Math

If A+B+C=pi then prove that cos3A. cos3B+cos3B. cos3C+cos3C. cos3A=1

Step-by-step explanation:

Given:

A+B+C= π

<=> 3A+3B+3C = 3π

<=> cos(3A+3B) = - cos3C

<=> cos3A.cos3B-sin3A.sin3B = - cos3C

<=> cos3A.cos3B = sin3A.sin3B  - cos3C (1)

similarly  apply for the other two angles, we have:

cos3B.cos3C = sin3B.sin3C  - cos3A  (2) cos3C.cos3A = sin3C.sin3A  - cos3B  (3)

Grouping three equations, (1) + (2) + (3), we have:

<=> cos3A.cos3B+cos3B.cos3C+cos3C.cos3A = sin3A.sin3B + sin3B.sin3C + sin3C.sin3A - ( cos3A  + cos3B + cos3C )

= 1  

Hope it can find you well.

Okay so for every triangle has three angles that always add up to 180 degrees. and to tell the difference between whether they are secure, right or obtuse is that acute: has one angle that is less than 90 degrees. right triangle means there is at least one angle that is exactly 90 degrees. and an obtuse angle has an angle that is bigger than 180 degrees. i’ll do #13 as an example: 13. 40 and 50 so if every triangle has 180 degrees 40+50= 90 degrees then180-90= 90 degrees so the third angle is 90 degrees and because this triangle has a 90 degree angle it means it is a right triangle

A: 0.26 would be the answer