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  • phi
what is angle B ?
  • phi
angle B= 180 - 24 can you find angle ACB knowing A is 16, and B is 156 ?
  • phi
remember 3 angles of a triangle add up to 180º

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Other answers:

  • phi
if you can find angle ACB, you can use the Law of Sines to find the side BC \[ \frac{BC}{\sin 16}= \frac{7600}{\sin( ACB)} \]
  • phi
can you be more specific ?
  • phi
yes, angle at C is 8º now use the Law of Sines to find the length of side BC
bc/sin16= 7600/ sin (16)(156)(8)
  • phi
**im not typing i dont know why its saying that** that's scary. what is this bc/sin16= 7600/ sin (16)(156)(8) ?? \[ \frac{BC}{\sin 16}= \frac{7600}{\sin( 8º)} \]
hah! iknow
  • phi
no, I was trying to say the angle at C (start at A go to C, then go to B) the Law of Sines see http://www.mathsisfun.com/algebra/trig-sine-law.html uses the ratio of the sin(angle) / side opposite in this case, we want to know side BC, so we need to know the angle across from BC. we do , it is 16º we know side AB= 7600, so we want to know the angle opposite that side. it is angle C=8
  • phi
if you work it through, you can write \[ \frac{BC}{\sin (16º)}= \frac{7600}{\sin( 8º)} \] multiply both sides by sin(16). \[ \frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º)\]
  • phi
on the left side, sin(16)/sin(16) "cancel" (become 1 so we can ignore) now get a calculator and figure out BC
  • phi
what were those 2's they are little º meaning "degrees"
  • phi
no. start with \[ \frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º) \]
  • phi
only on the left side you should remember that anything divided by itself is 1 so the left side \[ \frac{BC}{\sin (16º) }\cdot \sin(16º) = BC \cdot \frac{\sin(16)}{\sin(16)}= BC \cdot 1 = BC \]
  • phi
you need a calculator to figure out the right side
  • phi
yes, but the sin(16) is up top 7600*sin(16)/sin(8)
  • phi
yes. but I think they want to the nearest foot
  • phi
you can use the Law of Sines to do part (b)
  • phi
in triangle BCD , what side do you know ?
  • phi
are you looking at the same picture I am ? put you finger on B, go to C, down to D, over to B. that triangle. which of its 3 sides do you know ?
  • phi
yes, but what did you figure out in part (a) ?
  • phi
You found the length of side BC in part (a) BC= 15052.07
oh!
  • phi
do you know the angle that is opposite to side BC ?
no.. how do i find that?
  • phi
the angle opposite side BC is the angle that is not B and not C
im sorry but i just dont understand..
  • phi
what is the *angle* opposite side BC ? you have 3 choices: B , C or D. (and it's not B or C)
  • phi
|dw:1369784436661:dw|
i thought you meant side not angle
  • phi
that means we can write down \[ \frac{15052.07}{\sin(90)} = \] now you need a side (I would pick the one we are looking for, and its opposite angle)
right?
  • phi
yes, write CD over sin(24) and set that equal to the first ratio (up above) what do you get ?
  • phi
I don't know what that is. You should be writing down the equation that you get when you use the Law of Sines. In part (a) we used \[ \frac{BC}{\sin (16)}= \frac{7600}{\sin( 8)} \] your job is to write a similar equation for part (b) you already have both sides....
  • phi
*both sides of the equation \[ \frac{15052.07}{\sin(90)} = \] and ....
  • phi
yes, but use that in the equation... CD/sin(24) is part of the equation
  • phi
the idea is length/sin(angle) = length/ sin(angle)
  • phi
perfect. now "solve" for CD multiply both sides by sin(24) (this will "cancel" sin(24) on the right side)
  • phi
that is the right side of the equation. you should also show the other side. in other words, start with \[ \frac{15052.07}{\sin(90)} = \frac{CD}{\sin(24)} \] multiply both sides by sin(24). the right side will simplify to CD, just like you showed what do you get for the left side of the equation ?
  • phi
not after you multiply it by sin(24)
  • phi
\[ \frac{15052.07}{\sin(90)} = \frac{CD}{\sin(24)} \\ \frac{15052.07}{\sin(90)} \cdot \sin(24)= \frac{CD}{\sin(24)}\cdot \sin(24) \\\frac{15052.07}{\sin(90)} \cdot \sin(24)= CD \]
  • phi
now round to the nearest foot. that is the answer for part (b)
  • phi
This was a difficult problem. But it go faster if you try to learn some of this.
Thank you @phi! I really apperciate your help and patience

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