The sum of the angles in a triangle is 180^o:  a + b + c = 180

In this case, the average of a and b is 75:

(a + b)/2 = 75, then multiply both sides by 2

(a + b) = 150, then substitute into first equation

150 + c = 180

c = 30

In a triangle, there can only be one obtuse angle. Additionally, all the angle measures must add up to 180.

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:

\(\displaystyle x = 55\)

Since the sum of the angles of a triangle is \(\displaystyle 180^{\circ}\), we know that 

\(\displaystyle x^{\circ} + 4x^{\circ} + 30^{\circ} = 180^{\circ}\).

So 

\(\displaystyle 5x^{\circ} + 30^{\circ} = 180^{\circ}\)

\(\displaystyle 5x^\circ=180^\circ-30^\circ\)

\(\displaystyle 5x = 150\) 

and \(\displaystyle x = 30^{\circ}\).

To solve for \(\displaystyle x\), you must first solve for \(\displaystyle y\).

All triangles' angles add up to \(\displaystyle 180$^{\circ}$\).

So subtract \(\displaystyle (40+20)\) from \(\displaystyle 180\) to get \(\displaystyle 120\), the value of \(\displaystyle y\).

Angles \(\displaystyle y\) and \(\displaystyle 3x\) are supplementary, meaning they add up to \(\displaystyle 180\).

Subtract \(\displaystyle 180\) from \(\displaystyle 120\) to get \(\displaystyle 60$^{\circ}$\).

\(\displaystyle 3x=60$^{\circ}$\), so \(\displaystyle x=20$^{\circ}$\).

\(\displaystyle \bigtriangleup ABC\) is marked with three congruent sides, making it an equilateral triangle, so \(\displaystyle m \angle ABC = 60 ^{\circ }\). This is an exterior angle of \(\displaystyle \bigtriangleup BCD\), making its measure the sum of those of its remote interior angles; that is, 

\(\displaystyle m \angle BCD + m \angle BDC = m \angle ABC\)

\(\displaystyle \bigtriangleup BCD\) has congruent sides \(\displaystyle \overline{BC}\) and \(\displaystyle \overline{CD}\), so, by the Isosceles Triangle Theorem, \(\displaystyle m \angle BCD = m \angle BDC\). Substituting  \(\displaystyle m \angle BDC\) for \(\displaystyle m \angle BCD\) and \(\displaystyle 60 ^{\circ }\) for \(\displaystyle m \angle ABC\):

\(\displaystyle m \angle BDC + m \angle BDC = 60 ^{\circ }\)

\(\displaystyle 2 \cdot m \angle BDC = 60 ^{\circ }\)

\(\displaystyle m \angle BDC = 30 ^{\circ }\)

\(\displaystyle \angle BDC\) and \(\displaystyle \angle CDE\) form a linear pair and are therefore supplementary - that is, their degree measures total \(\displaystyle 180 ^{\circ }\). Setting up the equation 

\(\displaystyle m \angle BDC + m\angle CDE= 180 ^{\circ }\)

and substituting:

\(\displaystyle 30 ^{\circ } + m\angle CDE= 180 ^{\circ }\)

\(\displaystyle m\angle CDE= 150 ^{\circ }\)

\(\displaystyle \bigtriangleup ABC\) is an equilateral, so all of its angles - in particular,  \(\displaystyle \angle ACB\) - measure \(\displaystyle 60^{\circ }\). This angle is an exterior angle to \(\displaystyle \bigtriangleup DAC\),  and its measure is equal to the sum of those of its two remote interior angles, \(\displaystyle \angle DAC\) and \(\displaystyle \angle D\), so 

\(\displaystyle m\angle DAC + m\angle D = m \angle ACB\)

Setting \(\displaystyle m \angle ACB =60^{ \circ}\) and \(\displaystyle m \angle D = 42^{\circ }\), solve for \(\displaystyle m\angle DAC\):

\(\displaystyle m\angle DAC + 42^{\circ }= 60^{\circ }\)

\(\displaystyle m\angle DAC = 60^{\circ } - 42^{\circ }= 18^{\circ }\)