Respuesta :

[tex]\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means} \\\\\\ sin(\theta )=some\ value\qquad \textit{now, also bear in mind that} \\\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad % tangent tan(\theta)=\cfrac{opposite}{adjacent}\\\\ -------------------------------\\\\[/tex]

[tex]\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that} \\\\\\ sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]

[tex]\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a \\\\\\ \textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\ -------------------------------\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}} \\\\\\ \textit{and now, let's rationalize the denominator} \\\\\\ \cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21} [/tex]
ankitprmr2

[tex]\rm tan\theta = \dfrac{2}{\sqrt{21} }[/tex]

Step-by-step explanation:

Given :

[tex]\rm sin^-^1(\dfrac{2}{5})=\theta[/tex]

Calculation :

[tex]\rm sin\theta = \dfrac{perpendicular}{hypotenuse}[/tex]

According to Pythagorean theorem,

[tex]a^2+b^2=c^2[/tex]

here a = 2, c = 5. Then,

[tex]2^2+b^2=5^2[/tex]

[tex]b^2=25-4=21[/tex]

[tex]b=\pm\sqrt{21}[/tex]

Nothing is given in the question so we take positive

[tex]b = \sqrt{21}[/tex]

Therefore,

[tex]\rm tan\theta=\dfrac{perpendicular}{base}[/tex]

[tex]\rm tan\theta = \dfrac{2}{\sqrt{21} }[/tex]

For more information, refer the link given below

https://brainly.com/question/18163189?referrer=searchResults

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