which angle is an acute angle? ∠aoc ∠doc ∠eoc ∠boa

Question

Accepted Answer

# Explanation:
## Step1: Recall acute angle definition
An acute angle is less than $90^\circ$ (a right angle).

## Step2: Analyze each angle
- $\angle AOC$: Line $AO$ and $BO$ are perpendicular (right angle), $OC$ is between $BO$ and $DO$. So $\angle AOC$ is greater than $90^\circ$ (obtuse? Wait, no—wait, $AO$ and $BO$ are perpendicular (right angle, $90^\circ$), $OC$ is below $BO$, so $\angle AOC$ would be $90^\circ +$ some angle? Wait, no, maybe I missee. Wait, the diagram: $A$ and $B$ are on a horizontal line, $A$ top, $B$ right? Wait, no, the right angle is between $AO$ (vertical? Wait, the right angle symbol is between $AO$ and $BO$? Wait, the diagram: $A$ is on a vertical line (up), $B$ on horizontal (right), with a right angle at $O$ between $AO$ (vertical) and $BO$ (horizontal). Then $D$ is down on vertical, $E$ left on horizontal. $C$ is a point between $B$ (right horizontal) and $D$ (down vertical).

- $\angle DOC$: $DO$ is vertical down, $OC$ is between $BO$ (right) and $DO$ (down). So $\angle DOC$: $DO$ and $EO$ are horizontal? No, $DO$ is vertical. Wait, $E$ and $B$ are horizontal, $A$ and $D$ are vertical. So $\angle EOC$: $EO$ (left horizontal) and $OC$ – no, $E$ is left, $O$ center, $C$ is between $B$ (right) and $D$ (down). Wait, $\angle BOA$: $BO$ (right horizontal) and $AO$ (up vertical) – that's a right angle ($90^\circ$). $\angle EOC$: $EO$ (left horizontal) and $OC$ – but $OC$ is in the fourth quadrant? Wait, no, $E$ is left, $B$ right, $A$ up, $D$ down. So $OC$ is in the lower right (between $B$ and $D$). So $\angle DOC$: $DO$ (down vertical) and $OC$ – since $OC$ is between $B$ (right) and $D$ (down), $\angle DOC$ is between vertical down and $OC$, which is less than $90^\circ$? Wait, no, let's re-express:

Wait, the options:

- $\angle AOC$: $AO$ (up vertical) and $OC$. Since $OC$ is below $BO$ (right horizontal), $\angle AOC$ is $90^\circ + \angle BOC$, so greater than $90^\circ$ (obtuse).

- $\angle DOC$: $DO$ (down vertical) and $OC$. $OC$ is between $BO$ (right) and $DO$ (down), so the angle between down vertical and $OC$ – if $BO$ and $DO$ are perpendicular (right angle, $90^\circ$), then $OC$ is inside that right angle, so $\angle DOC$ is less than $90^\circ$? Wait, no, $BO$ is right, $DO$ is down, so the angle between $BO$ and $DO$ is $90^\circ$ (right angle). $OC$ is a ray from $O$ to $C$ inside that right angle, so $\angle DOC$ is the angle between $DO$ (down) and $OC$ – which is less than $90^\circ$ (acute). Wait, but let's check other options:

- $\angle EOC$: $EO$ (left horizontal) and $OC$. $EO$ is left, $OC$ is right-down, so the angle between left horizontal and $OC$ is greater than $90^\circ$ (obtuse, since it spans from left to right-down, crossing the vertical).

- $\angle BOA$: $BO$ (right horizontal) and $AO$ (up vertical) – that's a right angle ($90^\circ$), not acute.

