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3 years ago

For an angle 0 with the point (-20,-21) on its terminating side, what is the value of cosine ?

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

check the picture below.

$\bf (\stackrel{a}{-20},\stackrel{b}{-21})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{-20}\\ b=\stackrel{opposite}{-21}\\ \end{cases} \\\\\\ c=\sqrt{(-20)^2+(-21)^2}\implies c=\sqrt{841}\implies c=29 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-20}}{\stackrel{hypotenuse}{29}}~\hfill$

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2 years ago

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BaLLatris [955]

Answer:

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Step-by-step explanation:

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Apply both of these given to:

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$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

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$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

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ArbitrLikvidat [17]

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