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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

do have any reasons why you think it's B?
did you use the law of Cosines? cos A = \[\cos A = \frac{(b^2+c^2-a^2)}{2bc}\] \[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question