Question

mixed exercises
find the value of x. then find the measure of each angle.

1. 16. 17.

18.

Explanation:

Step1: Use the triangle angle - sum theorem

The sum of the interior angles of a triangle is \(180^{\circ}\). For the first triangle (problem 15), we have the equation \(2x + 3x+4x=180\).
Combining like terms: \(9x = 180\).
Dividing both sides by 9: \(x=\frac{180}{9}=20\).
The angles are: \(2x = 2\times20 = 40^{\circ}\), \(3x=3\times20 = 60^{\circ}\), \(4x = 4\times20=80^{\circ}\).

Step2: For the right - angled triangle (problem 16)

The sum of the interior angles of a triangle is \(180^{\circ}\), and one angle is \(90^{\circ}\). So, \(x + 2x+90=180\).
Combining like terms: \(3x+90 = 180\).
Subtracting 90 from both sides: \(3x=180 - 90=90\).
Dividing both sides by 3: \(x = 30\).
The angles are: \(x = 30^{\circ}\), \(2x=2\times30 = 60^{\circ}\), and \(90^{\circ}\).

Step3: For problem 17

Use the exterior - angle theorem. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
So, \(5x + 62=3x + 47+37\).
Simplify the right - hand side: \(5x+62=3x + 84\).
Subtract \(3x\) from both sides: \(5x-3x+62=3x - 3x+84\), \(2x+62 = 84\).
Subtract 62 from both sides: \(2x=84 - 62=22\).
Divide both sides by 2: \(x = 11\).
The angles: \(5x+62=5\times11 + 62=55 + 62 = 117^{\circ}\), \(3x + 47=3\times11+47=33 + 47=80^{\circ}\), and \(37^{\circ}\).

Answer:

1. \(x = 20\), angles are \(40^{\circ}\), \(60^{\circ}\), \(80^{\circ}\).
2. \(x = 30\), angles are \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\).
3. \(x = 11\), angles are \(37^{\circ}\), \(80^{\circ}\), \(117^{\circ}\).