SOLUTION: One angle of a triangle measures 10� more than the second. The measure of the third angle is twice the sum of the first two angles. Find the measure of each angle.

Question 388722: One angle of a triangle measures 10� more than the second. The measure of the third angle is twice the sum of the first two angles. Find the measure of each angle.
Found 2 solutions by Earlsdon, Jk22:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let the three angles be A, B, and C.
In any triangle, we know that A+B+C = 180 degrees.
The problem gives us:
A = B+10 and
C = 2(A+B) = 2((B+10)+B) = 2(2B+10) = 4B+20
Start with:
A+B+C = 180 Substitute A = B+10
(B+10)+B+C = 180 Substitute C = 4B+20
B+10+B+(4B+20) = 180 Simplify.
6B+30 = 180 Subtract 30 from both sides.
6B = 150 Divide both sides by 6.
B = 25 degrees.
A = B+10
A = 35 degrees.
C = 4B+20
C = 120 degrees.
Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!
The sum of the three angles is 180� and in this case the one angle plus this angle plus 10� plus twice (this angle and 10� and this angle), hence three times the angle and the angle plus 10�, hence twice the one angle plus 10� is hundred eighty divided by three, or sixty degrees. Sixty degree is twice the first angle plus ten, hence the first angle is twenty-five degrees.

The 2nd angle is thirty-five degrees, and the 3rd becomes hundred-eighty minus sixty, hence hundred-twenty.

The sum of the 2 first angles is sixty, which is indeed half of the third.

In formula this could be written as : let the angles be a,b,c,

the relationships are b=a+10, c=2(a+b)=2(a+a+10)=2(2a+10)=4a+20

and a+b+c=a+a+10+4a+20=6a+30 = 180

6a = 150 => a = 25�

b=35�

c=120�

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