anonymous
  • anonymous
What is the length of the part BC of the bridge?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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Diyadiya
  • Diyadiya
\[Cos \theta= \frac{adjacent \ side }{Hypotenuse}\] Can you try ?
anonymous
  • anonymous
Yeah I was hoping someone would give me a guide rather than a answer. Looking at the problem now

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Diyadiya
  • Diyadiya
\[\text {Adjacent side is the side adjacent to the angle here its BC}\]
anonymous
  • anonymous
so 1600/cos 30deg?
Diyadiya
  • Diyadiya
Nope \[Cos \theta= \frac{adjacent \ side }{Hypotenuse} \]We have to find adjacent side here \[\text{Adjacent side} = Cos\theta \times Hypotenuse \]
Diyadiya
  • Diyadiya
And you must know these common values http://cemca.org/ciet/Trigonometry/images/trig__mfc751a1.jpg
Diyadiya
  • Diyadiya
\[\text{did you understand how i got }\]\[\text{Adjacent side} = Cos\theta \times Hypotenuse\]
anonymous
  • anonymous
No, not at all actually. I dont understand half of any of this. I wasnt in class for half of the year and they just put me in here without any knowledge of geo.
Diyadiya
  • Diyadiya
Ok when we are given one of the sides of a triangle and an angle then you can find the other side using Trigonometric values like Sin,Cos,tan etc
anonymous
  • anonymous
Yeah, but what do Sin,Cos and Tan equal ?
Diyadiya
  • Diyadiya
\[\sin \theta= \frac{opposite \ side}{Hypotenuse}\]\[\cos \theta= \frac{adjacent \ side}{Hypotenuse}\]\[\tan \theta= \frac{sin\theta}{cos\theta }= \frac{opposite \ side}{adjacent \ side}\] \[Cosec \theta= \frac{1}{\sin \theta}=\frac{Hypotenuse}{Opposite \ side}\]\[Sec \theta= \frac{1}{\cos \theta}= \frac{Hypotenuse}{adjacent \ side}\]\[Cot \theta= \frac{1}{\tan \theta}= \frac{adjacent \ side }{opposite \ side }\]
Diyadiya
  • Diyadiya
http://www.mathsisfun.com/sine-cosine-tangent.html this might help you as well :)
Diyadiya
  • Diyadiya
Sohcahtoa an easy way to remember which side to divide by which.. Soh "S"ine = "O"pposite / "H"ypotenuse cah "C"osine = "A"djacent / "H"ypotenuse toa "T"angent = "O"pposite / "A"djacent And you must learn these common values http://cemca.org/ciet/Trigonometry/images/trig__mfc751a1.jpg
Diyadiya
  • Diyadiya
\[\text{Now you know }\]\[Cos \theta= \frac{adjacent \ side }{Hypotenuse}\] So how do you find adjacent side ?
Diyadiya
  • Diyadiya
If\(\large \ x=\frac{a}{b}\) then how do you find \(a\)? we mulltiply both sides by \(b\) \(\large xb=a\) same way\[Cos \theta= \frac{adjacent \ side }{Hypotenuse}\] So\[\text{Adjacent side} = Cos\theta \times Hypotenuse\] here \(\theta=30^o\)\[hypotenuse=1600m\]\[\cos 30^o=\frac{\sqrt{3}}{2}\] Now can you find ?
anonymous
  • anonymous
(sorry i was picking up my sister from day-care) and No i still dont understand it.
Diyadiya
  • Diyadiya
Its ok

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