BloomLocke367
  • BloomLocke367
In FGH, FH=9, FG=11 and angle F=65. What is the measure of angle G?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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BloomLocke367
  • BloomLocke367
it's the same process of last time, right?
BloomLocke367
  • BloomLocke367
law of sines?
geerky42
  • geerky42
|dw:1433257580394:dw| I don't think so. No pair of angle and its opposite side to use.

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More answers

geerky42
  • geerky42
let me think for a miin
BloomLocke367
  • BloomLocke367
okay
geerky42
  • 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?
geerky42
  • geerky42
\(g^2+h^2-2gh\cos F = f^2\)
BloomLocke367
  • BloomLocke367
I guess so.
BloomLocke367
  • BloomLocke367
give me some time to work that out
BloomLocke367
  • BloomLocke367
so far I have \(10\cos(65)=f^2\), is that right so far?
geerky42
  • geerky42
I have \(9^2+11^2-2(9)(11)\cos(65^\text o) = f^2\)
BloomLocke367
  • BloomLocke367
yes, then it's 81+121-192cos(65), right?
geerky42
  • geerky42
81+121-198cos(65)
geerky42
  • geerky42
Yes
BloomLocke367
  • BloomLocke367
okay, so now I have 10cos(65)=f^2. now what?
geerky42
  • geerky42
that number next to cos, it's 198, not 192. And you don't add number together; \(a^2+b^2-2ab\cos\theta~~~\neq~~~(a^2+b^2-2ab)\cos\theta\)
BloomLocke367
  • BloomLocke367
I wrote it down wrong from my calculator XD thanks
BloomLocke367
  • BloomLocke367
so it's 4, not 10
geerky42
  • 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
BloomLocke367
  • BloomLocke367
makes sense, order of operations. sorry.
geerky42
  • geerky42
just completely calculate left side, what do you get?
BloomLocke367
  • BloomLocke367
118.32
BloomLocke367
  • BloomLocke367
@geerky42
BloomLocke367
  • BloomLocke367
so the angle is 10.9
geerky42
  • geerky42
actually it's side opposite of angle F. it's 10.9 .
geerky42
  • geerky42
WE calculated for side f.
BloomLocke367
  • BloomLocke367
ohhhhh, duh. XD
BloomLocke367
  • BloomLocke367
sorry XD
geerky42
  • geerky42
NOW we can use law of sine here. \(\dfrac{\sin F}{f} = \dfrac{\sin G}{g}\)
BloomLocke367
  • BloomLocke367
sin65/10.9=sin(G)/9
BloomLocke367
  • BloomLocke367
9sin(65)=10.9sin(G)
geerky42
  • geerky42
Or you can just multiply both sides y 9
BloomLocke367
  • BloomLocke367
\(\Large\frac{9\sin(65)}{10.9}=\sin(G)\)
geerky42
  • geerky42
Yeah
BloomLocke367
  • BloomLocke367
then multiply it by arcsin, right?
BloomLocke367
  • BloomLocke367
I got 48.45
geerky42
  • geerky42
right \[\arcsin\left(\frac{9\sin(65)}{10.9}\right)=G\]
geerky42
  • geerky42
Same here
BloomLocke367
  • BloomLocke367
awesome, thanks for your help? Help with more? (sorry, I know I must be bothering you)
geerky42
  • geerky42
I need to go soon. One more.

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