Solved 100% Date: Per: Homework 2: Central Angles & Arc | Chegg.com - Free Printable
Educational worksheet: Solved 100% Date: Per: Homework 2: Central Angles & Arc | Chegg.com. Download and print for classroom or home learning activities.
JPG
700×511
50.9 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1371815
⭐
Show Answer Key & Explanations
Step-by-step solution for: Solved 100% Date: Per: Homework 2: Central Angles & Arc | Chegg.com
▼
Show Answer Key & Explanations
Step-by-step solution for: Solved 100% Date: Per: Homework 2: Central Angles & Arc | Chegg.com
Here are the step-by-step solutions for each problem on your homework sheet.
Key Concepts to Remember:
* Full Circle: A circle has $360^\circ$.
* Central Angle = Arc Measure: The measure of an arc is equal to the measure of its central angle (the angle at the center).
* Diameter/Straight Line: A straight line across the center creates a $180^\circ$ angle (a semicircle).
* Right Angle: A square symbol means $90^\circ$.
* Vertical Angles: Angles opposite each other where lines cross are equal.
---
* Given: Central angle $\angle JKL = 127^\circ$. $J, K, M$ form a straight line (diameter).
* Find $m\widehat{JL}$: This arc corresponds directly to the central angle $\angle JKL$.
* $m\widehat{JL} = 127^\circ$
* Find $m\widehat{JML}$: This is the major arc (the long way around) from $J$ to $L$. We subtract the minor arc from the full circle ($360^\circ$).
* $360^\circ - 127^\circ = 233^\circ$
* Given: Diameter $AB$. Central angle $\angle ADC = 164^\circ$.
* Find $m\widehat{BC}$: Since $AB$ is a straight line, angles on it add up to $180^\circ$. So, $\angle BDC = 180^\circ - 164^\circ = 16^\circ$. The arc equals the angle.
* $m\widehat{BC} = 16^\circ$
* Find $m\widehat{ABC}$: This arc starts at $A$, goes through $B$, and ends at $C$. It consists of the semicircle arc $AB$ ($180^\circ$) plus arc $BC$ ($16^\circ$).
* $180^\circ + 16^\circ = 196^\circ$
* Given: Diameter $DF$. Central angle $\angle DGE = 104^\circ$.
* Find $m\widehat{DE}$: Equals the central angle.
* $m\widehat{DE} = 104^\circ$
* Find $m\widehat{FE}$: $DF$ is a straight line ($180^\circ$). So, $\angle FGE = 180^\circ - 104^\circ = 76^\circ$.
* $m\widehat{FE} = 76^\circ$
* Find $m\widehat{DEF}$: This is the arc from $D$ through $E$ to $F$. Since $DF$ is a diameter, this is exactly half the circle.
* $m\widehat{DEF} = 180^\circ$
* Find $m\widehat{CFD}$: This is the arc from $C$ through $F$ to $D$. First, find arc $CD$. $\angle CGD$ and $\angle FGE$ are vertical angles, so they are equal ($76^\circ$). Arc $CF$ is a semicircle ($180^\circ$) minus arc $CD$? No, easier way: Arc $CFD$ is the whole circle minus arc $CD$? No, look at the letters. It goes $C \rightarrow F \rightarrow D$.
* Let's find arc $CD$ first. Vertical angle to $\angle FGE$ ($76^\circ$) is $\angle CGD$ ($76^\circ$). So arc $CD = 76^\circ$.
* Arc $CFD$ is the rest of the circle starting at C, going through F, ending at D. That is $360^\circ - m\widehat{CD}$.
* $360^\circ - 76^\circ = 284^\circ$.
* Find $m\widehat{DFE}$: Starts at $D$, goes through $F$, ends at $E$. This is semicircle $DF$ ($180^\circ$) + arc $FE$ ($76^\circ$).
* $180^\circ + 76^\circ = 256^\circ$
* Given: Diameter $QS$. $\angle UVT = 44^\circ$. $\angle RVS = 25^\circ$.
* Find $m\widehat{IQ}$: (Assuming "I" is a typo for "U" based on position, or just finding arc $UQ$). Let's assume the question asks for $m\widehat{UQ}$.
* $\angle UVT$ and $\angle QVR$ are vertical angles? No. $\angle UVT$ and $\angle QVP$? No.
* Let's look at vertical angles. $\angle UVT$ ($44^\circ$) is vertical to $\angle QVR$? No, $U-V-R$ is not a line. $Q-V-S$ is a line. $T-V-?$ is not a line. Wait, usually in these problems, lines that look straight are diameters. Let's assume $TR$ and $US$ are also lines? No, only $QS$ looks like a guaranteed diameter.
* Let's re-read carefully. Usually, intersecting chords create vertical angles. If $UR$ and $TS$ were lines, $\angle UVT$ would be vertical to $\angle QVR$? No.
* Let's assume standard vertical angles where lines cross. Line $QS$ crosses Line $TR$? If $TR$ is a straight line, then $\angle QVT$ and $\angle SVR$ are vertical. And $\angle QVR$ and $\angle TVS$ are vertical.
* Let's assume $TR$ is a straight line and $US$ is a straight line? No, let's look at Problem 4 again.
* Actually, usually all lines passing through center are diameters unless stated otherwise. Let's assume $QS$, $TR$, and $UP$... wait, there is no P. Let's assume $QS$, $TR$, and $UV$... no.
