The remedy is experience.. And I don't have any.. Help please ?

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The remedy is experience.. And I don't have any.. Help please ?

Mathematics
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First find the length of BD using △BCD
using Pythagorean theorem
\[BD^2=18^2+24^2\]\[\therefore BD=\sqrt{18^2+24^2}\] (you can remove the common factor of 18 and 24 out to make the math easier)\[\implies BD = 6\sqrt{3^2+4^2}=6\sqrt{25}=\Large 30 ft.\]

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Thanks :)
Then to find AD you need to know trigonometry ....OR at-least the ratios of the sides of 30-60-90 triangles.
Do what ?
|dw:1333147962445:dw| In 30-60-90 triangles the ratios of the sides is always \[\Large 1:\sqrt3:2\]
Cool... Now what do I do with that ?
△ABD is a 30-60-90 triangle.... if you know one side(BD) you can find the others
so do i do 30*the sqr rt of 3 ?
That would be 52 (rounded up)
AD is smaller than BD... ...If the side opposite the 30deg (smallest) angle is length 1(smallest side) ..........then the side opposite the 60deg angle is sqrt3
I'm confused again..
\[\Large {BD \over AD} = \frac {\sqrt3} 1\]
so would you do 30*the square root of 3 divided by 2 then ? That comes out to be 26 (rounded up)
Dude , I'm so confused . Can you just give me the answer ? Lol . I'll ask my teacher about all this crap on Monday .
\[{ 30 \over AD} = {\sqrt3 \over 1}\] \[AD = {30 \over \sqrt 3} = \Large 10\sqrt3\ ft.\]
17ish
17ish ?
17.3 (3sf)
thanks (what does 3sf mean ?)
3 significant figures (1, 7 and 3)
ignore significant figures if you don't know what it means... by seventeen-ish, I meant 10*sqrt3 is around 17
ok. 17.3 (1dp / 1 decimal point)

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