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Find the values of w and then x. Round lengths to the nearest tenth and angle measures to the nearest degree.

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you have multiple ways. one is to recognize that this is isoceles triangle ( the triangle shape with two equal sides)
i saw that to begin with

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Other answers:

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

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