The angles of a triangle must add to 180^o. In the triangle to the right, we know one angle and can find another using supplementary angles.

\(\displaystyle 57+(180-110)+x=180\)

Now we only need to solve for \(\displaystyle x\).

\(\displaystyle 57+70+x=180\)

\(\displaystyle x=180-(57+70)=180-127\)

\(\displaystyle x=53^o\)

All of the interior angles of a triangle add up to \(\displaystyle 180^{\circ}\).  

If \(\displaystyle \angle A=42^{\circ}\) and \(\displaystyle \angle B=76^{\circ}\), then 

\(\displaystyle \angle C= 180^{\circ}-(42^{\circ}+76^{\circ})\)

Therefore, \(\displaystyle \angle C=62^{\circ}\)

Now, \(\displaystyle \theta\) will equal\(\displaystyle 180^{\circ}-\angle C\) because \(\displaystyle \angle C\) and \(\displaystyle \theta\) form a straight line.  Therefore,

 \(\displaystyle \theta=180^{\circ}-62^{\circ}\)

\(\displaystyle \rightarrow118^{\circ}\)

Also, by definition, the angle of an exterior angle of a triangle is equal to the measure of the two interior angles opposite of it \(\displaystyle \dpi{100} (\theta =76^{\circ}+42^{\circ}\rightarrow 118^{\circ})\).

Interior angles of a triangle always add up to 180 degrees. 

Since the sum of the angles of a triangle is \(\displaystyle 180^{\circ}\), and given that the angles are in a ratio of 1:3:5, let the measure of the smallest angle be \(\displaystyle x\), then the following expression could be written:

\(\displaystyle x+3x+5x=180\)

\(\displaystyle 9x=180\)

\(\displaystyle x=20\)

If the smallest angle is 20 degrees, then given that the middle angle is in ratio of 1:3, the middle angle would be 3 times as large, or 60 degrees.

The sum of the measures of the angles of a triangle is 180.

Thus we set up the equation \(\displaystyle \dpi{100} \small 4x+3+2x+6+3x=180\)

After combining like terms and cancelling, we have \(\displaystyle \dpi{100} \small 9x=171\rightarrow x=19\)

Thus \(\displaystyle \dpi{100} \small m\angle BAC=3x=57\)

Every triangle has 180 degrees.  An isosceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\) = the vertex angle and \(\displaystyle 2x+5\) = the base angle

So the equation to solve becomes  \(\displaystyle x+(2x+5)+(2x+5)=180\)

Thus the vertex angle is 34 and the base angles are 73.

Every triangle contains 180 degrees.  An isosceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\) = vertex angle and \(\displaystyle 3x-15\) = base angle

So the equation to solve becomes \(\displaystyle x+(3x-15)+(3x-15)=180\).

Every triangle has 180 degrees.  An isosceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\) = vertex angle and \(\displaystyle 2x - 10\) = base angle

So the equation to solve becomes  \(\displaystyle x + (2x - 10) + (2x - 10) = 180\)

So the vertex angle is 40 and the base angles is 70

Every triangle has 180 degrees.  An isosceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\)= the vertex angle and \(\displaystyle 2x + 10\) = the base angle

So the equation to solve becomes \(\displaystyle x + (2x +10) + (2x +10) = 180\)

The vertex angle is 32 degrees and the base angle is 74 degrees

Every triangle has 180 degrees.  An isosceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\) = base angle and \(\displaystyle x - 15\) = vertex angle

So the equation to solve becomes \(\displaystyle (x - 15) + x + x = 180\)

Thus, 65 is the base angle and 50 is the vertex angle.