Question

triangle sum & exterior angle theorems
class work
in the given triangles, solve for the missing variable(s).
home work
in the given triangles, solve for the missing variable(s).

Explanation:

Step1: Recall triangle - sum theorem

The sum of interior angles of a triangle is 180°. For exterior - angle problems, an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.

Step2: Solve problem 50

Using the triangle - sum theorem: \(x + 12+87 = 180\). Combine like terms: \(x+99 = 180\). Subtract 99 from both sides: \(x=180 - 99=81\).

Step3: Solve problem 51

Using the triangle - sum theorem: \((3x - 14)+(2x + 15)+(x - 7)=180\). Combine like terms: \(3x+2x+x-14 + 15-7 = 180\), \(6x-6 = 180\). Add 6 to both sides: \(6x=186\). Divide by 6: \(x = 31\).

Step4: Solve problem 52

Using the triangle - sum theorem in a right - triangle (\(90^{\circ}\) angle): \(x-9+62 = 90\). Combine like terms: \(x + 53=90\). Subtract 53 from both sides: \(x=90 - 53 = 37\).

Step5: Solve problem 53

Using the triangle - sum theorem in a right - triangle: \((x + 4)+(3x-22)+90 = 180\). Combine like terms: \(x+3x+4-22 + 90=180\), \(4x + 72=180\). Subtract 72 from both sides: \(4x=108\). Divide by 4: \(x = 27\).

Step6: Solve problem 54

Using the exterior - angle theorem: \(6x-23=(3x - 18)+(2x + 27)\). Combine like terms on the right - hand side: \(6x-23=5x + 9\). Subtract \(5x\) from both sides: \(x-23 = 9\). Add 23 to both sides: \(x=32\).

Step7: Solve problem 55

Using the exterior - angle theorem: \(x=(y - 13)+29\) and using the fact that the non - exterior part of the angle with \(x\) and \(55^{\circ}\) and \(29^{\circ}\) forms a straight - line (\(180^{\circ}\)), we first find the non - exterior part of the angle with \(x\) is \(180-(55 + 29)=96^{\circ}\). Then, if we assume the triangle formed with \(x\) and \((y - 13)\) and \(29^{\circ}\), using the triangle - sum theorem in a different way, we know that \(x + 96=180\), so \(x = 84\).

Step8: Solve problem 56

Using the vertical - angles are equal and triangle - sum theorem. The vertical angle to \(66^{\circ}\) is \(66^{\circ}\). In the left - hand triangle, \(z+26 + 66=180\), \(z=180-(26 + 66)=88\). In the right - hand triangle, \(x + 35+66 = 180\), \(x=180-(35 + 66)=79\), \(y=z = 88\) (vertical angles).

Step9: Solve problem 57

In the left - hand sub - triangle: \(x+19+z = 90\). In the right - hand sub - triangle: \(y + 47+z=90\). Also, using the fact that the large triangle has a sum of \(180^{\circ}\). But from the left - hand sub - triangle \(z=90-(x + 19)\), from the right - hand sub - triangle \(z=90-(y + 47)\), so \(x + 19=y + 47\). However, if we consider the large triangle with the right - angle: \(x+y+(19 + 47)=90\). Let's use the left - hand sub - triangle first. If we assume we want to find \(x\) and \(y\) and \(z\) step - by - step. In the left - hand sub - triangle, if we assume \(z\) is the unknown for now, \(z=90-(x + 19)\). In the right - hand sub - triangle, \(y=90-(47 + z)\). Substituting \(z\) from the first equation into the second. First, in the left - hand sub - triangle with \(x\) and \(19^{\circ}\) and \(z\), \(x=90-(19 + z)\). In the right - hand sub - triangle with \(y\), \(47^{\circ}\) and \(z\), \(y=90-(47 + z)\). Since the large triangle has a right - angle, we know that \(x + y+66 = 90\). From the left - hand sub - triangle \(z = 34\), then \(x=90-(19 + 34)=37\), \(y=90-(47 + 34)=9\).

Step10: Solve problem 58

In the left - hand triangle: \(x+62+23 = 180\), \(x=180-(62 + 23)=95\). In the right - hand triangle: \(y+34 + 90=180\), \(y=180-(34 + 90)=56\). In the combined figure, \(z\) can be found using the fact that the angles around the non - overlapping part form a certain relationship. But if we consider the right - hand triangle only for \(z\), \(z = 90 - 34=56\).

Answer:

1. \(x = 81\)
2. \(x = 31\)
3. \(x = 37\)
4. \(x = 27\)
5. \(x = 32\)
6. \(x = 84\)
7. \(x = 79\), \(y = 88\), \(z = 88\)
8. \(x = 37\), \(y = 9\), \(z = 34\)
9. \(x = 95\), \(y = 56\), \(z = 56\)