = 0$

$\normalsize\ \implies\sf\ 3tan\theta = \sqrt{3}$

$\normalsize\ \implies\sf\ tan\theta = \dfrac{{\sqrt{3}}}{{3}}$

$\begin{lgathered}\normalsize\ \implies\sf\ tan\theta = \frac{1}{\sqrt{3}} \\ \\ \normalsize\ \implies\sf\ tan\theta = tan30^{\circ}\end{lgathered}$

$\begin{lgathered}\normalsize\ \implies\sf{tan} \theta = {tan} 60^{\circ} \\ \\ \normalsize\ \implies{\boxed{\bf{\blue{ \theta = 60^{\circ} }}}}\end{lgathered}$

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Answer :

The Value of theta is 60° or 30°

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Note :

$\sf{\: \: \: \: \: \: We\: know \:{tan30^{\circ} = \dfrac{1}{\sqrt{3}}} }$

$\sf{\: \: \: \: \: \: We\: know \:{tan60^{\circ} = \sqrt{3}} }$