anonymous
  • anonymous
Will Medal! The measure of theta, in degrees, is approximately 48.8 degrees? A. True B. False (Picture Below) ***My Answer: B***
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
UsukiDoll
  • UsukiDoll
do have any reasons why you think it's B?
UsukiDoll
  • UsukiDoll
did you use the law of Cosines? cos A = \[\cos A = \frac{(b^2+c^2-a^2)}{2bc}\] \[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]

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anonymous
  • anonymous
I used the formula: \[c ^{2} = a ^{2} + b ^{2} -2ab cosC\]
zeesbrat3
  • zeesbrat3
Couldn't you just use arccos?
UsukiDoll
  • UsukiDoll
hmmm but I want cos C to be by itself and it's easier to grab the angle A and B first.
anonymous
  • anonymous
Now that I see the formulas you gave me, I think I may be doing the wrong formula.
UsukiDoll
  • UsukiDoll
@zeesbrat3 arccos is later into the formula... you have to find angle A and angle B first .. since this is all S S S (side side side)
zeesbrat3
  • zeesbrat3
Fair enough
UsukiDoll
  • UsukiDoll
c is supposed to be the longest side a.k.a hypotenuse
anonymous
  • anonymous
Oh ok. The laws of cosines were getting me confused for a second.
anonymous
  • anonymous
So we would have a = 5.3 b = 4 and c = 7?
UsukiDoll
  • UsukiDoll
I need letters first... we need to redraw the triangle with A B C and then look across for our a b c |dw:1434936158533:dw| ok let's use those values
UsukiDoll
  • UsukiDoll
so that would mean ... a = 5.3 , b=4, and c = 7... now we use Cos A to find Angle A
anonymous
  • anonymous
Ok
UsukiDoll
  • UsukiDoll
\[\cos A = \frac{(b^2+c^2-a^2)}{2bc} \] you just have to plug in those values.. and simplify
anonymous
  • anonymous
Great! Thank you! So we would have cos A = 0.66?
anonymous
  • anonymous
Then in degrees it would be 37.8 degrees?
UsukiDoll
  • UsukiDoll
this is where you need to use \[\cos^{-1}(0.66) \] and then you use a calculator to find the degrees of it
UsukiDoll
  • UsukiDoll
Can I just double check this for a sec? ^_^ so fast
anonymous
  • anonymous
Oh ok! Yeah go ahead :)
UsukiDoll
  • UsukiDoll
\[\cos A = \frac{(4^2+7^2-5.3^2)}{2(4)(7)}\]
anonymous
  • anonymous
Oh nevermind I got 48.7 degrees
zeesbrat3
  • zeesbrat3
I got \[\cos \theta = .6835\]
UsukiDoll
  • UsukiDoll
\[\cos A = \frac{(16+49-28.09)}{56} \rightarrow \frac{36.91}{56} \rightarrow 0.659 \rightarrow 0.66\]
UsukiDoll
  • UsukiDoll
\[\cos^{-1}(0.66) \rightarrow 48.7\] I also got 48.7 degrees
zeesbrat3
  • zeesbrat3
My bad, I used 54 as the denominator, not 56... ignore the blonde...
UsukiDoll
  • UsukiDoll
ok cool... we got angle A = 48.7 degrees... Now we need angle B
anonymous
  • anonymous
Oh whoops for the the rest of the equation haha
UsukiDoll
  • UsukiDoll
let's try and solve further.
anonymous
  • anonymous
Ok
UsukiDoll
  • UsukiDoll
\[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \] c = 7, b = 4, a = 5.3
UsukiDoll
  • UsukiDoll
back the equation tool gave me a hard time
anonymous
  • anonymous
I got around 35.1 degrees (34.95 not rounded)
UsukiDoll
  • UsukiDoll
\[\cos B = \frac{(7^2+5.3^2-4^2)}{2(7)(5.3)}\] \[\cos B = \frac{(49+28.09-16)}{74.2}\] \[\cos B = \frac{61.09}{74.2}\] \[\cos B = 0.84\] \[\cos^{-1}(0.84) =31.95 -> 32\] huh that's odd... how did you get 35.1 degrees?
triciaal
  • triciaal
another approach when given only the 3 sides use cosine rule to find largest angle opposite longest side
UsukiDoll
  • UsukiDoll
oh wait I got the division wrong \[\frac{61.09}{74.2} = 0.823 \]
UsukiDoll
  • UsukiDoll
\[\cos^{-1}(0.823) = 34.58\]
UsukiDoll
  • UsukiDoll
anyway .. let's just assume that Angle B = 34.58 Angle A = 48.7 to find Angle C.. we use the fact that angles of a triangle must add up to 180 degrees Angle C = 180-48.7-34.58 I will also take the 35.1 into consideration for Angle B Angle C = 180-48.7-35.1
triciaal
  • triciaal
|dw:1434937131238:dw|
triciaal
  • triciaal
now use the sin rule
UsukiDoll
  • UsukiDoll
Assuming that Angle B is 34.58 Angle C = 180-48.7-34.58 = 96.72 Assuming that Angle B is 35.1 Angle C = 180-48.7-35.1=96.20 But I think ... in the drawing... only one angle is being considered.
triciaal
  • triciaal
|dw:1434937491614:dw|
triciaal
  • triciaal
With rounding error true
UsukiDoll
  • UsukiDoll
I had angle A almost close to 48.8 as well because when I divided \[\frac{36.91}{56} \]
UsukiDoll
  • UsukiDoll
48.76 since the number after the decimal point is greater than 5 I can round up and it does become 48.8
UsukiDoll
  • UsukiDoll
if rounding isn't required then it wouldn't be 48.8 but that's cutting it close..

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