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Помогите! Можно просто ответ!

table123

Ответ:

$sina=\dfrac{3}{\sqrt{13}}\\\\\\a\in \Big(\ 0\ ;\ \dfrac{\pi}{2}\Big)\ \ \ \Rightarrow \ \ \ cosa>0\ ,\ tga>0\\\\\\cosa=+\sqrt{1-sin^2a}=\sqrt{1-\dfrac{9}{13}}=\sqrt{\dfrac{4}{13}}=\dfrac{2}{\sqrt{13}}\\\\\\tga=\dfrac{sina}{cosa}=\dfrac{3}{\sqrt{13}}:\dfrac{2}{\sqrt{13}}=\dfrac{3}{2}=1,5$

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NNNLLL54

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Ответ: 1,5

Объяснение:

cos^2a=1-sin^2a=1-9/13=4/13, cosa=2/V13 V- корень

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