In question 5, straight lines AD, BE, and CF intersect at point O. (1) Three pairs of adjacent angles are ( COB, BOA), ( BOA, AOF), and ( COD, DOE). (2) Two linear pairs are ( COA, AOF) and ( EOB, BOD). (3) Vertically opposite angle pairs formed by the intersecting lines are ( AOF, COD), ( BOA, DOE), and ( COB, EOF). (4) The sum of angles COB + BOA + AOF = 180^° because ray OC and ray OF form a straight line CF.

In question 5, straight lines AD, BE, and CF intersect at point O. (1) Three pairs of adjacent angles are ( COB, BOA), ( BOA, AOF), and ( COD, DOE). (2) Two linear pairs are ( COA, AOF) and ( EOB, BOD). (3) Vertically opposite angle pairs formed by the intersecting lines are ( AOF, COD), ( BOA, DOE), and ( COB, EOF). (4) The sum of angles COB + BOA + AOF = 180^° because ray OC and ray OF form a straight line CF.

The three angles lie on a straight line, so their sum is 180^°. Combining like terms gives (2x + 4x + 3x) + (-9 + 5 + 4) = 9x + 0 = 180^°. Dividing both sides by 9 gives x = 20^°.

3. Can two acute angles be adjacent angles? 4. Can two acute angles form a linear pair? 5. Observe the given figure and answer these questions. [Diagram Description: Multiple lines or rays meet at a common point $O$ . The straight lines forming straight-line pairs through $O$ are line $CF$ , line $AD$ (passing from $A$ through $O$ to $D$ ), and line $BE$ (passing from $B$ through $O$ to $E$ ). The rays emanating from $O$ are labeled $\vec{OA}, \vec{OB}, \vec{OC}, \vec{OD}, \vec{OE}, \vec{OF}$ .] 1. Name any three pairs of adjacent angles. 2. Name any two linear pairs. 3. Name the pairs of vertically opposite angles. 4. Find the sum of angles $\angle COB$ , $\angle BOA$ and $\angle AOF$ . 6. Find $x$ in each figure. 1. [Diagram Description: A straight horizontal line with two rays originating from the same point on the line, dividing the straight angle of $180^\circ$ into three consecutive adjacent angles labeled from left to right as $2x - 9^\circ$ , $4x + 5^\circ$ , and $3x + 4^\circ$ .] 2. [Diagram Description: Two straight lines intersect each other. One of the angles at the intersection is labeled as $x$ , and the vertically opposite angle (or adjacent angle shown) has an arc with angle $x$ in the upper-right quadrant and an angle $2x$ in the lower-right quadrant adjacent along the vertical line.]

Answer: 3. Yes. 4. No. 5. (1) (∠COB, ∠BOA), (∠BOA, ∠AOF), (∠COD, ∠DOE); (2) (∠COA, ∠AOF), (∠EOB, ∠BOD); (3) (∠AOF, ∠COD), (∠BOA, ∠DOE), (∠COB, ∠EOF); (4) 180°. 6. (1) $x = 20°, (2) x = 60°$ .

Step-by-step solution

Step 1: Evaluate Questions 3 and 4 (Acute Angles)

For question 3, yes, two acute angles can be adjacent because they only need a common vertex and a common arm with non-overlapping interiors. For question 4, no, two acute angles cannot form a linear pair because each angle is strictly less than $90^\circ$ , meaning their sum is always strictly less than $180^\circ$ , whereas a linear pair must sum to exactly $180^\circ$ .

Step 2: Analyze Figure for Question 5 (Parts 1 to 4)

In question 5, straight lines $AD$ , $BE$ , and $CF$ intersect at point $O$ . (1) Three pairs of adjacent angles are $(\angle COB, \angle BOA)$ , $(\angle BOA, \angle AOF)$ , and $(\angle COD, \angle DOE)$ . (2) Two linear pairs are $(\angle COA, \angle AOF)$ and $(\angle EOB, \angle BOD)$ . (3) Vertically opposite angle pairs formed by the intersecting lines are $(\angle AOF, \angle COD)$ , $(\angle BOA, \angle DOE)$ , and $(\angle COB, \angle EOF)$ . (4) The sum of angles $\angle COB + \angle BOA + \angle AOF = 180^\circ$ because ray $OC$ and ray $OF$ form a straight line $CF$ .

Step 3: Solve Question 6, Part 1

The three angles lie on a straight line, so their sum is $180^\circ$ . Combining like terms gives $(2x + 4x + 3x) + (-9 + 5 + 4) = 9x + 0 = 180^\circ$ . Dividing both sides by $9$ gives $x = 20^\circ$ .

Step 4: Solve Question 6, Part 2

In the second diagram, the two angles $x$ and $2x$ form a linear pair along the straight line. Therefore, $x + 2x = 180^\circ$ , which gives $3x = 180^\circ$ . Dividing by $3$ yields $x = 60^\circ$ .

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