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Question Number 126700 by Eric002 last updated on 23/Dec/20

$$\theta ={sin}^{-1}\left(\frac{2}{5}\right)findcos\left(\theta \right)andtan\left(\theta \right)$$$$$$

Answered by akornes last updated on 23/Dec/20

$$cos\theta =\pm \frac{\sqrt{21}}{5}andtan\theta =\pm \frac{2\sqrt{21}}{21}$$

Commented by MJS_new last updated on 23/Dec/20

$$\mathrm{how}\mathrm{can}\mathrm{the}\cos \mathrm{and}\tan \mathrm{of}\mathrm{an}\mathrm{angle}\mathrm{in}\mathrm{the}$$$$\mathrm{first}\mathrm{quadrant}\mathrm{be}<0???$$

Answered by MJS_new last updated on 23/Dec/20

$$\theta ={\sin }^{-1}\frac{2}{5}\Rightarrow \theta \approx 23.58°$$$$\Rightarrow \cos \theta =\sqrt{1-{\left(\frac{2}{5}\right)}^2}=\frac{\sqrt{21}}{5};\tan \theta =\frac{\frac{2}{5}}{\sqrt{1-{\left(\frac{2}{5}\right)}^2}}=\frac{2\sqrt{21}}{21}$$

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