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i saw that to begin with

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

pythagoras

half of \(x\) is \(\sqrt{102^2-68^2}\)

can i have a fraction in the numerator

so \(x=2\sqrt{102^2-68^2}\)

http://www.wolframalpha.com/input/?i=2sqrt%28102^2-68^2%29

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

\[\cos (42)\rightarrow 65*102\]

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.

|dw:1376262142492:dw|

cos42=x/204

:( math is really hard for me right now yall

x=204*cos42.

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).

151.6