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Help me please! I literally have no idea what to do Attachment~
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after determing the length of the water hose; you would have to be familiar with properties of the 30-60-90 rt triangle
|dw:1331057342874:dw|
Find the length of the line B^D, which is sqrt(24^2+18^2) then, as this is the base to the triangle ABD, the length AB satisfies the following trig. equation: BD=ABcos(30) as 30 degrees is the angle between BD and AB. hence AB is given by AB=BD/(cos(30)
\[\frac{\sqrt{3}}{wh}=\frac{ad}{1}\]
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might have that a little off ...
|dw:1331057558232:dw|
Goodness me, I'm terrible, The length AD required is given by AD=BD/tan(60)
wh is to sqrt(3) as ad is to 1 wh/sqrt(3) = ad
yep :)
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Oh my! THANK YOU BOTH SO MUCH!
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