Question

1. triangle with points c, d, b, and line ab. angles: at b (exterior) 20x, at c 7x + 5, at d 60°. 30) triangle with exterior angle (6x - 9)° and interior angles (2x + 14)°, (3x + 2)°. 31) triangle with exterior angle 3x + 2° and interior angles x + 25°, x + 17°. 32) triangle with exterior angle (19x + 3)° and interior angles (9x + 16)°, (6x + 15)°. some handwritten marks crossed out.

Explanation:

Problem 29:

Step1: Identify exterior angle theorem

The exterior angle at \( B \) ( \( 20x \)) is equal to the sum of the two non - adjacent interior angles (\( 7x + 5\) and \( 60^\circ\)). So, \( 20x=(7x + 5)+60\).

Step2: Solve the equation

Simplify the right - hand side: \( 20x=7x+65\).
Subtract \( 7x \) from both sides: \( 20x - 7x=7x + 65-7x\), which gives \( 13x = 65\).
Divide both sides by 13: \( x=\frac{65}{13}=5\).

Step1: Identify exterior angle theorem

The exterior angle (\(6x - 9\)) is equal to the sum of the two non - adjacent interior angles (\(2x + 14\) and \(3x+2\)). So, \(6x-9=(2x + 14)+(3x + 2)\).

Step2: Solve the equation

Simplify the right - hand side: \(6x-9 = 5x+16\).
Subtract \(5x\) from both sides: \(6x-5x-9=5x + 16-5x\), which gives \(x-9 = 16\).
Add 9 to both sides: \(x=16 + 9=25\).

Step1: Identify exterior angle theorem

The exterior angle (\(3x + 2\)) is equal to the sum of the two non - adjacent interior angles (\(x + 25\) and \(x+17\)). So, \(3x+2=(x + 25)+(x + 17)\).

Step2: Solve the equation

Simplify the right - hand side: \(3x+2=2x + 42\).
Subtract \(2x\) from both sides: \(3x-2x+2=2x + 42-2x\), which gives \(x+2 = 42\).
Subtract 2 from both sides: \(x=42-2 = 40\).

Answer:

\( x = 5 \)

Problem 30: