SOLUTION: If all angles in a triangle equal 180, and the largest angle measures 10 degrees less than the sum of the other two angles; and a different angle is the average of the other two, w

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Question 215469: If all angles in a triangle equal 180, and the largest angle measures 10 degrees less than the sum of the other two angles; and a different angle is the average of the other two, what are the measurements of all three angles?
Answer by drj(1380)   (Show Source): You can put this solution on YOUR website!
If the sum of all angles in a triangle equal to 180, and the largest angle measures 10 degrees less than the sum of the other two angles; and a different angle is the average of the other two, what are the measurements of all three angles?

Step 1. Let x be one angle, y be the another angle

Step 2. Let since the largest angle measures 10 degrees less than the sum of the other two angles.

Step 3. Let since a different angle is the average of the other two. Note the largest angle is included in the average. So if we solve for x then the other angle is y is added to the largest angle when conducting the average. Simplifying yields or

Step 4. or since adding the three angles equals to 180

Step 5. Solve the system of equations shown below using substitution




Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
We'll use substitution. After moving -2*y to the right, we get:
, or . Substitute that
into another equation:
and simplify: So, we know that y=35. Since , x=60.

Answer: .


With x=60 and y=35 then 180-35-60=85.

Now check for consistency.

85=x+y-10=60+35-10 for the largest triangle which is a true statement

x=60=(35+60)/2 which is also a true statement.

Step 6. The three angles are 35, 60, and 85 degrees

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J

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