help?!?

- anonymous

help?!?

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- schrodinger

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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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## More 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

- anonymous

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) ??
\[ \frac{BC}{\sin 16}= \frac{7600}{\sin( 8º)} \]

- anonymous

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

- anonymous

oh!

- phi

do you know the angle that is opposite to side BC ?

- anonymous

no.. how do i find that?

- phi

the angle opposite side BC is the angle that is not B and not C

- anonymous

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|

- anonymous

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)

- anonymous

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.

- anonymous

Thank you @phi! I really apperciate your help and patience

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