* Let's assume $QS$ and $TR$ are diameters. Then $\angle QVT$ and $\angle SVR$ are vertical. But we don't know $\angle QVT$.
* Let's assume $QS$ and $UT$ are diameters? No.
* Let's look at the angles given: $44^\circ$ is $\angle UVT$. $25^\circ$ is $\angle RVS$ (or $\angle SVR$).
* If we assume $TR$ is a straight line (diameter), then $\angle UVT$ ($44^\circ$) and $\angle UVQ$? No.
* Let's assume $QS$, $TR$, and $UP$... wait, the point is V. Let's assume $QS$, $TR$, and $U...$ something are diameters.
* Looking at the diagram, $Q-V-S$ is a line. $T-V-R$ looks like a line. $U-V-$(point between R and S) looks like a line? No.
* Let's assume only $QS$ is a diameter. We can't solve it without more lines being straight.
* Standard convention for these worksheets: Lines appearing to pass through the center are diameters. So, $QS$, $TR$, and $U$(opposite point) are diameters. Let's call the point opposite $U$ as $W$ (not labeled). Or maybe $U-V-R$ is a line? No, $T-V-R$ looks straighter.
* Let's assume $QS$ and $TR$ are diameters.
* Then $\angle QVT$ and $\angle SVR$ are vertical angles. We don't have $\angle QVT$.
* We have $\angle UVT = 44^\circ$.
* We have $\angle SVR = 25^\circ$.
* If $TR$ is a diameter, $\angle TVS = 180^\circ$. Then $\angle TVU + \angle UVS = 180$? No.
* Let's try assuming $QS$, $TR$, and $U...$ let's say the line from U goes through V to a point between R and S. Let's call the point opposite U "X".
* Actually, let's look at vertical angles for $\angle UVT = 44^\circ$. The vertical angle is $\angle QV$(point opposite U). Let's assume the line segment is $U-V-R$? If $UR$ is a diameter, then $\angle UVT$ and $\angle RVQ$ are vertical? No.
* Let's assume the three lines are $QS$, $TR$, and $U$(point opposite). Let's assume the line containing U is a diameter. Let's call the other end $W$.
* Okay, let's look at $\angle RVS = 25^\circ$. Its vertical angle is $\angle QVT$. So $\angle QVT = 25^\circ$.
* Now look at $\angle UVT = 44^\circ$.
* Angle $\angle QU V$? No.
* $\angle QVU = \angle QVT + \angle TVU$? No, T is between Q and U? Looking at the circle order: Q, U, T, ...
* From the drawing: Order is Q, U, T, S, R.
* $\angle QVT = 25^\circ$ (vertical to $\angle SVR$).
* $\angle UVT = 44^\circ$.
* Therefore, $\angle QVU = \angle UVT - \angle QVT$? No, U is between Q and T?
* Visually, U is between Q and T. So $\angle QVT = \angle QVU + \angle UVT$.
* $25^\circ = \angle QVU + 44^\circ$? Impossible.
* So T must be between Q and U? Visually, Q-U-T.
* Let's re-evaluate vertical angles. Maybe $U-V-S$ is a line? If $US$ is a diameter:
* $\angle UVT = 44^\circ$. Vertical angle is $\angle SVR$? No, vertical to $\angle UVT$ is $\angle SV$(point opposite U).
* If $US$ is a diameter and $TR$ is a diameter:
* $\angle UVT = 44^\circ$. Vertical angle is $\angle SVR$? No, vertical angle is $\angle S V (\text{point opposite U})$. If the line is $TR$, the opposite ray to $VU$ is not drawn.
* Let's try the most common setup: $QS$, $TR$, and $U...$ let's assume the line from U passes through V and hits the circle between R and S. Let's call that point $W$.
* If $QS$ and $TR$ are diameters:
* $\angle SVR = 25^\circ$. Vertical angle $\angle QVT = 25^\circ$.
* We are given $\angle UVT = 44^\circ$.
* From the picture, Ray $VU$ is "above" Ray $VT$. Ray $VQ$ is "left".
* So $\angle QVU = \angle UVT - \angle QVT$? No.
* $\angle QVU = \angle QVT + \angle TVU$? No.
* Let's look at the positions. Q is top-left. U is top-left-ish. T is left.
* Angle $\angle QVT$ is the angle between radius Q and radius T. We found it is $25^\circ$.
* Angle $\angle UVT$ is the angle between radius U and radius T. Given as $44^\circ$.
* Since $44 > 25$, Ray $VU$ is further from $VT$ than $VQ$ is. So the order is T, Q, U? Or T, U, Q?
* If order is T, Q, U: $\angle TVU = \angle TVQ + \angle QVU$. $44 = 25 + \angle QVU$. So $\angle QVU = 19^\circ$.
* If order is T, U, Q: $\angle TVQ = \angle TVU + \angle UVQ$. $25 = 44 + ...$ Impossible.
* So, $\angle QVU = 19^\circ$.
* Now we can solve the arcs:
* $m\widehat{UQ}$: Equals central angle $\angle QVU = 19^\circ$. (Note: Question says $m\widehat{IQ}$, likely typo for $UQ$).
* $m\widehat{QR}$: Arc from Q to R. $\angle QVR$ is a straight line? No. $\angle QVS = 180$. $\angle SVR = 25$. So $\angle QVR = 180 - 25 = 155^\circ$.
* $m\widehat{QR} = 155^\circ$.
* $m\widehat{TS}$: Vertical angle to $\angle QVR$? No. $\angle TVS$ and $\angle QVR$ are vertical? No. $\angle TVS$ and $\angle QVR$?
* Line $TR$ and Line $QS$. $\angle TVS$ and $\angle QVR$ are NOT vertical. $\angle TVS$ and $\angle QVR$?
* Vertical to $\angle SVR$ ($25^\circ$) is $\angle QVT$ ($25^\circ$).
* $\angle TVS$ forms a linear pair with $\angle SVR$ on line $TR$? No, on line $QS$?
* $\angle TVS + \angle SVR = \angle TVR = 180^\circ$. So $\angle TVS = 180 - 25 = 155^\circ$.
* $m\widehat{TS} = 155^\circ$.
* $m\widehat{SQR}$: Major arc from S to R via Q.
* Circle is 360. Minor arc $SR = 25^\circ$.
* $360 - 25 = 335^\circ$.
* $m\widehat{RQT}$: Major arc from R to T via Q.
* Minor arc $RT$ is a semicircle ($180^\circ$) because TR is a diameter.
* Wait, $m\widehat{RQT}$ goes R -> Q -> T. Since R-V-T is a line, the arc is $180^\circ$.
*Let's double check the "I" in $m\widehat{IQ}$. Looking at crop 4, it definitely says $m\widehat{UQ}$? No, it looks like an I. But there is no I. There is a U. It is almost certainly a typo for U.*
* Given: Diameter $KM$ (straight line). Right angle at $\angle OPN$? No, square is at $\angle KPO$? No, square is at $\angle NPM$?
* Looking closely at Crop 5: The square symbol is between radius $PN$ and radius $PM$. So $\angle NPM = 90^\circ$.
* Also given: $\angle LPM = 67^\circ$.
* Find $m\widehat{KL}$:
* $KM$ is a line ($180^\circ$).
* $\angle KPL + \angle LPM = 180^\circ$.
* $\angle KPL + 67^\circ = 180^\circ \Rightarrow \angle KPL = 113^\circ$.
* $m\widehat{KL} = 113^\circ$.
* Find $m\widehat{LON}$:
* This is the arc from L, through O, to N.
* We need the angle $\angle LPN$.
* We know $\angle NPM = 90^\circ$ and $\angle LPM = 67^\circ$.
* $\angle LPN = \angle NPM - \angle LPM = 90^\circ - 67^\circ = 23^\circ$.
* Wait, is L inside the right angle? Yes, $67 < 90$.
* So arc $LN = 23^\circ$.
* Arc $LON$ is the major arc? No, "LON" implies direction L -> O -> N.
* O is on the bottom left.
* Let's find the positions.
* $\angle KPL = 113^\circ$.
* $\angle LPN = 23^\circ$.
* $\angle NPM = 90^\circ$.
* Check sum: $113 + 23 + 90 = 226 \neq 180$. Something is wrong.
* Ah, $\angle KPL$ and $\angle LPM$ are supplementary. $113 + 67 = 180$. Correct.
* $\angle NPM = 90$.
* Where is N? If $\angle NPM = 90$, and $\angle LPM = 67$, then Ray $PL$ is inside $\angle NPM$.
* So $\angle LPN = 90 - 67 = 23^\circ$.
* Where is O? We need more info. Is $LO$ a diameter? Is $KN$ a diameter?
* Usually, if not specified, we look for vertical angles or straight lines.
* Line $KM$ is a diameter.
* Is there another diameter? Line $LN$? No. Line $KO$? No.
* Look at the lines crossing. $K-P-M$ is a line. $L-P-O$? If $LO$ is a line, then $\angle KPO$ and $\angle LPM$ are vertical.
* If $LO$ is a diameter:
* $\angle KPO = \angle LPM = 67^\circ$ (vertical).
* $\angle OPN$? We know $\angle NPM = 90$. $\angle KPN = 180 - 90 = 90$.
* $\angle OPN = \angle KPN - \angle KPO = 90 - 67 = 23^\circ$.
* Let's check consistency. $\angle LPN = 23$. Vertical angle $\angle KPO$? No, vertical to $\angle LPN$ is $\angle KPO$? No, vertical to $\angle LPN$ is $\angle KPO$ only if $KN$ and $LP$ are lines.
* Let's assume $LO$ is a diameter.
* Then $\angle KPO = 67^\circ$ (vertical to $\angle LPM$).
* $\angle OPN$: We know $\angle KPN = 90^\circ$ (supplementary to $\angle NPM=90$).
* $\angle OPN = \angle KPN - \angle KPO = 90 - 67 = 23^\circ$.
* Arc $ON = 23^\circ$.
* Arc $LON$: Start L, go through O, end N.
* Arc $LO$ is semicircle ($180^\circ$).
* Arc $LON = 180 + 23 = 203^\circ$? Or is it just arc $LN$ via O?
* Usually 3 letters denote the path. L -> O -> N.
* Angle $\angle LON$ at center? No, arc measure.
* Arc $L \to K \to O \to N$? No.
* Arc $L \to O$ is $180^\circ$. Arc $O \to N$ is $23^\circ$. Total $203^\circ$.
* Alternatively, calculate the other way: Arc $L \to N$ (minor) is $23^\circ$. Major arc is $360 - 23 = 337^\circ$. Does the path L-O-N cover the major arc?
* L is top right. O is bottom left. N is bottom right (since $\angle NPM=90$).
* Path L -> O (bottom left) -> N (bottom right). Yes, this is the long way.
* So $m\widehat{LON} = 360^\circ - m\widehat{LN}$.
* $m\widehat{LN} = 23^\circ$.
* $m\widehat{LON} = 337^\circ$.
* Find $m\widehat{OM}$:
* Assuming $LO$ is diameter.
* $\angle LPM = 67^\circ$.
* $\angle OPM = 180 - 67 = 113^\circ$.
* $m\widehat{OM} = 113^\circ$.
* Find $m\widehat{KNL}$:
* Path K -> N -> L.
* Arc $KN$: $\angle KPN = 90^\circ$ (supp to $\angle NPM$). So arc $KN = 90^\circ$.
* Arc $NL$: We found $\angle LPN = 23^\circ$. So arc $NL = 23^\circ$.
* Total $m\widehat{KNL} = 90 + 23 = 113^\circ$.
* Find $m\widehat{NL}$:
* Just the minor arc calculated above.
* $m\widehat{NL} = 23^\circ$.
*(Self-Correction on Problem 5: What if LO is NOT a diameter? If LO is not a diameter, we cannot find the position of O. In geometry problems of this type, lines that appear straight through the center are diameters. $LO$ looks straight. $KN$ does not look straight. $KM$ is explicitly a diameter due to the straight line appearance and typical problem structure. I will proceed with $LO$ being a diameter.)*
* Given: $\angle YZU = 55^\circ$. $\angle UZV = 108^\circ$.
* Assume: $XZV$ and $YZW$ are diameters (straight lines).
* Find $m\widehat{YU}$:
* Equals central angle $\angle YZU$.
* $m\widehat{YU} = 55^\circ$.
* Find $m\widehat{XW}$:
* $\angle XZW$ and $\angle YZV$ are vertical? No.
* $\angle XZY$ and $\angle VZW$ are vertical.
* $\angle YZV = \angle YZU + \angle UZV = 55^\circ + 108^\circ = 163^\circ$.
* So $\angle XZW = 163^\circ$? No, vertical angles are opposite.
* Angle opposite $\angle YZV$ is $\angle XZW$.
* Wait, is $X-Z-V$ a line? Yes. Is $Y-Z-W$ a line? Yes.
* Then $\angle XZW$ is vertical to $\angle YZV$.
* $\angle YZV = 55 + 108 = 163^\circ$.
* So $\angle XZW = 163^\circ$.
* $m\widehat{XW} = 163^\circ$.
* Find $m\widehat{XVW}$:
* Path X -> V -> W.
* Since $XZV$ is a diameter, arc $XV$ is $180^\circ$? No, arc $XV$ is a semicircle.
* Wait, $m\widehat{XVW}$ starts at X, goes through V, ends at W.
* Arc $XV$ is a semicircle ($180^\circ$) because XV is a diameter? No, $X-Z-V$ is the line. So arc $XV$ (either side) is 180.
* But we need to stop at W.
* Arc $XVW$ = Arc $XV$ (semicircle) + Arc $VW$? No.
* Let's trace angles.
* Start at X. Go to V. That's $180^\circ$.
* From V, go to W.
* We need $\angle VZW$.
* $\angle YZV = 163^\circ$. $\angle YZW = 180^\circ$ (line).
* $\angle VZW = 180 - 163 = 17^\circ$.
* So Arc $VW = 17^\circ$.
* Total Arc $XVW = 180 + 17 = 197^\circ$.
* Alternative: Arc $XVW$ is the whole circle minus Arc $XW$ (minor)?
* Minor Arc $XW$? We found $\angle XZW = 163^\circ$?
* Wait. $\angle XZW$ and $\angle YZV$ are vertical. $\angle YZV = 163$. So $\angle XZW = 163$.
* So Minor Arc $XW = 163^\circ$.
* Major Arc $XW$ (which is $X-V-W$) = $360 - 163 = 197^\circ$. Matches.
* Find $m\widehat{VW}$:
* Calculated above as $17^\circ$.
* Check: $\angle UZV = 108$. $\angle YZU = 55$. $\angle YZV = 163$.
* $\angle VZW = 180 - 163 = 17^\circ$. Correct.
* Find $m\widehat{YWU}$:
* Path Y -> W -> U.
* This is the major arc connecting Y and U.
* Minor Arc $YU = 55^\circ$.
* Major Arc $YWU = 360 - 55 = 305^\circ$.
Final Answer:
1.
$m\widehat{JL} = 127^\circ$
$m\widehat{JML} = 233^\circ$
2.
$m\widehat{BC} = 16^\circ$
$m\widehat{ABC} = 196^\circ$
3.
$m\widehat{DE} = 104^\circ$
$m\widehat{FE} = 76^\circ$
$m\widehat{DEF} = 180^\circ$
$m\widehat{CFD} = 284^\circ$
$m\widehat{DFE} = 256^\circ$
4.
$m\widehat{UQ} = 19^\circ$ (Assuming 'I' is a typo for 'U')
$m\widehat{QR} = 155^\circ$
$m\widehat{TS} = 155^\circ$
$m\widehat{SQR} = 335^\circ$
$m\widehat{RQT} = 180^\circ$
5.
$m\widehat{KL} = 113^\circ$
$m\widehat{LON} = 337^\circ$
$m\widehat{OM} = 113^\circ$
$m\widehat{KNL} = 113^\circ$
$m\widehat{NL} = 23^\circ$
6.
$m\widehat{YU} = 55^\circ$
$m\widehat{XW} = 163^\circ$
$m\widehat{XVW} = 197^\circ$
$m\widehat{VW} = 17^\circ$
$m\widehat{YWU} = 305^\circ$
Key Concepts to Remember:
* Full Circle: A circle has $360^\circ$.
* Central Angle = Arc Measure: The measure of an arc is equal to the measure of its central angle (the angle at the center).
* Diameter/Straight Line: A straight line across the center creates a $180^\circ$ angle (a semicircle).
* Right Angle: A square symbol means $90^\circ$.
* Vertical Angles: Angles opposite each other where lines cross are equal.
---
Problem 1
* Given: Central angle $\angle JKL = 127^\circ$. $J, K, M$ form a straight line (diameter).
* Find $m\widehat{JL}$: This arc corresponds directly to the central angle $\angle JKL$.
* $m\widehat{JL} = 127^\circ$
* Find $m\widehat{JML}$: This is the major arc (the long way around) from $J$ to $L$. We subtract the minor arc from the full circle ($360^\circ$).
* $360^\circ - 127^\circ = 233^\circ$
Problem 2
* Given: Diameter $AB$. Central angle $\angle ADC = 164^\circ$.
* Find $m\widehat{BC}$: Since $AB$ is a straight line, angles on it add up to $180^\circ$. So, $\angle BDC = 180^\circ - 164^\circ = 16^\circ$. The arc equals the angle.
* $m\widehat{BC} = 16^\circ$
* Find $m\widehat{ABC}$: This arc starts at $A$, goes through $B$, and ends at $C$. It consists of the semicircle arc $AB$ ($180^\circ$) plus arc $BC$ ($16^\circ$).
* $180^\circ + 16^\circ = 196^\circ$
Problem 3
* Given: Diameter $DF$. Central angle $\angle DGE = 104^\circ$.
* Find $m\widehat{DE}$: Equals the central angle.
* $m\widehat{DE} = 104^\circ$
* Find $m\widehat{FE}$: $DF$ is a straight line ($180^\circ$). So, $\angle FGE = 180^\circ - 104^\circ = 76^\circ$.
* $m\widehat{FE} = 76^\circ$
* Find $m\widehat{DEF}$: This is the arc from $D$ through $E$ to $F$. Since $DF$ is a diameter, this is exactly half the circle.
* $m\widehat{DEF} = 180^\circ$
* Find $m\widehat{CFD}$: This is the arc from $C$ through $F$ to $D$. First, find arc $CD$. $\angle CGD$ and $\angle FGE$ are vertical angles, so they are equal ($76^\circ$). Arc $CF$ is a semicircle ($180^\circ$) minus arc $CD$? No, easier way: Arc $CFD$ is the whole circle minus arc $CD$? No, look at the letters. It goes $C \rightarrow F \rightarrow D$.
* Let's find arc $CD$ first. Vertical angle to $\angle FGE$ ($76^\circ$) is $\angle CGD$ ($76^\circ$). So arc $CD = 76^\circ$.
* Arc $CFD$ is the rest of the circle starting at C, going through F, ending at D. That is $360^\circ - m\widehat{CD}$.
* $360^\circ - 76^\circ = 284^\circ$.
* Find $m\widehat{DFE}$: Starts at $D$, goes through $F$, ends at $E$. This is semicircle $DF$ ($180^\circ$) + arc $FE$ ($76^\circ$).
* $180^\circ + 76^\circ = 256^\circ$
Problem 4
* Given: Diameter $QS$. $\angle UVT = 44^\circ$. $\angle RVS = 25^\circ$.
* Find $m\widehat{IQ}$: (Assuming "I" is a typo for "U" based on position, or just finding arc $UQ$). Let's assume the question asks for $m\widehat{UQ}$.
* $\angle UVT$ and $\angle QVR$ are vertical angles? No. $\angle UVT$ and $\angle QVP$? No.
* Let's look at vertical angles. $\angle UVT$ ($44^\circ$) is vertical to $\angle QVR$? No, $U-V-R$ is not a line. $Q-V-S$ is a line. $T-V-?$ is not a line. Wait, usually in these problems, lines that look straight are diameters. Let's assume $TR$ and $US$ are also lines? No, only $QS$ looks like a guaranteed diameter.
* Let's re-read carefully. Usually, intersecting chords create vertical angles. If $UR$ and $TS$ were lines, $\angle UVT$ would be vertical to $\angle QVR$? No.
* Let's assume standard vertical angles where lines cross. Line $QS$ crosses Line $TR$? If $TR$ is a straight line, then $\angle QVT$ and $\angle SVR$ are vertical. And $\angle QVR$ and $\angle TVS$ are vertical.
* Let's assume $TR$ is a straight line and $US$ is a straight line? No, let's look at Problem 4 again.
* Actually, usually all lines passing through center are diameters unless stated otherwise. Let's assume $QS$, $TR$, and $UP$... wait, there is no P. Let's assume $QS$, $TR$, and $UV$... no.
* Let's assume $QS$ and $TR$ are diameters. Then $\angle QVT$ and $\angle SVR$ are vertical. But we don't know $\angle QVT$.
* Let's assume $QS$ and $UT$ are diameters? No.
* Let's look at the angles given: $44^\circ$ is $\angle UVT$. $25^\circ$ is $\angle RVS$ (or $\angle SVR$).
* If we assume $TR$ is a straight line (diameter), then $\angle UVT$ ($44^\circ$) and $\angle UVQ$? No.
* Let's assume $QS$, $TR$, and $UP$... wait, the point is V. Let's assume $QS$, $TR$, and $U...$ something are diameters.
* Looking at the diagram, $Q-V-S$ is a line. $T-V-R$ looks like a line. $U-V-$(point between R and S) looks like a line? No.
* Let's assume only $QS$ is a diameter. We can't solve it without more lines being straight.
* Standard convention for these worksheets: Lines appearing to pass through the center are diameters. So, $QS$, $TR$, and $U$(opposite point) are diameters. Let's call the point opposite $U$ as $W$ (not labeled). Or maybe $U-V-R$ is a line? No, $T-V-R$ looks straighter.
* Let's assume $QS$ and $TR$ are diameters.
* Then $\angle QVT$ and $\angle SVR$ are vertical angles. We don't have $\angle QVT$.
* We have $\angle UVT = 44^\circ$.
* We have $\angle SVR = 25^\circ$.
* If $TR$ is a diameter, $\angle TVS = 180^\circ$. Then $\angle TVU + \angle UVS = 180$? No.
* Let's try assuming $QS$, $TR$, and $U...$ let's say the line from U goes through V to a point between R and S. Let's call the point opposite U "X".
* Actually, let's look at vertical angles for $\angle UVT = 44^\circ$. The vertical angle is $\angle QV$(point opposite U). Let's assume the line segment is $U-V-R$? If $UR$ is a diameter, then $\angle UVT$ and $\angle RVQ$ are vertical? No.
* Let's assume the three lines are $QS$, $TR$, and $U$(point opposite). Let's assume the line containing U is a diameter. Let's call the other end $W$.
* Okay, let's look at $\angle RVS = 25^\circ$. Its vertical angle is $\angle QVT$. So $\angle QVT = 25^\circ$.
* Now look at $\angle UVT = 44^\circ$.
* Angle $\angle QU V$? No.
* $\angle QVU = \angle QVT + \angle TVU$? No, T is between Q and U? Looking at the circle order: Q, U, T, ...
* From the drawing: Order is Q, U, T, S, R.
* $\angle QVT = 25^\circ$ (vertical to $\angle SVR$).
* $\angle UVT = 44^\circ$.
* Therefore, $\angle QVU = \angle UVT - \angle QVT$? No, U is between Q and T?
* Visually, U is between Q and T. So $\angle QVT = \angle QVU + \angle UVT$.
* $25^\circ = \angle QVU + 44^\circ$? Impossible.
* So T must be between Q and U? Visually, Q-U-T.
* Let's re-evaluate vertical angles. Maybe $U-V-S$ is a line? If $US$ is a diameter:
* $\angle UVT = 44^\circ$. Vertical angle is $\angle SVR$? No, vertical to $\angle UVT$ is $\angle SV$(point opposite U).
* If $US$ is a diameter and $TR$ is a diameter:
* $\angle UVT = 44^\circ$. Vertical angle is $\angle SVR$? No, vertical angle is $\angle S V (\text{point opposite U})$. If the line is $TR$, the opposite ray to $VU$ is not drawn.
* Let's try the most common setup: $QS$, $TR$, and $U...$ let's assume the line from U passes through V and hits the circle between R and S. Let's call that point $W$.
* If $QS$ and $TR$ are diameters:
* $\angle SVR = 25^\circ$. Vertical angle $\angle QVT = 25^\circ$.
* We are given $\angle UVT = 44^\circ$.
* From the picture, Ray $VU$ is "above" Ray $VT$. Ray $VQ$ is "left".
* So $\angle QVU = \angle UVT - \angle QVT$? No.
* $\angle QVU = \angle QVT + \angle TVU$? No.
* Let's look at the positions. Q is top-left. U is top-left-ish. T is left.
* Angle $\angle QVT$ is the angle between radius Q and radius T. We found it is $25^\circ$.
* Angle $\angle UVT$ is the angle between radius U and radius T. Given as $44^\circ$.
* Since $44 > 25$, Ray $VU$ is further from $VT$ than $VQ$ is. So the order is T, Q, U? Or T, U, Q?
* If order is T, Q, U: $\angle TVU = \angle TVQ + \angle QVU$. $44 = 25 + \angle QVU$. So $\angle QVU = 19^\circ$.
* If order is T, U, Q: $\angle TVQ = \angle TVU + \angle UVQ$. $25 = 44 + ...$ Impossible.
* So, $\angle QVU = 19^\circ$.
* Now we can solve the arcs:
* $m\widehat{UQ}$: Equals central angle $\angle QVU = 19^\circ$. (Note: Question says $m\widehat{IQ}$, likely typo for $UQ$).
* $m\widehat{QR}$: Arc from Q to R. $\angle QVR$ is a straight line? No. $\angle QVS = 180$. $\angle SVR = 25$. So $\angle QVR = 180 - 25 = 155^\circ$.
* $m\widehat{QR} = 155^\circ$.
* $m\widehat{TS}$: Vertical angle to $\angle QVR$? No. $\angle TVS$ and $\angle QVR$ are vertical? No. $\angle TVS$ and $\angle QVR$?
* Line $TR$ and Line $QS$. $\angle TVS$ and $\angle QVR$ are NOT vertical. $\angle TVS$ and $\angle QVR$?
* Vertical to $\angle SVR$ ($25^\circ$) is $\angle QVT$ ($25^\circ$).
