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frutty [35]
3 years ago
10

Pls answer 2,3,5,6 and thank you for helping and pls work it out

Mathematics
1 answer:
skad [1K]3 years ago
7 0

Answer:

2: add 4

3: 6

5: multiply by 2

6: 7

Step-by-step explanation:

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anzhelika [568]

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ b=\stackrel{opposite}{2}\\ \end{cases} \\\\\\ \pm\sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a\implies \stackrel{I~Quadrant}{+\sqrt{21}=a}


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\bf cos(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}


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5 0
3 years ago
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