Triangulation Help!

Joeyray's Bar
Please, I need help on this.
I cannot provide a reliable image, but I can provide a description of what I need.

-It is a triangle, can be any triangle; Right, Equalateral, Isosceles, or Scalene
-Angles "a" and "b" are known, angle "c" can be deduced from the other two
-Sides A and B are unknown, side C is constant and known
-Point S bisects line C
-Point T is located at the point where lines A and B intersect
-Lines A and B MUST intersect, aka. angles a and b cannot be both 90 degrees

-How do I find the length of a line starting at point S and ending at point T

Now for the somewhat reliable image

~~~~~~~~~~~~~~~~~Point T
~~~~~~~~~~~~~~~~~~/\
~~~~~~~~~~~~~~~~~!/!!!\
~~~~~~~~~~~~~~~~~/!!!!!\
~~~~~~~~~~~~~~~~!/!!!!!!!!\
~~~~~~~~~~~~~~~~/!!!!!!!!!!\
~~~~~~~~~~~~~~~!/_______\
~~~~~~~~~~~angle a````^````angle b
~~~~~~~~~~~~~~~~~line C


That might help a little

Just remember that this triangle won't always be this neat, and it won't always be a right triangle. Sometimes it will be a scalene triangle.

Just to recap, I need to find the distance between point S (Which bisects C) and point T

Can anybody give me a formula for that?
this is relatively simple geometry. first of all, take half of segment AT and make the triangle into two triangles, like so...
................A...A
.............../|...|\
............./..|...|.\
.........../....|...|..\
........./......|...|....\
......./........|...|.....\
...../..........|...|.......\
.B/_______|C.|_____\B
so here segment AC is your solution, and you can find it through first figuring out segment BC by cutting segment AB of your figure in half, and you can figure out segment BA by using sin, cosine, and tangent. remember when finding angle C that all of the angles of triangles add up to 180 degrees.
What if the triangle is scalene? What if angle ACB is not right?

I'm looking for a triangle that cannot be divided into right angles.
AH-HA! GOT IT!

If something bisects line C, it also bisects angle c!
In that case, i can use the law of sines or the law of cosines to find the answer.

So I know angles a and b, and the length of line C.
-Knowing angles a and b, I can find c.
-Next, Bisect line C, to get point S
-Then I take angle c and cut it in half, then form a new line between T and S
-This gives me a new Triangle!!
-I plug angles a, b, and c (bisected) into the law of sines
-LineA/Sine of angle a= LineB/Sine of andle b=LineC/Sine of angle c
-Knowing the length of line C, and all the angles, I can figure the Equasion fo ALL the lines. This gives me line B(New), which I want.

Sorry for all the trouble, I just needed to dig a little deeper.

Hey, at least you know how to describe Instinctive Eye Distance Calculation.

Thank You!
Oh my god.. math makes my head hurt. Least favorite subject of mine by far :(
Having people help you on math homework on the sc2 forums? That's a great idea!!!

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