Wait, maybe I made a mistake. Wait, the right angle is between $AO$ (vertical) and $BO$ (horizontal), so $\angle BOA = 90^\circ$ (right angle). $\angle AOC$: $AO$ (up) and $OC$ – $OC$ is below $BO$, so $\angle AOC = 90^\circ + \angle BOC$, so obtuse. $\angle DOC$: $DO$ (down) and $OC$ – $OC$ is between $BO$ (right) and $DO$ (down), so the angle between $DO$ (down) and $OC$ is equal to $\angle BOC$ (since $BO$ and $DO$ are perpendicular). Wait, no, $BO$ is right, $DO$ is down, so the angle between $BO$ and $DO$ is $90^\circ$. $OC$ is a ray from $O$ to $C$ inside that $90^\circ$ angle, so $\angle DOC$ is the angle between $DO$ (down) and $OC$, which is less than $90^\circ$ (acute). Wait, but maybe the correct answer is $\angle DOC$? Wait, no, let's check again. Wait, the options: the first option is $\angle AOC$ (maybe I misread the diagram). Wait, maybe the right angle is between $AO$ (horizontal) and $BO$ (vertical)? Wait, the diagram has a right angle symbol at $O$ between $AO$ and $BO$. Let's assume $AO$ is horizontal (left-right) and $BO$ is vertical (up-down)? No, the arrow for $A$ is up, $B$ is right, so $AO$ is vertical (up), $BO$ is horizontal (right), with right angle at $O$. Then $D$ is down (vertical), $E$ is left (horizontal). $C$ is a point between $B$ (right) and $D$ (down). So:

- $\angle AOC$: $AO$ (up vertical) and $OC$ (to $C$ in lower right). So the angle between up vertical and $OC$ – since $OC$ is in the fourth quadrant (relative to $AO$ as y-axis, $BO$ as x-axis), the angle between $AO$ (y-axis) and $OC$ is greater than $90^\circ$ (obtuse).

- $\angle DOC$: $DO$ (down vertical) and $OC$ (to $C$ in lower right). The angle between down vertical (y-axis negative) and $OC$ – since $OC$ is in the fourth quadrant (x positive, y negative), the angle between $DO$ (y negative) and $OC$ is the angle between negative y-axis and $OC$, which is less than $90^\circ$ (acute), because $OC$ is between negative y-axis and positive x-axis (which is $90^\circ$ angle between negative y and positive x).

- $\angle EOC$: $EO$ (left horizontal, x negative) and $OC$ (x positive, y negative). The angle between x negative and $OC$ is greater than $90^\circ$ (obtuse, since it spans from x negative to x positive, y negative, crossing the y-axis).

- $\angle BOA$: $BO$ (right horizontal, x positive) and $AO$ (up vertical, y positive) – right angle ($90^\circ$), not acute.

So the acute angle is $\angle DOC$? Wait, but maybe the correct answer is $\angle DOC$. Wait, but let's confirm: acute angle is less than $90^\circ$. $\angle DOC$ is between down vertical and $OC$, which is inside the right angle between $BO$ (right) and $DO$ (down), so yes, less than $90^\circ$. So the answer should be $\angle DOC$. Wait, but the first option is $\angle AOC$ – maybe I messed up the diagram. Wait, maybe $AO$ is horizontal (left-right), $BO$ is vertical (up-down). Let's reorient: $A$ left, $B$ up, $O$ center, right angle between $AO$ (left-right) and $BO$ (up-down). Then $D$ is down, $E$ is left? No, the arrows: $A$ is up, $B$ is right, $D$ is down, $E$ is left. So $AO$ is vertical (up), $BO$ is horizontal (right), $DO$ is vertical (down), $EO$ is horizontal (left). So the right angle is between $AO$ (vertical) and $BO$ (horizontal), so $\angle BOA = 90^\circ$. $OC$ is a ray from $O$ to $C$ in the lower right (between $B$ and $D$). So:

- $\angle AOC$: $AO$ (up) and $OC$ (lower right) – angle is $90^\circ + \angle BOC$ (obtuse).

- $\angle DOC$: $DO$ (down) and $OC$ (lower right) – angle is $\angle BOC$ (since $BO$ and $DO$ are perpendicular, $90^\circ$ between them, so $OC$ inside, so $\angle DOC < 90^\circ$ (acute).

- $\angle EOC$: $EO$ (left) and $OC$ (lower right) – angle is $180^\circ - \angle DOC$ (obtuse).

- $\angle BOA$: $90^\circ$ (right).

Thus, the acute angle is $\angle DOC$. Wait, but the options are: first $\angle AOC$, then $\angle DOC$, then $\angle EOC$, then $\angle BOA$. Wait, maybe the first option is a typo, but according to analysis, $\angle DOC$ is acute. Wait, but maybe I made a mistake. Wait, let's check again. If $OC$ is between $BO$ (right) and $DO$ (down), then the angle between $DO$ (down) and $OC$ is acute, yes. So the answer is $\angle DOC$.

# Answer:
$\angle DOC$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall acute angle definition

Step2: Analyze each angle

Answer: