Midnight K.
asked • 04/24/23Please show full steps. Thank you.
Dayv O. answered • 04/24/23
Caring Super Enthusiastic Knowledgeable Trigonometry Tutor
Most likely, A is opposite a, same with B & C
a2=b2+c2-(2bc)cosA
cosA=[5.42-102-7.22]/[-2*10*7.2)
cosA=.851....
A=.55 radians approximately=31.5 degrees approximately
Stephen H. answered • 04/24/23
Tutor of Math, Physics and Engineering ... available online
assume A is formed by intersection of a & c with b opposite. Use law of cosines (b2=a2+c2-2accos(A)) and solve for cos(A)=.537. Use calculator to find A=57.52 degrees
Stephen H.
04/24/23
Dayv O.
You from cuemath.com/trigonometry/law-of-cosine/ student help site,,,,Let a, b, and c be the lengths of the three sides of a triangle and A, B, and C be the three angles of the triangle. Then, the law of cosine states that: a^2 = b^2 + c^2 − 2bc·cosA. The issue I see is standards of notation. Angle A is opposite side a is the standard notation.04/24/23
Stephen H.
04/24/23
Dayv O.
I guess the conclusion is: there is no right answer for the way the question was posed because angle A can be opposite side a, or b, or c. In site engineeringtoolbox dot com, law of cosine: c^2=a^2+b^2-(2ab)cos(C)04/24/23
Dayv O.
which answer is correct A=57.5 or A=31.5?04/24/23
Stephen H.
04/24/23
Midnight K.
I think that Dayv O.'s answer is the one I'm looking for; it is usually assumed that angle A is opposite to side a. Thank you both for your time and answers!04/25/23
Dayv O.
No problem, hope you learned you can find angles in triangle using law of cosine given you have all side measurements,04/25/23
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Dayv O.
the equation is wrong as it is written in your answer --- b^2=a^2+c^2-(2ac)cos(B) --- not cos(A)04/24/23