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Question Number 102635 by ketto255 last updated on 10/Jul/20

Answered by bobhans last updated on 10/Jul/20

$$\sin ({90}^o-\theta )=\cos \theta =\pm \sqrt{1-\sin {}^2\theta }=$$$$\pm \sqrt{1-\frac{2}{25}}=\pm \frac{\sqrt{23}}{5}$$

Answered by Rasheed.Sindhi last updated on 10/Jul/20

$$\sin (90-\theta )=\cos \theta$$$${\sin }^2\theta +{\cos }^2\theta =1$$$${\left(\frac{\sqrt{2}}{5}\right)}^2+{\cos }^2\theta =1$$$$\cos \theta =\pm \sqrt{1-\frac{2}{25}}=\pm \frac{\sqrt{23}}{5}$$$$\sin (90-\theta )=\pm \frac{\sqrt{23}}{5}★$$

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