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lala2
help?!?
angle B= 180 - 24 can you find angle ACB knowing A is 16, and B is 156 ?
remember 3 angles of a triangle add up to 180º
if you can find angle ACB, you can use the Law of Sines to find the side BC \[ \frac{BC}{\sin 16}= \frac{7600}{\sin( ACB)} \]
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)
**im not typing i dont know why its saying that** that's scary. what is this bc/sin16= 7600/ sin (16)(156)(8) ?? \[ \frac{BC}{\sin 16}= \frac{7600}{\sin( 8º)} \]
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
if you work it through, you can write \[ \frac{BC}{\sin (16º)}= \frac{7600}{\sin( 8º)} \] multiply both sides by sin(16). \[ \frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º)\]
on the left side, sin(16)/sin(16) "cancel" (become 1 so we can ignore) now get a calculator and figure out BC
what were those 2's they are little º meaning "degrees"
no. start with \[ \frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º) \]
only on the left side you should remember that anything divided by itself is 1 so the left side \[ \frac{BC}{\sin (16º) }\cdot \sin(16º) = BC \cdot \frac{\sin(16)}{\sin(16)}= BC \cdot 1 = BC \]
you need a calculator to figure out the right side
yes, but the sin(16) is up top 7600*sin(16)/sin(8)
yes. but I think they want to the nearest foot
you can use the Law of Sines to do part (b)
in triangle BCD , what side do you know ?
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 ?
yes, but what did you figure out in part (a) ?
You found the length of side BC in part (a) BC= 15052.07
do you know the angle that is opposite to side BC ?
the angle opposite side BC is the angle that is not B and not C
im sorry but i just dont understand..
what is the *angle* opposite side BC ? you have 3 choices: B , C or D. (and it's not B or C)
i thought you meant side not angle
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)
yes, write CD over sin(24) and set that equal to the first ratio (up above) what do you get ?
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....
*both sides of the equation \[ \frac{15052.07}{\sin(90)} = \] and ....
yes, but use that in the equation... CD/sin(24) is part of the equation
the idea is length/sin(angle) = length/ sin(angle)
perfect. now "solve" for CD multiply both sides by sin(24) (this will "cancel" sin(24) on the right side)
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 ?
not after you multiply it by sin(24)
\[ \frac{15052.07}{\sin(90)} = \frac{CD}{\sin(24)} \\ \frac{15052.07}{\sin(90)} \cdot \sin(24)= \frac{CD}{\sin(24)}\cdot \sin(24) \\\frac{15052.07}{\sin(90)} \cdot \sin(24)= CD \]
now round to the nearest foot. that is the answer for part (b)
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