## lala2 Group Title help?!? one year ago one year ago

1. phi Group Title

what is angle B ?

2. phi Group Title

angle B= 180 - 24 can you find angle ACB knowing A is 16, and B is 156 ?

3. phi Group Title

remember 3 angles of a triangle add up to 180º

4. phi Group Title

if you can find angle ACB, you can use the Law of Sines to find the side BC $\frac{BC}{\sin 16}= \frac{7600}{\sin( ACB)}$

5. phi Group Title

can you be more specific ?

6. phi Group Title

yes, angle at C is 8º now use the Law of Sines to find the length of side BC

7. lala2 Group Title

bc/sin16= 7600/ sin (16)(156)(8)

8. phi Group Title

**im not typing i dont know why its saying that** that's scary. what is this bc/sin16= 7600/ sin (16)(156)(8) ?? $\frac{BC}{\sin 16}= \frac{7600}{\sin( 8º)}$

9. lala2 Group Title

hah! iknow

10. phi Group Title

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

11. phi Group Title

if you work it through, you can write $\frac{BC}{\sin (16º)}= \frac{7600}{\sin( 8º)}$ multiply both sides by sin(16). $\frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º)$

12. phi Group Title

on the left side, sin(16)/sin(16) "cancel" (become 1 so we can ignore) now get a calculator and figure out BC

13. phi Group Title

what were those 2's they are little º meaning "degrees"

14. phi Group Title

no. start with $\frac{BC}{\sin (16º) }\cdot \sin(16º)= \frac{7600}{\sin( 8º)} \cdot \sin(16º)$

15. phi Group Title

only on the left side you should remember that anything divided by itself is 1 so the left side $\frac{BC}{\sin (16º) }\cdot \sin(16º) = BC \cdot \frac{\sin(16)}{\sin(16)}= BC \cdot 1 = BC$

16. phi Group Title

you need a calculator to figure out the right side

17. phi Group Title

yes, but the sin(16) is up top 7600*sin(16)/sin(8)

18. phi Group Title

yes. but I think they want to the nearest foot

19. phi Group Title

you can use the Law of Sines to do part (b)

20. phi Group Title

in triangle BCD , what side do you know ?

21. phi Group Title

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 ?

22. phi Group Title

yes, but what did you figure out in part (a) ?

23. phi Group Title

You found the length of side BC in part (a) BC= 15052.07

24. lala2 Group Title

oh!

25. phi Group Title

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

26. lala2 Group Title

no.. how do i find that?

27. phi Group Title

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

28. lala2 Group Title

im sorry but i just dont understand..

29. phi Group Title

what is the *angle* opposite side BC ? you have 3 choices: B , C or D. (and it's not B or C)

30. phi Group Title

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31. lala2 Group Title

i thought you meant side not angle

32. phi Group Title

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)

33. lala2 Group Title

right?

34. phi Group Title

yes, write CD over sin(24) and set that equal to the first ratio (up above) what do you get ?

35. phi Group Title

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....

36. phi Group Title

*both sides of the equation $\frac{15052.07}{\sin(90)} =$ and ....

37. phi Group Title

yes, but use that in the equation... CD/sin(24) is part of the equation

38. phi Group Title

the idea is length/sin(angle) = length/ sin(angle)

39. phi Group Title

perfect. now "solve" for CD multiply both sides by sin(24) (this will "cancel" sin(24) on the right side)

40. phi Group Title

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 ?

41. phi Group Title

not after you multiply it by sin(24)

42. phi Group Title

$\frac{15052.07}{\sin(90)} = \frac{CD}{\sin(24)} \\ \frac{15052.07}{\sin(90)} \cdot \sin(24)= \frac{CD}{\sin(24)}\cdot \sin(24) \\\frac{15052.07}{\sin(90)} \cdot \sin(24)= CD$

43. phi Group Title

now round to the nearest foot. that is the answer for part (b)

44. phi Group Title

This was a difficult problem. But it go faster if you try to learn some of this.

45. lala2 Group Title

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