Rangpur Higher Secondary 2011/6
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Rangpur Higher Secondary 2011/6
Problem 6:
In the figure $AB = 12$ is the diameter of the circle. $MNAB$ and $\angle BAN = 15^{\circ}$. Find the length of the arc $MN$.

 Posts: 78
 Joined: Thu Jan 20, 2011 10:46 am
Re: Rangpur Higher Secondary 2011/6
Let the centre is O.Connect O,N.ON=OA.Angle OAN=angleONA=15.AB is parallel to MN.Then angle ANM=15.SO that
angleONM=angleOMN=30.So angleMON=120.Then the length of the arc MN is (120/360).2.pi.6=4pi
angleONM=angleOMN=30.So angleMON=120.Then the length of the arc MN is (120/360).2.pi.6=4pi
 Tahmid Hasan
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 Joined: Thu Dec 09, 2010 5:34 pm
 Location: Khulna,Bangladesh.
Re: Rangpur Higher Secondary 2011/6
arc MN is (120/360).2.pi.6=4pi[/quote]
is it a generalized formula????/please explain it.
is it a generalized formula????/please explain it.
Re: Rangpur Higher Secondary 2011/6
Mehfuj Zahir wrote:.Then the length of the arc MN is (120/360).2.pi.6=4pi
is it a generalized formula??please someone explain it.
Re: Rangpur Higher Secondary 2011/6
yes the length of an arc= angle(radian)* radius(910 higher math book droshtobbo)