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Subject: "Constucting a triangle instructions"     Previous Topic | Next Topic
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Conferences The CTK Exchange Middle school Topic #132
Reading Topic #132
Gerald B.
guest
May-20-08, 12:15 PM (EST)
 
"Constucting a triangle instructions"
 
   Dear Alex,

I’ve been enjoying your math-focused website. I ran across this explanation for triangle construction given an angle “A”, an opposite side “a” and a height (at “A”) of “h”. I’m trying to follow the instructions but keep getting stuck at the first part. I just can’t figure out how to “Construct an arc (part of a circle) subtending segment BC = a such that for every point A on it, angle BAC equals the given angle A” How do I choose the radius of the arc or its starting point?

Any hints?

Insructions from the site:


  • A, a, ha

Construct an arc (part of a circle) subtending segment BC = a such that for every point A on it, angle BAC equals the given angle A. Draw a line parallel to BC at the distance ha. There could be 0,1, or 2 points of intersection of the line with the arc. In the latter case solutions are mirror images of each other; so even then there is at most one solution.


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alexbadmin
Charter Member
2228 posts
May-20-08, 12:20 PM (EST)
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1. "RE: Constucting a triangle instructions"
In response to message #0
 
   Since angle A is given, you can draw a circle of an arbitrary radius R' and find a chord B'C' = a' that subtends the required angle. Now, you have to solve a proportion

R/a = R'/a'

for radius R.


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Gerlad B.
guest
May-20-08, 01:31 PM (EST)
 
2. "RE: Constucting a triangle instructions"
In response to message #1
 
   OK I think I get it. Say angle A = 30 degrees. (Forget a and h for the moment). On a clock face, I find that if I put A’ at the centre, legs coming out at 30 degree angles cross the other edge of the clock one hour apart (say 6 and 7 o’clock). What I take from this is the radius of the clock and the linear distance between 6 and 7 (not in hours!). That gives me a ratio. Now, I am given the linear distance BC=a, so I can scale up my clock so that B’C’: BC = R’:R


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alexbadmin
Charter Member
2228 posts
May-20-08, 01:32 PM (EST)
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3. "RE: Constucting a triangle instructions"
In response to message #2
 
   That is almost correct. The central angle is twice the inscribed angle. So that if your A = 30 degrees, you may want to consider the arc between 6 and 8 o'clock instead of 6 and 7.

The rest is OK.


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