May 18th, 2014, 01:11 AM |
#1 |

Newbie Joined: May 2014 From: U.K Posts: 1 Thanks: 0 Math Focus: Calculus! | I need help. |

May 18th, 2014, 02:02 PM |
#2 |

Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 |
C: 108 No trigonometry required, just calculating angles is sufficient. Start with angle OKL. Notice 3 isosceles triangles: MOK, JOK and JOM (all sides equal to radius). So angleOKL = angleOMK, angleOKJ = angleOJK, angleOJM = angleOMJ. Also evident that angle KJM = 180-28-46 = 106. |

May 18th, 2014, 07:20 PM |
#3 |

Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond |
$\displaystyle \text{By the inscribed angle theorem }\angle{JOM}=2\angle{JKM}=56^{\circ}$ $\displaystyle \text{Hence }56+2(46+\angle{KMO})=180\text{ and }\angle{KMO}=\angle{MKO}=16^{\circ}$ $\displaystyle x=\angle{KLJ}=180-28-44=108^{\circ}$ |

May 19th, 2014, 07:43 AM |
#4 |

Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 |
Thanks Greg; how do you remember all them theorems? This problem can be done in 3 steps without using that theorem, agree? x = KMO = MKO ; indirectly given that KJM=106 Step 1 MJO = 46+x, KJO = 28+x; so 46+x+28+x = 106 : x = 16 Step2 MOJ = 180 - 2(46+x) = 180 - 2(46+16) = 56 Etape numero trois: MLO = 180 - x - MOJ = 180 - 16 - 56 = 108 Think the Habs can beat the Rangers? |

May 19th, 2014, 10:23 AM |
#5 |

Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond |
Agree. Price is out for the remainder of the conference final. Looks grim ... |

May 19th, 2014, 10:57 AM |
#6 |

Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 | |