Using the law of cosines, write an algebraic proof to show that the angles in an equilateral triangle must equal 60°. Use "∧" to indicate exponents. For example, type a2 as a∧2. (Hint: Let s be a length of each side and x be the angle measure; then use these variables in the law of cosines.)
|dw:1432175229785:dw|
Personally I would use the fact that the sum of the three angles of a triangle equal 180 degrees (3x = 180) thus x = 60 degrees. But alas, we are told to use the Law of Cosines.
so using the law of cosines how would that work? @radar
Law of Cosines, "c^2 = a^2 + b^2 - 2abCos c In this case a = s, b = s, and c=s substituting in the equation we now have s^2 = s^2 + s^2 - 2 ssCosx s^2 = 2s^2 -2s^2Cos x -2s^2Cosx=s^2 - 2s^2 -2s^2Cosx = -s^2 Cosx =-s^2/-2s^2 = 1/2 = .5 Leaving the last step as an exercise for @Valerie709
Just take the inverse Cos of .5 to get the angle.
I just plug that in my calculator using the cos-1 button right? @radar
Yes, on mine I have to use the shift button, then the Cos button.
\[Cos ^{-1}\]
What did you get?
and that's it nothing else? the problem is complete with that? @radar
I would think so, using the law of cosines with a triangle that has 3 equal sides, the angles will be 60 degrees, we only did it once, but doing it for each angle x would appear identical.
What did you get for the value of x in degrees???
okay can you help me with this one please? @radar In a ABC, if B = 60° and the ratio of two sides is a : c = 2 : √3 + 1, then A= 45°. True or False?
Were there instructions on the procedures or what law is used to solve this?
no. @radar but my lesson is tittled law of cosines so I'm assuming law of cosines?
Probably a good assumption. O.K. let me ponder this to see how to employ the law of cosines to verify that angle A is 45 degree.....
okay thank you! @radar
I haven't figured out how to use that a/c ratio to determine angle A. And it is pass my bedtime here in the Ozarks of MO. Good luck on this, and suggest you repost it as a new problem.
alright well thank you so much for even giving it a try! @radar
Join our real-time social learning platform and learn together with your friends!