Triangle ABC has side lengths a,b and c. If these lengths satisfy a^2=a+2b+2c and -3=a+2b-2c, what is the measure (in degrees) of the largest angle?
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To find the measure of the largest angle in triangle ABC, we need to first solve for the side lengths a, b, and c using the given equations. Once we have the side lengths, we can apply the cosine rule to find the angles.
Let's start by solving the given equations:
Equation 1: a^2 = a + 2b + 2c
Equation 2: -3 = a + 2b - 2c
To eliminate one variable, let's subtract Equation 2 from Equation 1:
(a^2) - (-3) = (a + 2b + 2c) - (a + 2b - 2c)
a^2 + 3 = 4c
Now, let's solve Equation 1 for a:
a^2 - a - 2b - 2c = 0
This is a quadratic equation. We can use the quadratic formula to solve for a:
a = (-(-1) ± √((-1)^2 - 4(1)(-2b - 2c))) / (2(1))
a = (1 ± √(1 + 8(b + c))) / 2
Now, let's substitute this value of a into Equation 2 to solve for b and c:
-3 = a + 2b - 2c
-3 = ((1 ± √(1 + 8(b + c))) / 2) + 2b - 2c
Simplifying the equation above will give two possible solutions for b and c.
Once we have the values of a, b, and c, we can use the cosine rule to find the angles in triangle ABC. The cosine rule states:
cos(A) = (b^2 + c^2 - a^2) / (2bc)
cos(B) = (a^2 + c^2 - b^2) / (2ac)
cos(C) = (a^2 + b^2 - c^2) / (2ab)
Of the three angles A, B, and C, the largest angle will have the smallest value of its cosine. So, calculate the values of cos(A), cos(B), and cos(C), and find the angle with the smallest value of cosine.
Once you have the value of the smallest cosine, use the inverse cosine function (cos^(-1)) to find the measure of the largest angle.
I hope this explanation helps you understand the process of finding the largest angle in triangle ABC.