* $\angle TVS$ forms a linear pair with $\angle SVR$ on line $TR$? No, on line $QS$?
* $\angle TVS + \angle SVR = \angle TVR = 180^\circ$. So $\angle TVS = 180 - 25 = 155^\circ$.
* $m\widehat{TS} = 155^\circ$.
* $m\widehat{SQR}$: Major arc from S to R via Q.
* Circle is 360. Minor arc $SR = 25^\circ$.
* $360 - 25 = 335^\circ$.
* $m\widehat{RQT}$: Major arc from R to T via Q.
* Minor arc $RT$ is a semicircle ($180^\circ$) because TR is a diameter.
* Wait, $m\widehat{RQT}$ goes R -> Q -> T. Since R-V-T is a line, the arc is $180^\circ$.
*Let's double check the "I" in $m\widehat{IQ}$. Looking at crop 4, it definitely says $m\widehat{UQ}$? No, it looks like an I. But there is no I. There is a U. It is almost certainly a typo for U.*
Problem 5
* Given: Diameter $KM$ (straight line). Right angle at $\angle OPN$? No, square is at $\angle KPO$? No, square is at $\angle NPM$?
* Looking closely at Crop 5: The square symbol is between radius $PN$ and radius $PM$. So $\angle NPM = 90^\circ$.
* Also given: $\angle LPM = 67^\circ$.
* Find $m\widehat{KL}$:
* $KM$ is a line ($180^\circ$).
* $\angle KPL + \angle LPM = 180^\circ$.
* $\angle KPL + 67^\circ = 180^\circ \Rightarrow \angle KPL = 113^\circ$.
* $m\widehat{KL} = 113^\circ$.
* Find $m\widehat{LON}$:
* This is the arc from L, through O, to N.
* We need the angle $\angle LPN$.
* We know $\angle NPM = 90^\circ$ and $\angle LPM = 67^\circ$.
* $\angle LPN = \angle NPM - \angle LPM = 90^\circ - 67^\circ = 23^\circ$.
* Wait, is L inside the right angle? Yes, $67 < 90$.
* So arc $LN = 23^\circ$.
* Arc $LON$ is the major arc? No, "LON" implies direction L -> O -> N.
* O is on the bottom left.
* Let's find the positions.
* $\angle KPL = 113^\circ$.
* $\angle LPN = 23^\circ$.
* $\angle NPM = 90^\circ$.
* Check sum: $113 + 23 + 90 = 226 \neq 180$. Something is wrong.
* Ah, $\angle KPL$ and $\angle LPM$ are supplementary. $113 + 67 = 180$. Correct.
* $\angle NPM = 90$.
* Where is N? If $\angle NPM = 90$, and $\angle LPM = 67$, then Ray $PL$ is inside $\angle NPM$.
* So $\angle LPN = 90 - 67 = 23^\circ$.
* Where is O? We need more info. Is $LO$ a diameter? Is $KN$ a diameter?
* Usually, if not specified, we look for vertical angles or straight lines.
* Line $KM$ is a diameter.
* Is there another diameter? Line $LN$? No. Line $KO$? No.
* Look at the lines crossing. $K-P-M$ is a line. $L-P-O$? If $LO$ is a line, then $\angle KPO$ and $\angle LPM$ are vertical.
* If $LO$ is a diameter:
* $\angle KPO = \angle LPM = 67^\circ$ (vertical).
* $\angle OPN$? We know $\angle NPM = 90$. $\angle KPN = 180 - 90 = 90$.
* $\angle OPN = \angle KPN - \angle KPO = 90 - 67 = 23^\circ$.
* Let's check consistency. $\angle LPN = 23$. Vertical angle $\angle KPO$? No, vertical to $\angle LPN$ is $\angle KPO$? No, vertical to $\angle LPN$ is $\angle KPO$ only if $KN$ and $LP$ are lines.
* Let's assume $LO$ is a diameter.
* Then $\angle KPO = 67^\circ$ (vertical to $\angle LPM$).
* $\angle OPN$: We know $\angle KPN = 90^\circ$ (supplementary to $\angle NPM=90$).
* $\angle OPN = \angle KPN - \angle KPO = 90 - 67 = 23^\circ$.
* Arc $ON = 23^\circ$.
* Arc $LON$: Start L, go through O, end N.
* Arc $LO$ is semicircle ($180^\circ$).
* Arc $LON = 180 + 23 = 203^\circ$? Or is it just arc $LN$ via O?
* Usually 3 letters denote the path. L -> O -> N.
* Angle $\angle LON$ at center? No, arc measure.
* Arc $L \to K \to O \to N$? No.
* Arc $L \to O$ is $180^\circ$. Arc $O \to N$ is $23^\circ$. Total $203^\circ$.
* Alternatively, calculate the other way: Arc $L \to N$ (minor) is $23^\circ$. Major arc is $360 - 23 = 337^\circ$. Does the path L-O-N cover the major arc?
* L is top right. O is bottom left. N is bottom right (since $\angle NPM=90$).
* Path L -> O (bottom left) -> N (bottom right). Yes, this is the long way.
* So $m\widehat{LON} = 360^\circ - m\widehat{LN}$.
* $m\widehat{LN} = 23^\circ$.
* $m\widehat{LON} = 337^\circ$.
* Find $m\widehat{OM}$:
* Assuming $LO$ is diameter.
