let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$

5 0

3 years ago

What is the median distance jumped 7.48, 7.60, 7.15, 7.44, 7.73, 7.72

irga5000 [103]

It would be 7.54 because you fist have to put the numbers ordered smallest to biggest amount 7.15, 7.44, 7.48,7.60, 7.72, 7.73 so the numbers that are in the middle are 7.48 and 7.60 you add them up, and the result you divide by 2 which gives you 7.54.

3 0

3 years ago

Need answers ASAP.

mafiozo [28]

Answer: If you have solved to get x=6 just plug that back into the problem. You get 50

Step-by-step explanation:

7 0

2 years ago

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