The measures of the three angles are x, y, and z. Because the sum of the measures of the angles in any triangle must be 180 degrees, we know that x + y + z = 180. We can use this equation, along with the other two equations given, to form this system of equations:

x + y + z = 180

y = 2z

z = 0.5x - 30

If we can solve for both y and x in terms of z, then we can substitute these values into the first equation and create an equation with only one variable.

Because we are told already that y = 2z, we alreay have the value of y in terms of z.

We must solve the equation z = 0.5x - 30 for x in terms of z.

Add thirty to both sides.

z + 30 = 0.5x

Mutliply both sides by 2

2(z + 30) = 2z + 60 = x

x = 2z + 60

Now we have the values of x and y in terms of z. Let's substitute these values for x and y into the equation x + y + z = 180.

(2z + 60) + 2z + z = 180

5z + 60 = 180

5z = 120

z = 24

Because y = 2z, we know that y = 2(24) = 48. We also determined earlier that x = 2z + 60, so x = 2(24) + 60 = 108.

Thus, the measures of the three angles of the triangle are 24, 48, and 108. The question asks for the largest of these measures, which is 108.

The answer is 108. 

The answer is 18

We know that the sum of all the angles is 180. Using the rest of the information given we can write the other two equations:

x + y + z = 180      

x = 3y      

2x = z

We can solve for y and z in the second and third equations and then plug into the first to solve.

x + (1/3)x + 2x = 180

3[x + (1/3)x + 2x = 180]

3x + x + 6x = 540

10x = 540

x = 54

y = 18

z = 108

A straight line segment has 180 degrees. Therefore, the angle that is not labelled must have:

\(\displaystyle 180-110 = 70^{\circ}\)

We know that the sum of the angles in a triangle is 180 degrees. As a result, we can set up the following algebraic equation:

\(\displaystyle 70 + x + x = 180\)

Subtract 70 from both sides:

\(\displaystyle 2x = 110\)

Divide by 2:

\(\displaystyle x = 55\)

For this problem, remember that the sum of the degrees in a triangle is \(\displaystyle 180^\circ\).

This means that \(\displaystyle a+b+c=180^\circ\).

Plug in our given values to solve:

\(\displaystyle 38^\circ+b+82^\circ=180^\circ\)

\(\displaystyle 120^\circ+b=180^\circ\)

\(\displaystyle c=60^\circ\)

The sum of the two smallest sides is less than the greatest side:

\(\displaystyle 26 + 23 = 49 < 51\)

By the Triangle Inequality, this triangle cannot exist.

The key to solving this problem lies in the geometric fact that a triangle possesses a total of \(\displaystyle 180^{\circ}\) between its interior angles.  Therefore, one can calculate the measure of \(\displaystyle \angle C\) and then find the measure of its supplementary angle, \(\displaystyle \angle\theta\).

\(\displaystyle 180^{\circ}=\angle A + \angle B + \angle C\)

\(\displaystyle \angle C= 180^{\circ}- 52^{\circ} - 43^{\circ}\)

\(\displaystyle \angle C= 85^{\circ}\)

\(\displaystyle \angle C\) and \(\displaystyle \angle\theta\) are supplementary, meaning they form a line with a measure of \(\displaystyle 180^{\circ}\).

\(\displaystyle \angle\theta =180^{\circ} - 85^{\circ}\)

\(\displaystyle \rightarrow 95^{\circ}\)

One could also solve this problem with the knowledge that the sum of the exterior angle of a triangle is equal to the sum of the two interior angles opposite of it.

The key to solving this problem lies in the geometric fact that a triangle possesses a total of \(\displaystyle 180^{\circ}\) between its interior angles.  Therefore, one can calculate the measure of \(\displaystyle \angle C\) and then find the measure of its supplementary angle, \(\displaystyle \angle\theta\).

\(\displaystyle 180^{\circ}=\angle A + \angle B + \angle C\)

\(\displaystyle \angle C= 180^{\circ}- 56^{\circ} - 47^{\circ}\)

\(\displaystyle \angle C=77^{\circ}\)

\(\displaystyle \angle C\) and \(\displaystyle \angle\theta\) are supplementary, meaning they form a line with a measure of \(\displaystyle 180^{\circ}\).

\(\displaystyle \angle\theta =180^{\circ} - 77^{\circ}\)

\(\displaystyle \rightarrow 103^{\circ}\)

One could also solve this problem with the knowledge that the sum of the exterior angle of a triangle is equal to the sum of the two interior angles opposite of it.

The sum of the angles of any triangle is always \(\displaystyle 180\) degrees. Since the third angle will make up the difference between \(\displaystyle 180\) and the sum of the other two angles, add the other two angles together and subtract this sum from \(\displaystyle 180\).

Sum of the two given angles: \(\displaystyle 114 + 36 = 150\) degrees

Difference between \(\displaystyle 180\) and this sum: \(\displaystyle 180 - 150 = 30\) degrees

  If angle A is 41°, then angle C must also be 41°, since AB=BC.  So, the sum of these 2 angles is:

41° + 41° = 82°

Since the sum of the angles in a triangle is 180°, you can find out the measure of the remaining angle by subtracting 82 from 180:

180° - 82° = 98°

Recall that the sum of the angles of a triangle is \(\displaystyle 180\) degrees. Since we are given two angles, we can then find the third. Call our missing angle \(\displaystyle x\). 

\(\displaystyle 60 + 80 + x = 180\)

We combine the like terms on the left. 

\(\displaystyle 140 + x = 180\)

Subtract \(\displaystyle 140\) from both sides.

\(\displaystyle x = 40\)

Thus, we have that our missing angle is \(\displaystyle 40\) degrees.