Question

geometry name date feb 4th period 1st practice: central and inscribed angles (center/on) use the figure to the right to find the indicated measures. 1. \\(\widehat{rs}\\) 2. \\(\widehat{rps}\\) 3. \\(\widehat{pqr}\\) 4. \\(\widehat{qs}\\) 5. \\(\widehat{qsp}\\) 6. \\(\angle qtr\\) use the figure to the right to find the indicated measures. 7. \\(\widehat{kl}\\) 8. \\(\widehat{mn}\\) 9. \\(\widehat{lnk}\\) 10. \\(\widehat{mkn}\\) 11. \\(\widehat{njk}\\) 12. \\(\angle mql\\) 13. \\(\widehat{ml}\\) 14. \\(\angle jqn\\) 15. \\(\widehat{jm}\\) find the indicated measures. 16. \\(m\angle acb\\) 17. \\(m\widehat{adb}\\) 18. \\(m\angle acb\\)

Explanation:

Step1: Solve first circle (center T)

A full circle is $360^\circ$. First find missing arc $\overset{\frown}{PT}$:
$m\overset{\frown}{PT} = 360^\circ - 120^\circ - 60^\circ - 40^\circ = 140^\circ$

1. $\overset{\frown}{RS}$: Equal to central angle $\angle RTS$

$m\overset{\frown}{RS} = 60^\circ$

1. $\overset{\frown}{RPS}$: Sum of $\overset{\frown}{RP}+\overset{\frown}{PS}$

$m\overset{\frown}{RPS} = (40^\circ+140^\circ) + 120^\circ = 300^\circ$

1. $\overset{\frown}{PQR}$: Sum of $\overset{\frown}{PQ}+\overset{\frown}{QR}$

$m\overset{\frown}{PQR} = 140^\circ + 40^\circ = 180^\circ$

1. $\overset{\frown}{QS}$: Sum of $\overset{\frown}{QR}+\overset{\frown}{RS}$

$m\overset{\frown}{QS} = 40^\circ + 60^\circ = 100^\circ$

1. $\overset{\frown}{QSP}$: Sum of $\overset{\frown}{QS}+\overset{\frown}{SP}$

$m\overset{\frown}{QSP} = 100^\circ + 120^\circ = 220^\circ$

1. $\angle QTR$: Equal to arc $\overset{\frown}{QR}$

$m\angle QTR = 40^\circ$

Step2: Solve second circle (center Q)

First find missing arcs: $\angle NJM=55^\circ$, so $\overset{\frown}{NM}=55^\circ$; $\angle KQL=60^\circ$, so $\overset{\frown}{KL}=60^\circ$.
Missing arc $\overset{\frown}{LJ} = 360^\circ - 180^\circ - 55^\circ - 60^\circ = 65^\circ$ (since NK is diameter, $180^\circ$)

1. $\overset{\frown}{KL}$: Equal to central angle $\angle KQL$

$m\overset{\frown}{KL} = 60^\circ$

1. $\overset{\frown}{MN}$: Equal to central angle $\angle MQN$

$m\overset{\frown}{MN} = 55^\circ$

1. $\overset{\frown}{LNK}$: Sum of $\overset{\frown}{LN}+\overset{\frown}{NK}$

$m\overset{\frown}{LNK} = (65^\circ+180^\circ) = 245^\circ$

1. $\overset{\frown}{MKN}$: Sum of $\overset{\frown}{MK}+\overset{\frown}{KN}$

$m\overset{\frown}{MKN} = (55^\circ+180^\circ) = 235^\circ$

1. $\overset{\frown}{NJK}$: Sum of $\overset{\frown}{NJ}+\overset{\frown}{JK}$

$m\overset{\frown}{NJK} = 180^\circ$ (NK is diameter)

1. $\angle MQL$: Sum of $\angle MQN+\angle NQJ+\angle JQL$

$m\angle MQL = 55^\circ + 180^\circ - 60^\circ = 175^\circ$

1. $\overset{\frown}{ML}$: Sum of $\overset{\frown}{MN}+\overset{\frown}{NK}+\overset{\frown}{KL}$

$m\overset{\frown}{ML} = 55^\circ + 180^\circ + 60^\circ = 295^\circ$

1. $\angle JQN$: Semicircle minus $\angle JQK$

$m\angle JQN = 180^\circ - (60^\circ+65^\circ) = 55^\circ$

1. $\overset{\frown}{JM}$: Sum of $\overset{\frown}{JN}+\overset{\frown}{NM}$

$m\overset{\frown}{JM} = 180^\circ - 65^\circ + 55^\circ = 170^\circ$

Step3: Solve third set of circles

1. $m\angle ACB$: Central angle equal to arc $\overset{\frown}{AB}$

$m\angle ACB = 165^\circ$

1. $m\overset{\frown}{ADB}$: Full circle minus $\overset{\frown}{AB}$

$m\overset{\frown}{ADB} = 360^\circ - 90^\circ = 270^\circ$

1. $m\angle ACB$: Central angle equal to arc $\overset{\frown}{AB}$

$m\angle ACB = 180^\circ$

Answer:

1. $m\overset{\frown}{RS} = 60^\circ$
2. $m\overset{\frown}{RPS} = 300^\circ$
3. $m\overset{\frown}{PQR} = 180^\circ$
4. $m\overset{\frown}{QS} = 100^\circ$
5. $m\overset{\frown}{QSP} = 220^\circ$
6. $m\angle QTR = 40^\circ$
7. $m\overset{\frown}{KL} = 60^\circ$
8. $m\overset{\frown}{MN} = 55^\circ$
9. $m\overset{\frown}{LNK} = 245^\circ$
10. $m\overset{\frown}{MKN} = 235^\circ$
11. $m\overset{\frown}{NJK} = 180^\circ$
12. $m\angle MQL = 175^\circ$
13. $m\overset{\frown}{ML} = 295^\circ$
14. $m\angle JQN = 55^\circ$
15. $m\overset{\frown}{JM} = 170^\circ$
16. $m\angle ACB = 165^\circ$
17. $m\overset{\frown}{ADB} = 270^\circ$
18. $m\angle ACB = 180^\circ$