anonymous
  • anonymous
I need help on this, please? The vertices of a triangle are A(7, 5), B(4, 2), and C(9, 2). What is m? A. 30° B. 45° C. 56.31° D. 78.69°
Geometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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misty1212
  • misty1212
HI!!
anonymous
  • anonymous
hey
misty1212
  • misty1212
that is a good question, what is \(m\)? not clear at all since none of the vertices are labelled "m"

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anonymous
  • anonymous
misty1212
  • misty1212
here is a nice picture of your triangle i don't see any m in it though http://www.wolframalpha.com/input/?i=triangle+%287%2C+5%29%2C+%284%2C+2%29%2C+%289%2C+2%29
anonymous
  • anonymous
yeah, that doesn't help me.. I don't even understand what the question is tell me nor Geometry it self?
misty1212
  • misty1212
look carefully at your question you are given the vertices of a triangle, with points labelled A, B, C but then you ask for an angle m there is no m in sight, only A, B, C
anonymous
  • anonymous
but thats all it gives me?? hmm?
misty1212
  • misty1212
m usually stands for the measure of some angle but that would be "find m
jim_thompson5910
  • jim_thompson5910
\[\Large m\angle ABC\] means "angle ABC" where B is the vertex of the angle something like this |dw:1436923873975:dw|
anonymous
  • anonymous
Here's the question, maybe this is better?
jim_thompson5910
  • jim_thompson5910
First you use the distance formula to find the distance from A to B. That will give you the length of AB. Do the same for BC and AC
jim_thompson5910
  • jim_thompson5910
tell me what distances you get
anonymous
  • anonymous
Alrighty ^-^
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
I don't agree with your last distance of 13
jim_thompson5910
  • jim_thompson5910
you forgot to take the square root of 13
anonymous
  • anonymous
oh
anonymous
  • anonymous
Square root of 13 is 3.60
jim_thompson5910
  • jim_thompson5910
|dw:1436929152566:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1436929181793:dw|
jim_thompson5910
  • jim_thompson5910
a = 5 b = 3.60555 c = 4.24264 use these three values in the law of cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B) to solve for the value of B
anonymous
  • anonymous
-173.8115277? Thats what I got?
jim_thompson5910
  • jim_thompson5910
b^2 = a^2 + c^2 - 2*a*c*cos(B) 3.60555^2 = 5^2 + 4.24264^2 - 2*5*4.24264*cos(B) are you able to isolate cos(B) ? If not, try replacing "cos(B)" with "x" ie, solve for x in 3.60555^2 = 5^2 + 4.24264^2 - 2*5*4.24264*x
anonymous
  • anonymous
For 5^2+4.24264^2-2*5*4.24264*x=42.99999417 then 3.60555^2 =12.9999908 and then I'm stuck, If I'm getting any of this right??
jim_thompson5910
  • jim_thompson5910
`5^2+4.24264^2-2*5*4.24264*x` will turn into `42.9999941696 - 42.4264*x`
jim_thompson5910
  • jim_thompson5910
42.9999941696 - 42.4264*x = 12.9999908025 -42.4264*x = 12.9999908025 - 42.9999941696 -42.4264*x = -30.0000033671 x = -30.0000033671/(-42.4264) x = ???
anonymous
  • anonymous
Dont understand
jim_thompson5910
  • jim_thompson5910
where are you stuck at?
anonymous
  • anonymous
all of this.. I get .7071069751?
jim_thompson5910
  • jim_thompson5910
yes x = 0.7071069751
anonymous
  • anonymous
oh so I was doing all that right?
jim_thompson5910
  • jim_thompson5910
you agree that 5^2 = 25 3.60555^2 = 12.9999908025 4.24264^2 = 17.9999941696 2*5*4.24264 = 42.4264 right?
jim_thompson5910
  • jim_thompson5910
yeah if you got to x = 0.7071069751, then you are doing it right
anonymous
  • anonymous
yeah and awesome
jim_thompson5910
  • jim_thompson5910
recall that I replaced "cos(B)" with "x" to make things simpler
jim_thompson5910
  • jim_thompson5910
if x = 0.7071069751 then cos(B) = 0.7071069751
anonymous
  • anonymous
oooh theres more work???
jim_thompson5910
  • jim_thompson5910
to find B itself, we apply the arccosine function to both sides cos(B) = 0.7071069751 arccos(cos(B)) = arccos(0.7071069751) B = ???
jim_thompson5910
  • jim_thompson5910
just these last steps and you're done
anonymous
  • anonymous
your killing me man, your killing me haha
jim_thompson5910
  • jim_thompson5910
lol what is `arccos(0.7071069751)` equal to ?
jim_thompson5910
  • jim_thompson5910
use a calculator like this one if you don't have a calculator with arccos http://web2.0calc.com/
anonymous
  • anonymous
I got .7602444711?
jim_thompson5910
  • jim_thompson5910
http://web2.0calc.com/#arccos(0.7071069751)
jim_thompson5910
  • jim_thompson5910
once you click the link, hit enter to have it evaluate
anonymous
  • anonymous
round it up its 45. Cool
jim_thompson5910
  • jim_thompson5910
yeah 44.999984287488 ---> 45
anonymous
  • anonymous
awesome, Its correct. Thank you, alot
jim_thompson5910
  • jim_thompson5910
you're welcome
anonymous
  • anonymous
Means a lot ^-^
jim_thompson5910
  • jim_thompson5910
I'm glad to be of help
anonymous
  • anonymous
Yay! lol
ali2x2
  • ali2x2
Close the question, But not before you give me a medal :3

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