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In the figure, if $MR=MK$, the measure of arc $MK$ is 130 degrees, and measure of arc $MQ$ is 28 degrees, then what is $\angle RPK$, in degrees?

AdminMod2  Aug 26, 2017

Best Answer 

 #1
avatar+4476 
+3

Since   \(MR = MK\)   and   \(\overset{\frown}{MK} = 130°\)   ,   \(\overset{\frown}{MR} = 130°\)

 

\(\overset{\frown}{MK} +\overset{\frown}{KR} +\overset{\frown}{MR} =360°\)

 

Replace  \(\overset{\frown}{MK}\)  with  130°  and  \(\overset{\frown}{MR}\)  with  130°

 

\(130°+\overset{\frown}{KR}+130°=360° \\~\\ \overset{\frown}{KR}=360°-130°-130° \\~\\ \overset{\frown}{KR}=100°\)

 

 

And..

\(m\angle RPK=\frac{\overset{\frown}{KR}+\overset{\frown}{MQ}}{2}\\~\\ m\angle RPK=\frac{100°+28°}{2} \\~\\ m\angle RPK=64°\)

hectictar  Aug 26, 2017
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8+0 Answers

 #1
avatar+4476 
+3
Best Answer

Since   \(MR = MK\)   and   \(\overset{\frown}{MK} = 130°\)   ,   \(\overset{\frown}{MR} = 130°\)

 

\(\overset{\frown}{MK} +\overset{\frown}{KR} +\overset{\frown}{MR} =360°\)

 

Replace  \(\overset{\frown}{MK}\)  with  130°  and  \(\overset{\frown}{MR}\)  with  130°

 

\(130°+\overset{\frown}{KR}+130°=360° \\~\\ \overset{\frown}{KR}=360°-130°-130° \\~\\ \overset{\frown}{KR}=100°\)

 

 

And..

\(m\angle RPK=\frac{\overset{\frown}{KR}+\overset{\frown}{MQ}}{2}\\~\\ m\angle RPK=\frac{100°+28°}{2} \\~\\ m\angle RPK=64°\)

hectictar  Aug 26, 2017
 #2
avatar+76094 
+2

 

 

Draw RK

 

Since minor arc MK = 130°....then angle KRM  = (1/2)  of this = 65° 

 

But  MR  = MK...so angle RKM  also  = 65°

 

And since minor arc MQ = 28°...then angle QKM  = (1/2)  of this = 14°

 

Therefore, angle RKP  =  m< RKM - m< QKM  = 65 - 14  = 51°

 

Then, in triangle RKP....angle KRP = angle KRM  = 65°

 

And angle RKP  = 51°

 

Therefore, angle  RPK  = 180 - 65 - 51  = 64°

 

 

 

cool cool cool

CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
 #3
avatar+76094 
+1

 

Rats...!!!...hectictar beat me to it!!!

 

Her method is better, anyway.....no lines had to be drawn and she just dealt with arc measures....much more efficient.....

 

 

cool cool cool

CPhill  Aug 26, 2017
 #4
avatar+4476 
+3

Hahah!! Well, maybe it's a little more efficient, but it isn't as explanatory  :)

hectictar  Aug 26, 2017
edited by hectictar  Aug 26, 2017
 #7
avatar+675 
+1

Lancelot described many of Heureka’s works as “Songs Without Words.”

This also describes some of your presentations. The others are “Songs With Words.”smiley 

GingerAle  Aug 26, 2017
 #5
avatar+76094 
+2

 

Here's one other method to solve this  :

 

Since chord KM = chord RM....then  arc KMR =  130 + 130  = 260°

 

So....minor arc RK  = 360 - 260  = 100°

 

And angle RMK is an inscribed angle subtending this arc...so its measure  = 50°

 

And angle QKM subtends minor arc QM = 28°....so  angle QKM  = 14°

 

And by  the exterior angle theorem angle RPK  = RMK + QKM   = 50 + 14  = 64° 

 

[ Note...in essence....this is the same proof as hectictar's   ....!!!  ]

 

 

cool cool cool

CPhill  Aug 26, 2017
edited by CPhill  Aug 26, 2017
 #6
avatar+4476 
+3

Hey...here's another way!  laugh

 

∠KRM  =  130° / 2  =  65°

 

∠RKM  =  130° / 2  =  65°

 

∠QKM  =  28° / 2  =  14°

 

∠RKQ  =  65° - 14°  =  51°

 

And since there are 180° in triangle KPR....     ∠RPK  =  180° - 65° - 51°  =  64°

 

*edit*

Wait a minute...this isn't any different than the second one!!! Ah..sorry about that!

hectictar  Aug 26, 2017
edited by hectictar  Aug 26, 2017
 #8
avatar+675 
+1

I would describe this as your variation on a theme by Sir CPhill.

GingerAle  Aug 26, 2017

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