anonymous
  • anonymous
I am really stuck someone help
Geometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Find the value of x. Round to the nearest tenth. The diagram is not drawn to scale.
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anonymous
  • anonymous
Trigonometric functions are Sin, Cos, and Tan defined as: Sin=opposite/hypotenuse Cos=adjacent/hypotenuse Tan=opposite/adjacent Which of these three is best suited for the diagram? (which pieces do you know?)
anonymous
  • anonymous
sin?

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anonymous
  • anonymous
Cos, because we know the length of the leg adjacent to our angle(11) and we want to find the hypotenuse. So to solve this we setup the problem as follows: \[\cos(24)=\frac{11}{x}\]
anonymous
  • anonymous
then we solve for x, right?...
anonymous
  • anonymous
Yeah, we move x over to the other side by multiplying both sides by x. What happens when you evaluate: \[x(\cos(24))=(\frac{11}{x})*x\]
anonymous
  • anonymous
Yes, but be careful when multiplying trig functions. cos(24x) is very different from cos(24)*x or x*cos(24). So we have x*cos(24)=11. How do we get x by itself?
anonymous
  • anonymous
Multiply what by both sides? we have \[x*\cos(24)=11\] How do get x by itself?
anonymous
  • anonymous
What should we divide by?
anonymous
  • anonymous
24 is a part of the cos function, so they're tied together. Divide by cos(24) and what do you end up with?
anonymous
  • anonymous
Not quite. \[\frac{1}{\cos(24)}*x*\cos(24)=11*\frac{1}{\cos(24)}\]
anonymous
  • anonymous
It's okay. What happens when we simplify that expression? Can we cancel anything out?
anonymous
  • anonymous
Yes! Exactly. The cos(24) cancels out on the left leaving us with x=11/cos(24).
anonymous
  • anonymous
Mmhmm. If you're allowed a calculator or you're asked to give an actual number, just evaluate in your calculator and you get 12.04, rounded. If you're just asked to find x, I would say leave it exact in the form of 11/cos(24)
anonymous
  • anonymous
Yeah, just post another question and tag me in a comment.
anonymous
  • anonymous
New question. Where it says "ask a question..."

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