Question

lesson 4 - 2

1. p lies in the interior of \\( \angle rst \\). \\( m \angle rsp = 40 ^ { \circ } \\) and \\( m \angle tsp = 10 ^ { \circ } \\). \\( m \angle rst = \frac {? } { } \\)

a. \\( 100 ^ { \circ } \\) b. \\( 50 ^ { \circ } \\)
c. \\( 30 ^ { \circ } \\) d. \\( 10 ^ { \circ } \\)

1. \\( \overline { qs } \\) bisects \\( \angle pqr \\). if \\( m \angle pqs = 3 x \\) and \\( m \angle rqs = 2 x + 6 \\), then \\( m \angle pqr = \frac {? } { } \\)

a. \\( 18 ^ { \circ } \\) b. \\( 36 ^ { \circ } \\)
c. \\( 30 ^ { \circ } \\) d. \\( 6 ^ { \circ } \\)

1. \\( \angle p \\) and \\( \angle q \\) are supplementary. \\( m \angle p = 5 x + 3 \\) and \\( m \angle q = x + 3 \\). \\( x = \frac {? } { } \\)

a. 14 b. 0
c. 29 d. 30

Explanation:

Step1: Solve problem 17

Since \(P\) lies in the interior of \(\angle RST\), by the angle - addition postulate \(m\angle RST=m\angle RSP + m\angle TSP\).
Substitute \(m\angle RSP = 40^{\circ}\) and \(m\angle TSP=10^{\circ}\) into the formula: \(m\angle RST=40^{\circ}+ 10^{\circ}\).

Step2: Solve problem 18

Since \(\overrightarrow{QS}\) bisects \(\angle PQR\), then \(m\angle PQS=m\angle RQS\).
Set \(3x = 2x + 6\). Subtract \(2x\) from both sides: \(3x-2x=2x + 6-2x\), so \(x = 6\).
Then \(m\angle PQR=m\angle PQS+m\angle RQS\). Substitute \(x = 6\) into \(m\angle PQS = 3x\) and \(m\angle RQS=2x + 6\). \(m\angle PQS=3\times6 = 18^{\circ}\), \(m\angle RQS=2\times6+6=18^{\circ}\). So \(m\angle PQR=18^{\circ}+18^{\circ}=36^{\circ}\).

Step3: Solve problem 19

Since \(\angle P\) and \(\angle Q\) are supplementary, \(m\angle P+m\angle Q = 180^{\circ}\).
Substitute \(m\angle P = 5x + 3\) and \(m\angle Q=x + 3\) into the formula: \((5x + 3)+(x + 3)=180\).
Combine like terms: \(6x+6 = 180\). Subtract \(6\) from both sides: \(6x+6-6=180 - 6\), so \(6x=174\). Divide both sides by \(6\): \(x=\frac{174}{6}=29\).

Answer:

1. B. \(50^{\circ}\)
2. B. \(36^{\circ}\)
3. C. \(29\)