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vampirchik [111]
3 years ago
10

I have 183 feet of fencing. The length of the court is 32 feet more than the width. One of the shorter sides will be 5 feet shor

ter than the other one to leave room for a gate. The 4th wall is used by the gym
Mathematics
1 answer:
Aliun [14]3 years ago
7 0
Okay so the total is 183. So we can say 183 = 
Let's call the width (the longer side) x.
We know that we have x, x + 32, and x - 5.
We can say 183 = x + (x + 32) + (x - 5).
Let's solve for x.
x = 52. 
So we know the length is 84 feet, the shorter side has 47 feet, and the longer side has 52 feet.
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First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \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=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill

\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}

\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}

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Sure What do u need help with I am always here
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