Question

the exterior angle theorem
find the measure of each angle indicated.
1)
2)
3)
4)
5)
6)
7)
8)
solve for x.
9)

Explanation:

Step1: Recall the Exterior Angle Theorem

The Exterior Angle Theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles. For a triangle, if we have an exterior angle at a vertex, it is equal to the sum of the other two interior angles. Also, a linear pair of angles (angles on a straight line) sum up to \(180^{\circ}\).

Step2: Solve for the first triangle (Problem 1)

In triangle \(UTV\), we have an exterior angle \(\angle RVU = 120^{\circ}\) and an interior angle \(\angle T=50^{\circ}\). Let the unknown angle be \(\angle U\).
By the Exterior Angle Theorem, \(\angle RVU=\angle U + \angle T\)? Wait, no. Wait, \(\angle RVU\) and \(\angle UVT\) are a linear pair, so \(\angle UVT = 180^{\circ}- 120^{\circ}=60^{\circ}\). Then, in triangle \(UTV\), the sum of interior angles is \(180^{\circ}\). So \(\angle U+\angle T+\angle UVT = 180^{\circ}\). We know \(\angle T = 50^{\circ}\) and \(\angle UVT=60^{\circ}\). So \(\angle U=180^{\circ}-\angle T-\angle UVT=180 - 50 - 60=70^{\circ}\)? Wait, maybe a better way: The exterior angle at \(V\) (the angle adjacent to \(120^{\circ}\)) is supplementary to \(120^{\circ}\), so the interior angle at \(V\) is \(60^{\circ}\). Then, using the fact that the sum of angles in a triangle is \(180^{\circ}\), \(\angle U=180 - 50 - 60 = 70^{\circ}\). Wait, maybe the original approach was wrong. Let's re - do:
The exterior angle theorem: The exterior angle is equal to the sum of the two non - adjacent interior angles. So the exterior angle \(\angle RVU = 120^{\circ}\) is equal to \(\angle U+\angle T\). Wait, \(\angle T = 50^{\circ}\), so \(\angle U=120^{\circ}-\angle T=120 - 50 = 70^{\circ}\). Yes, that's correct. Because the exterior angle is equal to the sum of the two remote (non - adjacent) interior angles.

Step3: Solve for the second triangle (Problem 2)

In triangle \(UVT\), we have an exterior angle \(\angle TFP = 115^{\circ}\) (wait, the exterior angle at \(T\)) and an interior angle \(\angle V = 50^{\circ}\). Let the unknown angle be \(\angle U\).
The interior angle at \(T\) (adjacent to \(115^{\circ}\)) is \(180 - 115=65^{\circ}\). Then, in triangle \(UVT\), the sum of angles is \(180^{\circ}\). So \(\angle U=180-\angle V-\angle T=180 - 50 - 65 = 65^{\circ}\). Or using the exterior angle theorem: The exterior angle \(\angle TFP = 115^{\circ}=\angle U+\angle V\)? Wait, no, \(\angle V = 50^{\circ}\), so \(\angle U=115^{\circ}-\angle V=115 - 50 = 65^{\circ}\).

Step4: Solve for the third triangle (Problem 3)

In triangle \(STU\), we have interior angles \(\angle S = 70^{\circ}\) and \(\angle T = 50^{\circ}\). The exterior angle at \(U\) ( \(\angle UYT\)) is equal to the sum of \(\angle S\) and \(\angle T\) by the exterior angle theorem. So \(\angle UYT=70 + 50=120^{\circ}\).

Step5: Solve for the fourth triangle (Problem 4)

In triangle \(STR\), we have interior angles \(\angle S = 25^{\circ}\) and \(\angle T = 80^{\circ}\). The exterior angle at \(R\) ( \(\angle RTP\)) is equal to the sum of \(\angle S\) and \(\angle T\) by the exterior angle theorem. So \(\angle RTP=25 + 80 = 105^{\circ}\). Then, the interior angle at \(R\) (adjacent to \(\angle RTP\)) is \(180 - 105=75^{\circ}\).

Step6: Solve for the fifth triangle (Problem 5)

In triangle \(ECD\), we have an exterior angle \(\angle TCE = 140^{\circ}\) and an interior angle \(\angle D = 45^{\circ}\). The interior angle at \(C\) (adjacent to \(140^{\circ}\)) is \(180 - 140 = 40^{\circ}\). Then, in triangle \(ECD\), the sum of interior angles is \(180^{\circ}\). So \(\angle E=180-\angle D-\angle C=180 - 45 - 40 = 95^{\circ}\). Or using the exterior angle theorem: \(\angle TCE=\angle E+\angle D\), so \(\angle E=\angle TCE-\angle D=140 - 45 = 95^{\circ}\).

Step7: Solve for the sixth triangle (Problem 6)

In triangle \(TSU\), we have an exterior angle \(\angle JS T = 110^{\circ}\) and an interior angle \(\angle T = 80^{\circ}\). The interior angle at \(S\) (adjacent to \(110^{\circ}\)) is \(180 - 110 = 70^{\circ}\). Then, in triangle \(TSU\), the sum of interior angles is \(180^{\circ}\). So \(\angle U=180-\angle T-\angle S=180 - 80 - 70 = 30^{\circ}\). Or using the exterior angle theorem: \(\angle JS T=\angle T+\angle U\), so \(\angle U=\angle JS T-\angle T=110 - 80 = 30^{\circ}\).

Step8: Solve for the seventh triangle (Problem 7)

In triangle \(EFG\), we have interior angles \(\angle E = 28^{\circ}\) and \(\angle F = 58^{\circ}\). The exterior angle at \(G\) (the unknown angle) is equal to the sum of \(\angle E\) and \(\angle F\) by the exterior angle theorem. So the unknown angle \(\angle G=28^{\circ}+58^{\circ}=86^{\circ}\).

Step9: Solve for the eighth triangle (Problem 8)

In triangle \(RPQ\), we have interior angles \(\angle R = 33^{\circ}\) and \(\angle Q = 95^{\circ}\). The exterior angle at \(P\) (the unknown angle) is equal to the sum of \(\angle R\) and \(\angle Q\) by the exterior angle theorem. So the unknown angle \(=33^{\circ}+95^{\circ}=128^{\circ}\).

Step10: Solve for \(x\) in Problem 9

In triangle \(PRQ\), we have an exterior angle \(\angle R H P=8 + 6x\) and the two non - adjacent interior angles are \(\angle P = 30^{\circ}\) and \(\angle Q=4x + 2\). By the Exterior Angle Theorem, \(\angle R H P=\angle P+\angle Q\).
So \(8 + 6x=30+(4x + 2)\)
Simplify the right - hand side: \(30 + 4x+2=4x + 32\)
So the equation is \(8 + 6x=4x + 32\)
Subtract \(4x\) from both sides: \(8 + 6x-4x=4x + 32-4x\)
\(8 + 2x=32\)
Subtract 8 from both sides: \(2x=32 - 8=24\)
Divide both sides by 2: \(x = 12\)

Answer:

1. \(\boldsymbol{70^{\circ}}\)
2. \(\boldsymbol{65^{\circ}}\)
3. \(\boldsymbol{120^{\circ}}\)
4. \(\boldsymbol{75^{\circ}}\)
5. \(\boldsymbol{95^{\circ}}\)
6. \(\boldsymbol{30^{\circ}}\)
7. \(\boldsymbol{86^{\circ}}\)
8. \(\boldsymbol{128^{\circ}}\)
9. \(\boldsymbol{x = 12}\)