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DanielleElmas [232]
3 years ago
14

Find x and y. please help​

Mathematics
1 answer:
anygoal [31]3 years ago
8 0

Answer:

x = 4 \sqrt{2}

y = 4 \sqrt{6}

Step-by-step explanation:

\sin 45 \degree =  \frac{x}{8}  \\  \\ x = 8  \sin 45 \degree \\  \\ x = 8 \times  \frac{1}{ \sqrt{2} }  \\  \\ x =  \frac{8 \sqrt{2} }{2}  \\  \\ x = 4 \sqrt{2}  \\  \\   \\  \tan 30 \degree =  \frac{x}{y}  \\  \\  \frac{1}{ \sqrt{3} }  =  \frac{4 \sqrt{2} }{y}  \\  \\ y = 4 \sqrt{6}

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Step-by-step explanation:

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SSSSS [86.1K]

Answer:

Part 4) r=84\ units

Part 9) sin(\theta)=-\frac{\sqrt{5}}{3}

Part 10) sin(\theta)=-\frac{9\sqrt{202}}{202}

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

C=2\pi r

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\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}

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cos^2(\theta)+sin^2(\theta)=1

we have

cos(\theta)=-\frac{2}{3}

substitute the given value

(-\frac{2}{3})^2+sin^2(\theta)=1

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square root both sides

sin(\theta)=\pm\frac{\sqrt{5}}{3}

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

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see the attached figure to better understand the problem

In the right triangle ABC of the figure

sin(\theta)=\frac{BC}{AC}

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substitute the given values

AC^2=11^2+9^2

AC^2=202

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so

sin(\theta)=\frac{9}{\sqrt{202}}

simplify

sin(\theta)=\frac{9\sqrt{202}}{202}

Remember that      

The point (11,-9) lies in Quadrant IV

then      

The value of sin(∅) is negative

therefore

sin(\theta)=-\frac{9\sqrt{202}}{202}

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3 years ago
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