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

#1**+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

#1**+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**+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°

CPhill Aug 26, 2017

#3**+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.....

CPhill Aug 26, 2017

#5**+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 ....!!! ]

CPhill Aug 26, 2017

#6**+3 **

Hey...here's another way!

__∠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