Will Medal!
The measure of theta, in degrees, is approximately 48.8 degrees?
A. True
B. False
(Picture Below)
***My Answer: B***

- anonymous

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- schrodinger

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- anonymous

##### 1 Attachment

- UsukiDoll

do have any reasons why you think it's B?

- 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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## More answers

- anonymous

I used the formula:
\[c ^{2} = a ^{2} + b ^{2} -2ab cosC\]

- zeesbrat3

Couldn't you just use arccos?

- UsukiDoll

hmmm but I want cos C to be by itself
and it's easier to grab the angle A and B first.

- anonymous

Now that I see the formulas you gave me, I think I may be doing the wrong formula.

- 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

Fair enough

- UsukiDoll

c is supposed to be the longest side a.k.a hypotenuse

- anonymous

Oh ok. The laws of cosines were getting me confused for a second.

- anonymous

So we would have a = 5.3 b = 4 and c = 7?

- 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

so that would mean ... a = 5.3 , b=4, and c = 7...
now we use Cos A to find Angle A

- anonymous

Ok

- UsukiDoll

\[\cos A = \frac{(b^2+c^2-a^2)}{2bc} \]
you just have to plug in those values.. and simplify

- anonymous

Great! Thank you! So we would have cos A = 0.66?

- anonymous

Then in degrees it would be 37.8 degrees?

- 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

Can I just double check this for a sec? ^_^ so fast

- anonymous

Oh ok! Yeah go ahead :)

- UsukiDoll

\[\cos A = \frac{(4^2+7^2-5.3^2)}{2(4)(7)}\]

- anonymous

Oh nevermind I got 48.7 degrees

- zeesbrat3

I got \[\cos \theta = .6835\]

- UsukiDoll

\[\cos A = \frac{(16+49-28.09)}{56} \rightarrow \frac{36.91}{56} \rightarrow 0.659 \rightarrow 0.66\]

- UsukiDoll

\[\cos^{-1}(0.66) \rightarrow 48.7\] I also got 48.7 degrees

- zeesbrat3

My bad, I used 54 as the denominator, not 56... ignore the blonde...

- UsukiDoll

ok cool... we got angle A = 48.7 degrees...
Now we need angle B

- anonymous

Oh whoops for the the rest of the equation haha

- UsukiDoll

let's try and solve further.

- anonymous

Ok

- UsukiDoll

\[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]
c = 7, b = 4, a = 5.3

- UsukiDoll

back the equation tool gave me a hard time

- anonymous

I got around 35.1 degrees (34.95 not rounded)

- 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

another approach
when given only the 3 sides use cosine rule to find largest angle opposite longest side

- UsukiDoll

oh wait I got the division wrong
\[\frac{61.09}{74.2} = 0.823 \]

- UsukiDoll

\[\cos^{-1}(0.823) = 34.58\]

- 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

|dw:1434937131238:dw|

- triciaal

now use the sin rule

- 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

|dw:1434937491614:dw|

- triciaal

With rounding error true

- UsukiDoll

I had angle A almost close to 48.8 as well because when I divided
\[\frac{36.91}{56} \]

##### 2 Attachments

- 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

if rounding isn't required then it wouldn't be 48.8 but that's cutting it close..

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