Question

1. triangle with angles (11x - 1)°, (20x - 3)°, and an exterior angle 151°. 11. triangle with angles (14x - 13)°, (4x + 13)°, (6x + 2)° and an exterior angle (14x - 13)° (image shows two triangle angle problems with variables x).

Explanation:

Step1: Solve Problem 10

We know that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. For the first triangle, the exterior angle is \(151^{\circ}\), and the two non - adjacent interior angles are \((11x - 1)^{\circ}\) and \((20x-3)^{\circ}\)? Wait, no, actually, the adjacent interior angle to the \(151^{\circ}\) exterior angle is \(180 - 151=29^{\circ}\). Then, by the angle - sum property of a triangle, \((11x - 1)+(20x - 3)+29 = 180\)? Wait, no, the exterior angle theorem states that the exterior angle is equal to the sum of the two remote interior angles. So \(151=(11x - 1)+(20x-3)\)? Wait, no, the angle adjacent to \(151^{\circ}\) is \(180 - 151 = 29^{\circ}\). Then the sum of the other two angles \((11x - 1)\) and \((20x-3)\) should be equal to \(151^{\circ}\) (exterior angle theorem: exterior angle = sum of two non - adjacent interior angles). So:

\((11x - 1)+(20x-3)=151\)

\(11x+20x-1 - 3=151\)

\(31x-4 = 151\)

\(31x=151 + 4=155\)

\(x=\frac{155}{31}=5\)

Step2: Solve Problem 11

By the exterior angle theorem, the exterior angle \((14x - 13)^{\circ}\) is equal to the sum of the two non - adjacent interior angles \((4x + 13)^{\circ}\) and \((6x+2)^{\circ}\). So:

\(14x-13=(4x + 13)+(6x + 2)\)

\(14x-13=4x+6x + 13 + 2\)

\(14x-13=10x+15\)

\(14x-10x=15 + 13\)

\(4x=28\)

\(x = 7\)

Answer:

For problem 10, \(x = 5\); for problem 11, \(x=7\)