Question

13.
triangle qpr with angles: ∠q = (4x - 22)°, ∠p = (10x - 4)°, ∠r = (x + 11)°.
find x = _, m∠p = _, m∠q = _, m∠r = _.
14.
triangle abc with exterior angle ∠dcb = (18x - 15)°, ∠cab = (13x - 11)°, ∠abc = (4x + 1)°.
find x = _, m∠cab = _, m∠abc = _, m∠acb = _, m∠dcb = ___.

Explanation:

Step1: Solve for \( x \) in problem 13 (Triangle \( PQR \))

The sum of angles in a triangle is \( 180^\circ \). So, \( (4x - 22) + (10x - 4) + (x + 11) = 180 \).
Combine like terms: \( 4x + 10x + x - 22 - 4 + 11 = 180 \) → \( 15x - 15 = 180 \).
Add 15 to both sides: \( 15x = 195 \).
Divide by 15: \( x = 13 \).

Step2: Find \( m\angle P \), \( m\angle Q \), \( m\angle R \)

• \( m\angle P = 10x - 4 = 10(13) - 4 = 130 - 4 = 126^\circ \)
• \( m\angle Q = 4x - 22 = 4(13) - 22 = 52 - 22 = 30^\circ \)
• \( m\angle R = x + 11 = 13 + 11 = 24^\circ \)

Step3: Solve for \( x \) in problem 14 (Triangle \( ABC \) with exterior angle)

The exterior angle \( \angle DCB \) equals the sum of the two non - adjacent interior angles. So, \( 18x - 15 = (13x - 11) + (4x + 1) \).
Simplify right side: \( 13x + 4x - 11 + 1 = 17x - 10 \).
Equation: \( 18x - 15 = 17x - 10 \).
Subtract \( 17x \) from both sides: \( x - 15 = - 10 \).
Add 15 to both sides: \( x = 5 \).

Step4: Find \( m\angle CAB \), \( m\angle ABC \), \( m\angle ACB \), \( m\angle DCB \)

• \( m\angle CAB = 13x - 11 = 13(5) - 11 = 65 - 11 = 54^\circ \)
• \( m\angle ABC = 4x + 1 = 4(5) + 1 = 20 + 1 = 21^\circ \)
• \( m\angle ACB = 180 - 54 - 21 = 105^\circ \) (or using exterior angle property, \( m\angle DCB = 18x - 15 = 18(5)-15 = 90 - 15 = 75^\circ \), and \( m\angle ACB = 180 - 75 = 105^\circ \))
• \( m\angle DCB = 18x - 15 = 75^\circ \)

Answer:

Problem 13:

\( x = 13 \)
\( m\angle P = 126^\circ \)
\( m\angle Q = 30^\circ \)
\( m\angle R = 24^\circ \)

Problem 14:

\( x = 5 \)
\( m\angle CAB = 54^\circ \)
\( m\angle ABC = 21^\circ \)
\( m\angle ACB = 105^\circ \)
\( m\angle DCB = 75^\circ \)