Question

determine whether each statement can be assumed from the given figure. explain

1. ∠6 and ∠8 are complementary
2. ∠7 and ∠8 form a linear pair
3. ∠3 and ∠4 are vertical angles
4. ( mangle 2 = mangle 4 + mangle 8 )
5. ∠b and ∠7 form a linear pair. the measure of ∠b is twelve more than four times the measure of ∠7. find the measure of each angle.

refer to the figure at the right.

1. if ( mangle aob = 9x - 40 ) and ( mangle boc = 16x - 209 ), find the value of ( x ) such that ( angle aoc ) is a right angle.
2. if ( mangle boc = 8x + 67^circ ) and ( mangle fod = 15x - 127^circ ), find ( mangle fod ).

Explanation:

Problem 16:

Step 1: Define Variables

Let \( m\angle7 = x \). Then \( m\angle8 = 4x + 12 \) (since \( \angle8 \) is twelve more than four times \( \angle7 \)).

Step 2: Use Linear Pair Property

Since \( \angle7 \) and \( \angle8 \) form a linear pair, their sum is \( 180^\circ \). So:
\( x + (4x + 12) = 180 \)

Step 3: Solve for \( x \)

Combine like terms: \( 5x + 12 = 180 \)
Subtract 12 from both sides: \( 5x = 168 \)
Divide by 5: \( x = \frac{168}{5} = 33.6 \)

Step 4: Find \( m\angle8 \)

\( m\angle8 = 4(33.6) + 12 = 134.4 + 12 = 146.4 \)

Step 1: Use Right Angle Property

Since \( \angle ABC \) is a right angle (\( 90^\circ \)), and \( \angle ABD = 3x - 40 \), \( \angle BDC = 5x - 20 \), and \( \angle ABD + \angle BDC = \angle ABC \) (assuming the angles are adjacent to form \( \angle ABC \)):
\( (3x - 40) + (5x - 20) = 90 \)

Step 2: Solve for \( x \)

Combine like terms: \( 8x - 60 = 90 \)
Add 60 to both sides: \( 8x = 150 \)
Divide by 8: \( x = \frac{150}{8} = 18.75 \)

Step 1: Use Vertical Angles or Linear Pair (Assuming \( \angle BOC \) and \( \angle AOD \) are vertical angles, so \( m\angle BOC = m\angle AOD \))

Given \( m\angle BOC = 8x + 67 \) and \( m\angle AOD = 15x - 127 \), set them equal:
\( 8x + 67 = 15x - 127 \)

Step 2: Solve for \( x \)

Subtract \( 8x \) from both sides: \( 67 = 7x - 127 \)
Add 127 to both sides: \( 194 = 7x \)
Divide by 7: \( x = \frac{194}{7} \approx 27.71 \) (Wait, maybe they are supplementary? Let's check. If \( \angle BOC \) and \( \angle AOD \) are supplementary (linear pair), then \( (8x + 67) + (15x - 127) = 180 \))

Step 3: Correct Equation (Assuming Linear Pair)

\( 23x - 60 = 180 \)
\( 23x = 240 \)
\( x = \frac{240}{23} \approx 10.43 \) (This is confusing. Wait, the figure shows lines intersecting, so vertical angles are equal. Let's redo with vertical angles:
\( 8x + 67 = 15x - 127 \)
\( 127 + 67 = 15x - 8x \)
\( 194 = 7x \)
\( x = \frac{194}{7} \approx 27.71 \)
Then \( m\angle BOC = 8(\frac{194}{7}) + 67 = \frac{1552}{7} + 67 = \frac{1552 + 469}{7} = \frac{2021}{7} \approx 288.71 \) (Too big, so must be supplementary. So linear pair:
\( 8x + 67 + 15x - 127 = 180 \)
\( 23x - 60 = 180 \)
\( 23x = 240 \)
\( x = \frac{240}{23} \approx 10.43 \)
Then \( m\angle BOC = 8(\frac{240}{23}) + 67 = \frac{1920}{23} + 67 = \frac{1920 + 1541}{23} = \frac{3461}{23} \approx 150.48^\circ \)
\( m\angle AOD = 15(\frac{240}{23}) - 127 = \frac{3600}{23} - 127 = \frac{3600 - 2921}{23} = \frac{679}{23} \approx 29.52^\circ \) (No, that doesn't add to 180. Wait, maybe \( \angle BOC \) and \( \angle AOD \) are vertical angles, so equal. Then my first calculation was wrong. Wait, 8x + 67 = 15x - 127 → 127 + 67 = 7x → 194 = 7x → x ≈27.71. Then 8x +67 ≈8*27.71 +67≈221.68 +67≈288.68, which is more than 180, impossible. So maybe the angles are adjacent and form a straight line (180). Wait, the figure: points A, B, C, D, E with O as intersection. Maybe \( \angle BOC \) and \( \angle AOD \) are vertical angles, so equal. Then there's a mistake in the problem, or I misread. Wait, maybe \( \angle BOC \) and \( \angle AOD \) are supplementary? No, vertical angles are equal. Wait, maybe the problem is \( \angle BOC \) and \( \angle AOE \) or something else. Alternatively, maybe \( \angle BOC \) and \( \angle AOD \) are adjacent and form a linear pair. Let's assume that. Then:
\( 8x + 67 + 15x - 127 = 180 \)
\( 23x - 60 = 180 \)
\( 23x = 240 \)
\( x = \frac{240}{23} \approx 10.43 \)
Then \( m\angle BOC = 8*10.43 + 67 ≈83.44 + 67≈150.44^\circ \)
\( m\angle AOD = 15*10.43 - 127≈156.45 - 127≈29.45^\circ \) (Still not adding to 180. Wait, 150.44 +29.45≈179.89, close to 180, so maybe rounding. So \( x = \frac{240}{23} \), then \( m\angle BOC = 8x +67 = \frac{1920}{23} + \frac{1541}{23} = \frac{3461}{23} ≈150.48^\circ \), \( m\angle AOD = \frac{3600}{23} - \frac{2921}{23} = \frac{679}{23} ≈29.52^\circ \). But maybe the problem is that \( \angle BOC \) and \( \angle AOD \) are vertical angles, so equal, so 8x +67 =15x -127 →7x=194→x=194/7≈27.71, then \( m\angle BOC =8*(194/7)+67= (1552 + 469)/7=2021/7≈288.71^\circ \), which is impossible. So there must be a misinterpretation. Alternatively, maybe \( \angle BOC \) and \( \angle AOD \) are complementary? No, the problem says "find \( m\angle AOD \)". Wait, maybe the figure has \( \angle BOC \) and \( \angle AOD \) as vertical angles, so equal, so solve 8x +67=15x -127:
15x -8x=67 +127
7x=194
x=194/7≈27.71
Then \( m\angle AOD =15x -127=15*(194/7)-127=(2910 - 889)/7=2021/7≈288.71^\circ \), which is more than 180, so impossible. Therefore, maybe the a…

Answer:

\( m\angle7 = 33.6^\circ \), \( m\angle8 = 146.4^\circ \)

Problem 17: