7148
/www/nrich/html/content/id/7148/
1
2011-02-01T00:00:01
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<mdoxml version="1.0"><p>The three angle bisectors of triangle $LMN$ meet at a point $O$ as shown. <br></br>
<br></br>
$\angle LNM$ is $68^{\circ}$. What is the size of $\angle LOM$?<br></br>
<br></br>
<mdo:image alt="" height="284" src="01%202011.PNG" width="314"></mdo:image><br></br>
<br></br>
If you liked this problem, <a href="http://nrich.maths.org/526&part=">here is an NRICH task</a> which challenges you to use similar mathematical ideas.</p></mdoxml>
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<mdoxml version="1.0"><br></br>
<mdo:image width="314" height="284" src="01%202011a.PNG" alt=""></mdo:image><br></br>
<br></br>
<br></br>
Let<br></br>
<div style="margin-left: 40px;">$\angle OLM = \angle OLN =
a^{\circ},$</div>
<div style="margin-left: 40px;">$\angle OML = \angle OMN = b^{\circ}$
and</div>
<div style="margin-left: 40px;">$\angle LOM = c^{\circ}$</div>
<br></br>
Angles in a triangle add up to $180^{\circ}$, so from $\triangle
LMN$, $$2a^{\circ}+2b^{\circ}+68^{\circ} = 180^{\circ}$$ which
gives $$ 2(a^{\circ}+b^{\circ})=112^{\circ}$$ In other words
$$a^{\circ}+b^{\circ}=56^{\circ}$$<br></br>
<br></br>
<br></br>
Also, from $\triangle LOM$,
$$a^{\circ}+b^{\circ}+c^{\circ}=180^{\circ}$$ and so<br></br>
$$ \eqalign{<br></br>
c^{\circ}&= 180^{\circ} - (a^{\circ}+b^{\circ})\cr &=
180^{\circ}-56^{\circ}\cr &=124^{\circ}}$$<br></br>
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<br></br>
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Weekly Problem 1 - 2011
Weekly Problem 1 - 2011
2D Geometry, Shape and Space
Angle properties of shapes
2D Geometry, Shape and Space
Bisection
2D Geometry, Shape and Space
Triangle theorems
Secondary Mapping Document
Geometrical reasoning US