Question

directions: find the missing measures in each figure. keep the angle relate

1. diagram: intersecting lines, 112° and x°
2. diagram: right angle, x° and 68°
3. diagram: transversal, y°, x°, z°, 43°
4. partially visible diagram
5. ∠1 and ∠2 are vertical angles. if the measure of ∠2 is 105°, find the measure of ∠1.
6. partially visible: ∠a and ∠... measure o
7. ∠p and ∠q are supplementary angles. if the measure of ∠q is 64°, find the measure of ∠p.
8. partially visible: ∠1 and ∠... of ∠1 is

using algebra to solve for mis

1. if m∠pqt = 3x + 47 and m∠sqr = 6x − 25, find the me diagram: intersecting lines p-t and s-r at q
2. if $overleftrightarrow{ab} perp overleftrightarrow{cd}$, $mangle dce = 7x + 2$ and $mangle ecb = x + 8$, fin diagram: ab horizontal, cd vertical, e a ray from c
3. if $mangle knm = 8x - 5$ and $mangle mnj = 4x - 19$, find the diagram partially visible

Explanation:

Let's solve problem 6 first:

Step1: Recall vertical angles property

Vertical angles are equal, so \( m\angle1 = m\angle2 \).

Step2: Substitute the given measure

Given \( m\angle2 = 105^\circ \), so \( m\angle1 = 105^\circ \).

Now problem 8:

Step1: Recall supplementary angles definition

Supplementary angles add up to \( 180^\circ \), so \( m\angle P + m\angle Q = 180^\circ \).

Step2: Solve for \( m\angle P \)

Given \( m\angle Q = 64^\circ \), then \( m\angle P = 180^\circ - 64^\circ = 116^\circ \).

Problem 1:

Step1: Recall vertical angles property

Vertical angles are equal, so \( x = 112 \) (since the angle of \( 112^\circ \) and \( x^\circ \) are vertical angles).

Problem 2 (assuming it's a right angle, so \( x + 68 = 90 \)):

Step1: Recall right angle sum

Right angle is \( 90^\circ \), so \( x + 68 = 90 \).

Step2: Solve for \( x \)

\( x = 90 - 68 = 22 \).

Problem 4:

Step1: Vertical angles for \( x \) and \( 43^\circ \)? Wait, no, \( x \) and \( z \) are vertical? Wait, the angle \( 43^\circ \) and \( y \) are vertical? Wait, let's see: the straight line, so \( x + 43 = 180 \) (supplementary), so \( x = 180 - 43 = 137^\circ \). Then \( y = 43^\circ \) (vertical with \( 43^\circ \)), and \( z = x = 137^\circ \) (vertical with \( x \)).

Problem 10: Assuming \( \angle PQT \) and \( \angle SQR \) are vertical angles, so they are equal.

Step1: Set the expressions equal

\( 3x + 47 = 6x - 25 \).

Step2: Solve for \( x \)

\( 47 + 25 = 6x - 3x \)
\( 72 = 3x \)
\( x = 24 \).

Step3: Find the measure

Substitute \( x = 24 \) into \( 3x + 47 \): \( 3(24) + 47 = 72 + 47 = 119^\circ \).

Problem 11: Since \( AB \perp CD \), \( \angle DCE + \angle ECB = 90^\circ \) (right angle).

Step1: Set up the equation

\( 7x + 2 + x + 8 = 90 \).

Step2: Combine like terms

\( 8x + 10 = 90 \).

Step3: Solve for \( x \)

\( 8x = 80 \)
\( x = 10 \).

Step4: Find the angles

\( m\angle DCE = 7(10) + 2 = 72^\circ \), \( m\angle ECB = 10 + 8 = 18^\circ \).

(Note: Problem 12 is partially visible, but assuming similar logic with supplementary or vertical angles. If \( \angle KNM \) and \( \angle MNJ \) are supplementary, \( 8x - 5 + 4x - 19 = 180 \), \( 12x - 24 = 180 \), \( 12x = 204 \), \( x = 17 \), then \( m\angle KNM = 8(17) - 5 = 131^\circ \), \( m\angle MNJ = 4(17) - 19 = 49^\circ \).)

Answer:

s:

1. \( x = 112^\circ \)
2. \( x = 22^\circ \)
3. \( x = 137^\circ \), \( y = 43^\circ \), \( z = 137^\circ \)
4. \( m\angle1 = 105^\circ \)
5. \( m\angle P = 116^\circ \)
6. \( x = 24 \), \( m\angle PQT = 119^\circ \)
7. \( x = 10 \), \( m\angle DCE = 72^\circ \), \( m\angle ECB = 18^\circ \)
8. (assuming supplementary) \( x = 17 \), \( m\angle KNM = 131^\circ \), \( m\angle MNJ = 49^\circ \) (depends on angle relationship)