## BloomLocke367 one year ago In FGH, FH=9, FG=11 and angle F=65. What is the measure of angle G?

1. BloomLocke367

it's the same process of last time, right?

2. BloomLocke367

law of sines?

3. geerky42

|dw:1433257580394:dw| I don't think so. No pair of angle and its opposite side to use.

4. geerky42

let me think for a miin

5. BloomLocke367

okay

6. geerky42

I believe way to do it is to use law of cosines first, to solve for GH, then from there, we can use law of sines to solve for G. Good plan?

7. geerky42

$$g^2+h^2-2gh\cos F = f^2$$

8. BloomLocke367

I guess so.

9. BloomLocke367

give me some time to work that out

10. BloomLocke367

so far I have $$10\cos(65)=f^2$$, is that right so far?

11. geerky42

I have $$9^2+11^2-2(9)(11)\cos(65^\text o) = f^2$$

12. BloomLocke367

yes, then it's 81+121-192cos(65), right?

13. geerky42

81+121-198cos(65)

14. geerky42

Yes

15. BloomLocke367

okay, so now I have 10cos(65)=f^2. now what?

16. geerky42

that number next to cos, it's 198, not 192. And you don't add number together; $$a^2+b^2-2ab\cos\theta~~~\neq~~~(a^2+b^2-2ab)\cos\theta$$

17. BloomLocke367

I wrote it down wrong from my calculator XD thanks

18. BloomLocke367

so it's 4, not 10

19. geerky42

$\large a^2+b^2-2ab\cos\theta~~~\neq~~~(a^2+b^2-2ab)\cos\theta$ When calculating -2ab part, you need to multiply it by cos first before add them to other numbers

20. BloomLocke367

makes sense, order of operations. sorry.

21. geerky42

just completely calculate left side, what do you get?

22. BloomLocke367

118.32

23. BloomLocke367

@geerky42

24. BloomLocke367

so the angle is 10.9

25. geerky42

actually it's side opposite of angle F. it's 10.9 .

26. geerky42

WE calculated for side f.

27. BloomLocke367

ohhhhh, duh. XD

28. BloomLocke367

sorry XD

29. geerky42

NOW we can use law of sine here. $$\dfrac{\sin F}{f} = \dfrac{\sin G}{g}$$

30. BloomLocke367

sin65/10.9=sin(G)/9

31. BloomLocke367

9sin(65)=10.9sin(G)

32. geerky42

Or you can just multiply both sides y 9

33. BloomLocke367

$$\Large\frac{9\sin(65)}{10.9}=\sin(G)$$

34. geerky42

Yeah

35. BloomLocke367

then multiply it by arcsin, right?

36. BloomLocke367

I got 48.45

37. geerky42

right $\arcsin\left(\frac{9\sin(65)}{10.9}\right)=G$

38. geerky42

Same here

39. BloomLocke367

awesome, thanks for your help? Help with more? (sorry, I know I must be bothering you)

40. geerky42

I need to go soon. One more.