Question

12
a) what is the value of x?
x = \boxed{} °
b) if \overrightarrow{ot} bisects \angle aod, then m\angle aot = \boxed{} °

Explanation:

Part (a)

Step1: Identify right angle

From the diagram, \( \angle DOB \) is a right angle? Wait, no, looking at the angles around point \( O \), the angles \( 55^\circ \), \( 7x^\circ \), and the right angle? Wait, actually, the angles \( \angle DOB \) should be \( 55^\circ + 7x^\circ = 90^\circ \)? Wait, no, maybe the straight line? Wait, no, the diagram shows that \( \angle DOB \) is composed of \( 55^\circ \) and \( 7x^\circ \), and since \( OD \) is perpendicular to something? Wait, actually, the sum of \( 55^\circ \) and \( 7x^\circ \) should be \( 90^\circ \)? Wait, no, maybe it's a right angle. Wait, let's re-examine. The angle between \( OD \) and \( OB \) is \( 55^\circ + 7x^\circ \), and if \( OD \) is perpendicular to \( OA \) or something? Wait, no, the key is that \( 55 + 7x = 90 \)? Wait, no, maybe \( 55 + 7x = 90 \)? Wait, let's solve:

\( 55 + 7x = 90 \)

Step2: Solve for \( x \)

Subtract 55 from both sides:

\( 7x = 90 - 55 \)

\( 7x = 35 \)

Divide both sides by 7:

\( x = \frac{35}{7} = 5 \)

Step1: Find \( \angle AOD \)

First, we know that \( \angle AOD \) is a right angle? Wait, from the diagram, \( \angle AOD \) is \( 90^\circ \)? Wait, no, let's see. If \( x = 5 \), then \( 7x = 35^\circ \), and \( \angle DOB = 55 + 35 = 90^\circ \), so \( \angle AOD \) is \( 90^\circ \)? Wait, no, actually, \( \angle AOD \) is adjacent to \( \angle DOB \) on a straight line? Wait, no, the diagram shows that \( OA \) and \( OB \) are a straight line? Wait, no, maybe \( OA \) and \( OB \) are a straight line, so \( \angle AOB = 180^\circ \). Then \( \angle AOD + 55^\circ + 7x^\circ = 180^\circ \)? Wait, no, earlier we thought \( 55 + 7x = 90 \), which would mean \( \angle DOB = 90^\circ \), so \( \angle AOD = 90^\circ \) because \( OA \) and \( OB \) are a straight line (180°), so \( \angle AOD + \angle DOB = 180^\circ \)? Wait, no, that can't be. Wait, maybe the diagram has \( OD \) perpendicular to \( OA \), so \( \angle AOD = 90^\circ \). Wait, let's check with \( x = 5 \), \( 7x = 35 \), so \( 55 + 35 = 90 \), so \( \angle DOB = 90^\circ \), so \( \angle AOD = 90^\circ \) (since \( OA \) and \( OB \) are a straight line, 180°, so \( \angle AOD + \angle DOB = 180^\circ \)? No, that would be 90 + 90 = 180, so yes, \( \angle AOD = 90^\circ \).

Step2: Use angle bisector

Since \( \overrightarrow{OT} \) bisects \( \angle AOD \), then \( m\angle AOT = \frac{1}{2} m\angle AOD \)

We found \( m\angle AOD = 90^\circ \), so:

\( m\angle AOT = \frac{1}{2} \times 90^\circ = 45^\circ \)

Wait, but wait, maybe \( \angle AOD \) is not 90°? Wait, let's re-examine the diagram. The diagram shows \( OD \) with a 55° angle to \( OQ \) (or \( OB \)) and \( 7x \) angle. Wait, maybe \( OA \) and \( OB \) are a straight line, so \( \angle AOB = 180^\circ \). Then \( \angle AOD + 55^\circ + 7x^\circ = 180^\circ \). But if \( 55 + 7x = 90 \), then \( \angle AOD = 180 - 90 = 90^\circ \), which matches. So \( \angle AOD = 90^\circ \), and \( OT \) bisects it, so \( \angle AOT = 45^\circ \).

Answer:

\( x = 5 \)

Part (b)