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:

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.

step-by-step explanation:

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

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