Question

1. if \\(\overrightarrow{qs}\\) bisects \\(\angle pqr\\), find \\(m\angle pqr\\).

1. \\(\angle c\\) and \\(\angle d\\) are supplementary angles. if \\(m\angle d\\) is nine less than twice \\(m\angle c\\), find \\(m\angle d\\).

Explanation:

Problem 10:

Step1: Use Angle Bisector Definition

Since \( \overrightarrow{QS} \) bisects \( \angle PQR \), \( \angle PQS = \angle SQR \). So, \( 4x - 25 = 2x + 7 \).

Step2: Solve for \( x \)

Subtract \( 2x \) from both sides: \( 2x - 25 = 7 \). Add 25 to both sides: \( 2x = 32 \). Divide by 2: \( x = 16 \).

Step3: Find \( m\angle PQR \)

\( \angle PQR = \angle PQS + \angle SQR = (4x - 25) + (2x + 7) \). Substitute \( x = 16 \): \( (4(16) - 25) + (2(16) + 7) = (64 - 25) + (32 + 7) = 39 + 39 = 78^\circ \).

Step1: Define Supplementary Angles

Supplementary angles sum to \( 180^\circ \), so \( m\angle C + m\angle D = 180^\circ \).

Step2: Express \( m\angle D \) in Terms of \( m\angle C \)

Given \( m\angle D = 2m\angle C - 9 \). Let \( m\angle C = c \), then \( m\angle D = 2c - 9 \).

Step3: Solve for \( c \)

Substitute into supplementary equation: \( c + (2c - 9) = 180 \). Combine like terms: \( 3c - 9 = 180 \). Add 9: \( 3c = 189 \). Divide by 3: \( c = 63 \).

Step4: Find \( m\angle D \)

Substitute \( c = 63 \) into \( m\angle D = 2c - 9 \): \( 2(63) - 9 = 126 - 9 = 117^\circ \).

Answer:

\( 78^\circ \)

Problem 12: