<![CDATA[Math Is Fun Forum / In triangle ABC]]> 2022-04-25T08:43:58Z FluxBB https://www.mathisfunforum.com/viewtopic.php?id=27119 <![CDATA[Re: In triangle ABC]]> The given paragraphs deal with 2 topics, measurement of Square Meter and Exterior angle bisector of the triangle. The exterior angle bisector of a triangle divides the other facet outwardly within the quantitative relation of the perimeters containing the angle.
In a triangle MNO, MP is that the external bisector of angle M meeting NO created at P. IF MN = ten cm, MO = 6 cm, NO - 12 cm, then realize OP.
Solution:
(NP/OP)  = (MN/MO)
NP = NO + OP= twelve + OP
(12 + OP)/OP = 10/6
6 (12 + OP) = ten OP
72 + six OP = ten OP
72 = ten OP - six OP
4 OP = seventy-two
OP = 72/4
= 18 cm
I get few details of calculation of square meter from https://ksa.mytutorsource.com/blog/how-to-calculate-square-meter/.
Some tools used for measurements like metric tape, meter stick, ruler, etc. Use mensuration tape or ruler from one corner to the opposite corner. You'll be able to modify centimeters into meters by inserting the percentage point to the left by 2 numbers (1 centimeter = zero.01 meters). If measurements are in sq. feet then they will be born again from sq. feet to face meters by multiplying the previous by zero.093 (1 area unit = zero.093 sq. meters). If measurements are in sq. yards, then you may get to multiply them by zero.84. If you have got a fancy form, break it into smaller shapes like triangles and rectangles.

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https://www.mathisfunforum.com/profile.php?id=238300 2022-04-25T08:43:58Z https://www.mathisfunforum.com/viewtopic.php?pid=424901#p424901
<![CDATA[Re: In triangle ABC]]> hi bob
There is an important theorem in the external bisector of an angle in a triangle

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https://www.mathisfunforum.com/profile.php?id=8775 2022-02-12T22:27:39Z https://www.mathisfunforum.com/viewtopic.php?pid=423473#p423473
<![CDATA[Re: In triangle ABC]]> hi tony123

Thanks for posting another interesting puzzle for us.  I'd not met the term external bisector before so I googled it.  Here's what I found: https://archive.lib.msu.edu/crcmath/mat … e/e428.htm

Is that what you mean?

If that's right then the internal and external lines are the same line.  In my diagram it is the internal part that intersects the circle.  I tried moving C to see if I could find a position that requires the external intersection, but I could not. I did notice that F is fixed, wherever I place C, so I guess that's a part of this puzzle.  May I drop the words "An external" and replace it with "The" ?

Bob

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https://www.mathisfunforum.com/profile.php?id=67694 2022-02-12T08:29:18Z https://www.mathisfunforum.com/viewtopic.php?pid=423460#p423460
<![CDATA[In triangle ABC]]> In ∆ ABC, AB =7, AC=5.An external bisector of
∠BAC intersects the circumcircle of
∆ABC at E. Let F be the foot of
perpendicular from E to line AB.
find AF

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https://www.mathisfunforum.com/profile.php?id=8775 2022-02-11T17:14:52Z https://www.mathisfunforum.com/viewtopic.php?pid=423449#p423449