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

1. phi

what is angle B ?

2. phi

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

3. phi

remember 3 angles of a triangle add up to 180º

4. phi

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

5. phi

can you be more specific ?

6. phi

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

7. lala2

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

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

9. lala2

hah! iknow

10. 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

11. phi

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

12. phi

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

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

14. phi

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

15. phi

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

16. phi

you need a calculator to figure out the right side

17. phi

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

18. phi

yes. but I think they want to the nearest foot

19. phi

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

20. phi

in triangle BCD , what side do you know ?

21. 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 ?

22. phi

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

23. phi

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

24. lala2

oh!

25. phi

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

26. lala2

no.. how do i find that?

27. phi

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

28. lala2

im sorry but i just dont understand..

29. phi

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

30. phi

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31. lala2

i thought you meant side not angle

32. 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)

33. lala2

right?

34. phi

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

35. 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....

36. phi

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

37. phi

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

38. phi

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

39. phi

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

40. 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 ?

41. phi

not after you multiply it by sin(24)

42. phi

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

43. phi

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

44. phi

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

45. lala2

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