highschoolmom2010
  • highschoolmom2010
Find the values of w and then x. Round lengths to the nearest tenth and angle measures to the nearest degree.
Mathematics
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highschoolmom2010
  • highschoolmom2010
Find the values of w and then x. Round lengths to the nearest tenth and angle measures to the nearest degree.
Mathematics
jamiebookeater
  • jamiebookeater
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highschoolmom2010
  • highschoolmom2010
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anonymous
  • anonymous
you have multiple ways. one is to recognize that this is isoceles triangle ( the triangle shape with two equal sides)
highschoolmom2010
  • highschoolmom2010
i saw that to begin with

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anonymous
  • anonymous
which makes the right triangle have internal sum of 90+ 42+w= 180.
anonymous
  • anonymous
or does w mean the height?
highschoolmom2010
  • highschoolmom2010
i believe w is height
highschoolmom2010
  • highschoolmom2010
im not entirely sure
anonymous
  • anonymous
then sin42= w/102
anonymous
  • anonymous
and cos 42= (2/x)/102
anonymous
  • anonymous
cos42= (x/2)/102****
highschoolmom2010
  • highschoolmom2010
\[(102)*\sin 42=\frac{ w }{ 102 }*102\] \[102 \sin 42=w\] w\[\approx 68\]
highschoolmom2010
  • highschoolmom2010
so how now that i have found w how do i get x?
anonymous
  • anonymous
cos42= (x/2)/102
anonymous
  • anonymous
pythagoras
anonymous
  • anonymous
half of \(x\) is \(\sqrt{102^2-68^2}\)
highschoolmom2010
  • highschoolmom2010
can i have a fraction in the numerator
anonymous
  • anonymous
so \(x=2\sqrt{102^2-68^2}\)
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=2sqrt%28102^2-68^2%29
jdoe0001
  • jdoe0001
@highschoolmom2010 you can always just use the cosine of 42 degrees you know don't spare functions with your SOH CAH TOA
anonymous
  • anonymous
Satelite using wolfram alpha. LOL hes definitely not highschool student.
anonymous
  • anonymous
cos42=x/204= (x/2)/102 (the one i typed few mins ago)
highschoolmom2010
  • highschoolmom2010
^^confused -_-
highschoolmom2010
  • highschoolmom2010
@ganeshie8 any help
jdoe0001
  • jdoe0001
\(\bf sin(42^o) = \cfrac{\textit{opposite side}}{\textrm{hypotenuse}} \implies \cfrac{w}{102} \implies \color{blue}{sin(42^o) \times 102} = w\\ cos(42^o) = \cfrac{\textit{adjacent side}}{\textrm{hypotenuse}} \implies \cfrac{\textit{half of "x"}}{102} \implies \color{green}{cos(42^o) \times 102} = \textit{half of "x"}\)
highschoolmom2010
  • highschoolmom2010
\[\cos (42)\rightarrow 65*102\]
jdoe0001
  • jdoe0001
well, if "half of x" = 65, yes
highschoolmom2010
  • highschoolmom2010
ive done confused myself again
jdoe0001
  • jdoe0001
then again, if you already know that x= 65 * 2, you don't have to do the cosine part :)
highschoolmom2010
  • highschoolmom2010
aw heck imma use @satellite73 s method
highschoolmom2010
  • highschoolmom2010
\[\sqrt{108^2+68^2}=108-68=40\] \[\sqrt{40}=2\sqrt{10}\]
highschoolmom2010
  • highschoolmom2010
but that doesnt look right
jdoe0001
  • jdoe0001
using the cosine, I get ahemm, something else for "x"
highschoolmom2010
  • highschoolmom2010
i dont understand now to use it with this problem
highschoolmom2010
  • highschoolmom2010
for x
anonymous
  • anonymous
okay wild katherine has appeared.
highschoolmom2010
  • highschoolmom2010
|dw:1376262142492:dw|
anonymous
  • anonymous
cos42=x/204
highschoolmom2010
  • highschoolmom2010
:( math is really hard for me right now yall
anonymous
  • anonymous
x=204*cos42.
highschoolmom2010
  • highschoolmom2010
ya know that makes it a whole lot easier
anonymous
  • anonymous
now the number should be above 150, which is so much more amazing than just sqrt(40).
highschoolmom2010
  • highschoolmom2010
151.6
highschoolmom2010
  • highschoolmom2010

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