MaryJSmith
The remedy is experience.. And I don't have any..
Help please ?
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PaxPolaris
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First find the length of BD using △BCD
PaxPolaris
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using Pythagorean theorem
PaxPolaris
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\[BD^2=18^2+24^2\]\[\therefore BD=\sqrt{18^2+24^2}\]
(you can remove the common factor of 18 and 24 out to make the math easier)\[\implies BD = 6\sqrt{3^2+4^2}=6\sqrt{25}=\Large 30 ft.\]
MaryJSmith
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Thanks :)
PaxPolaris
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Then to find AD you need to know trigonometry ....OR at-least the ratios of the sides of 30-60-90 triangles.
MaryJSmith
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Do what ?
PaxPolaris
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|dw:1333147962445:dw|
In 30-60-90 triangles the ratios of the sides is always \[\Large 1:\sqrt3:2\]
MaryJSmith
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Cool... Now what do I do with that ?
PaxPolaris
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△ABD is a 30-60-90 triangle.... if you know one side(BD) you can find the others
MaryJSmith
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so do i do 30*the sqr rt of 3 ?
MaryJSmith
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That would be 52 (rounded up)
PaxPolaris
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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
MaryJSmith
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I'm confused again..
PaxPolaris
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\[\Large {BD \over AD} = \frac {\sqrt3} 1\]
MaryJSmith
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so would you do 30*the square root of 3 divided by 2 then ? That comes out to be 26 (rounded up)
MaryJSmith
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Dude , I'm so confused . Can you just give me the answer ? Lol . I'll ask my teacher about all this crap on Monday .
PaxPolaris
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\[{ 30 \over AD} = {\sqrt3 \over 1}\]
\[AD = {30 \over \sqrt 3} = \Large 10\sqrt3\ ft.\]
PaxPolaris
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17ish
MaryJSmith
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17ish ?
AsianDuck
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17.3 (3sf)
MaryJSmith
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thanks (what does 3sf mean ?)
AsianDuck
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3 significant figures (1, 7 and 3)
PaxPolaris
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ignore significant figures if you don't know what it means...
by seventeen-ish, I meant 10*sqrt3 is around 17
AsianDuck
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ok. 17.3 (1dp / 1 decimal point)