* $\angle LPM = 67^\circ$.
* $\angle OPM = 180 - 67 = 113^\circ$.
* $m\widehat{OM} = 113^\circ$.
* Find $m\widehat{KNL}$:
* Path K -> N -> L.
* Arc $KN$: $\angle KPN = 90^\circ$ (supp to $\angle NPM$). So arc $KN = 90^\circ$.
* Arc $NL$: We found $\angle LPN = 23^\circ$. So arc $NL = 23^\circ$.
* Total $m\widehat{KNL} = 90 + 23 = 113^\circ$.
* Find $m\widehat{NL}$:
* Just the minor arc calculated above.
* $m\widehat{NL} = 23^\circ$.
*(Self-Correction on Problem 5: What if LO is NOT a diameter? If LO is not a diameter, we cannot find the position of O. In geometry problems of this type, lines that appear straight through the center are diameters. $LO$ looks straight. $KN$ does not look straight. $KM$ is explicitly a diameter due to the straight line appearance and typical problem structure. I will proceed with $LO$ being a diameter.)*
Problem 6
* Given: $\angle YZU = 55^\circ$. $\angle UZV = 108^\circ$.
* Assume: $XZV$ and $YZW$ are diameters (straight lines).
* Find $m\widehat{YU}$:
* Equals central angle $\angle YZU$.
* $m\widehat{YU} = 55^\circ$.
* Find $m\widehat{XW}$:
* $\angle XZW$ and $\angle YZV$ are vertical? No.
* $\angle XZY$ and $\angle VZW$ are vertical.
* $\angle YZV = \angle YZU + \angle UZV = 55^\circ + 108^\circ = 163^\circ$.
* So $\angle XZW = 163^\circ$? No, vertical angles are opposite.
* Angle opposite $\angle YZV$ is $\angle XZW$.
* Wait, is $X-Z-V$ a line? Yes. Is $Y-Z-W$ a line? Yes.
* Then $\angle XZW$ is vertical to $\angle YZV$.
* $\angle YZV = 55 + 108 = 163^\circ$.
* So $\angle XZW = 163^\circ$.
* $m\widehat{XW} = 163^\circ$.
* Find $m\widehat{XVW}$:
* Path X -> V -> W.
* Since $XZV$ is a diameter, arc $XV$ is $180^\circ$? No, arc $XV$ is a semicircle.
* Wait, $m\widehat{XVW}$ starts at X, goes through V, ends at W.
* Arc $XV$ is a semicircle ($180^\circ$) because XV is a diameter? No, $X-Z-V$ is the line. So arc $XV$ (either side) is 180.
* But we need to stop at W.
* Arc $XVW$ = Arc $XV$ (semicircle) + Arc $VW$? No.
* Let's trace angles.
* Start at X. Go to V. That's $180^\circ$.
* From V, go to W.
* We need $\angle VZW$.
* $\angle YZV = 163^\circ$. $\angle YZW = 180^\circ$ (line).
* $\angle VZW = 180 - 163 = 17^\circ$.
* So Arc $VW = 17^\circ$.
* Total Arc $XVW = 180 + 17 = 197^\circ$.
* Alternative: Arc $XVW$ is the whole circle minus Arc $XW$ (minor)?
* Minor Arc $XW$? We found $\angle XZW = 163^\circ$?
* Wait. $\angle XZW$ and $\angle YZV$ are vertical. $\angle YZV = 163$. So $\angle XZW = 163$.
* So Minor Arc $XW = 163^\circ$.
* Major Arc $XW$ (which is $X-V-W$) = $360 - 163 = 197^\circ$. Matches.
* Find $m\widehat{VW}$:
* Calculated above as $17^\circ$.
* Check: $\angle UZV = 108$. $\angle YZU = 55$. $\angle YZV = 163$.
* $\angle VZW = 180 - 163 = 17^\circ$. Correct.
* Find $m\widehat{YWU}$:
* Path Y -> W -> U.
* This is the major arc connecting Y and U.
* Minor Arc $YU = 55^\circ$.
* Major Arc $YWU = 360 - 55 = 305^\circ$.
Final Answer:
1.
$m\widehat{JL} = 127^\circ$
$m\widehat{JML} = 233^\circ$
2.
$m\widehat{BC} = 16^\circ$
$m\widehat{ABC} = 196^\circ$
3.
$m\widehat{DE} = 104^\circ$
$m\widehat{FE} = 76^\circ$
$m\widehat{DEF} = 180^\circ$
$m\widehat{CFD} = 284^\circ$
$m\widehat{DFE} = 256^\circ$
4.
$m\widehat{UQ} = 19^\circ$ (Assuming 'I' is a typo for 'U')
$m\widehat{QR} = 155^\circ$
$m\widehat{TS} = 155^\circ$
$m\widehat{SQR} = 335^\circ$
$m\widehat{RQT} = 180^\circ$
5.
$m\widehat{KL} = 113^\circ$
$m\widehat{LON} = 337^\circ$
$m\widehat{OM} = 113^\circ$
$m\widehat{KNL} = 113^\circ$
$m\widehat{NL} = 23^\circ$
6.
$m\widehat{YU} = 55^\circ$
$m\widehat{XW} = 163^\circ$
$m\widehat{XVW} = 197^\circ$
$m\widehat{VW} = 17^\circ$
$m\widehat{YWU} = 305^\circ$
Parent Tip: Review the logic above to help your child master the concept of central angles and arcs worksheet.
