SOLUTION: hi can someone please help me. Thanks in advance. The sum of the measures of the angeles of any triangle is 180 degress. In triangle ABC, angle A measures 100 degress less than

Question 74520This question is from textbook intermediate algebra
: hi can someone please help me. Thanks in advance.
The sum of the measures of the angeles of any triangle is 180 degress. In triangle ABC, angle A measures 100 degress less than the sum of the measures of angle B and angles C, and the measure of C is 40 degres less than the twice the measure of angeles B. Find the measure of each angle of the triangle. This question is from textbook intermediate algebra

Answer by ptaylor(1894) (Show Source):
You can put this solution on YOUR website!
Let x= the measure of angle B
Then (2x-40)=the measure of angle C
{x+(2x-40)}=the sum of the measures of angles B and C
So {x+(2x-40)}-100=the measure of angle A
Then our equation to solve is
A+B+C=180 or
x+(2x-40)+{x+(2x-40)}-100=180 get rid of parens and brackets
x+2x-40+x+2x-40-100=180 collect like terms
6x-180=180 add 180 to both sides
6x=360 divide both sides by 6
x=60 degrees---------------------------the measure of angle B
2x-40=120-40=80 degrees---------------------the measure of angle C
x+2x-40-100=180-140=40 degrees ---------------the measure of angle A
CK
60+80+40=180
180=180
measure of C is 40 degrees less than the twice the measure of angle B.
2*60-40=80
80=80
angle A measures 100 degress less than the sum of the measures of angle B and angle C
60+80-100=40
40=40

Hoe this helps----ptaylor