MEDAL+FAN,help please! In the figure, if CD = 15 cm, then BC ≈_____cm 6.6 11 21.6 34.7
https://cdn.ple.platoweb.com/EdAssets/8f746af703b945dc88cb674fd21d2a2d?ts=635369963867900000
Hi!! is this trig?
yup, well partly geometry but yea! :)
so the first two make no sense
Do you think you could help me out please?
yeah i think so
alright, well yea i looked at the angle so it has to be greater than 15
lets call CD \(x\)
no sorry CD = 15, lets call BC \(x\)
and \(AB\) lets call \(y\)
so \[\tan(40)=\frac{x}{y}\] or \[y=\frac{x}{\tan(40)}\]
tan40 =.8390996312
oops i am messing this up sorry \[\tan(40)=\frac{y}{x}\] so \[y=x\tan(40)\]
no worries, were human, we make mistakes ^^
also \[tan(70)=\frac{y}{15}\] so \[y=15\tan(70)\]
since it is the same \(y\) for both it must be the case that \[x\tan(40)=15\tan(70)\]
15tan(70) = 41.21216129
yeah i need to go back with pencil and paper
lol its okay, i get confused on just computer to
\[\tan(40)=\frac{y}{x}\]\[y=x\tan(40)\] \[\tan(70)=\frac{y}{15}\] \[y=15\tan(70)\]
so \[\tan(40)=\frac{15\tan(70)}{x}\]
getting the same answer as before \[x=\frac{15\tan(70)}{\tan(40)}\]
i got 130.093641
ok i see my mistake
\[\tan(40)=\frac{y}{15+x}\]
i think your the most human helper here, everyones like a robot that dont make mistakes
and \[y=15\tan(70)\] so \[tan(40)=\frac{15\tan(70)}{15+x}\]
we gotta solve that for \(x\) no problem
\[\frac{15+x}{15\tan(70)}=\frac{1}{\tan(40)}\] \[15+x=\frac{15\tan(70)}{\tan(40)}\] \[x=\frac{15\tan(70)}{\tan(40)}-15\]
i hope 34 is an answer choice http://www.wolframalpha.com/input/?i=15tan(70)%2Ftan(40)-15
34.7! lol
ok i guess that will do it unless i made another mistake
alright thank you! :)
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