SOLUTION: The largest angle in a triangle is twice the degree measure of the second largest angle. One-third of the largest angle is 10 degrees larger than the difference of the other two..

Question 168503: The largest angle in a triangle is twice the degree measure of the second largest angle. One-third of the largest angle is 10 degrees larger than the difference of the other two.. Whaat is the measure in degreesof the smallest angle?
Found 2 solutions by gonzo, miguelmcp:
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
a = largest angle
b = second largest angle
c = smallest angle
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a = 2*b
(1/3)*a = b - c + 10
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c formula becomes:
c = b - (1/3)*a + 10
which becomes:
c = b - (2/3)*b + 10
which becomes
c = (1/3)*b + 10
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a + b + c = 180 (sum of angles of a triangle = 180)
2*b + b + (1/3)*b + 10 = 180
2*b + b + (1/3)*b = 170
3*b + (1/3)*b = 170
multiply both sides by 3:
9*b + b = 510
10*b = 510
b = 51
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2*b = 102 = a
c = b - (1/3)*a + 10
c = 51 - (1/3)*102 + 10
c = 27
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a = 102
b = 51
c = 27
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a + b + c = 180
102 + 51 + 27 = 180
180 = 180
ok
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a = 2 * b
102 = 2 * 51
102 = 102
ok
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(1/3)*a = (b-c) + 10
1/3)*102 = (51 - 27) + 10
34 = 30
ok
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Answer by miguelmcp(1) (Show Source): You can put this solution on YOUR website!
Hello there, this is the answer for this problem, perhaps more algebraic than geometrical:
We have the largest angle designated as A, the second largest as B and the smallest as C.
Every step must be notated in order to get the equations.
We know that the largest angle is twice the second largest which means: A=2B
Also we know that one-third of the largest angle is 10 degrees larger than the difference of the other two, so, we have: (1/3)A=B-C+10
As our last equation, we know that the sum of all interior angles in a triangle is 180, that is: A+B+C=180
We have our three equations:
A=2B
(1/3)A=B-C+10
A+B+C=180
From there we solve them, results are:
A=102
B=51
C=27
Which, in this case, the one we need is the smallest angle, that is C, which is 27 degrees.
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