A post from 7-17-07 generated some wonderful comments and solutions. I decided to bring it back using a slightly different diagram. This time I'm asking for a solution path using exterior angles, a tool students often overlook. The most efficient approach may still be the one tc used, involving the sum of the interior angles of a quadrilateral, but the challenge here is to find another way...
Heres' the problem:
Assume Q, S and T are collinear. Determine the value of a+b+c.
Note: Again, there are many wonderful approaches here. Try to use the suggested one...
Saturday, December 8, 2007
The Kite Problem Revisited -A View From the 'Exterior'
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11 comments:
The supplement to the 40-degree angle is 140 degrees. If you draw in QT, then you know that the triangle SQT has an angle sum of 180, so the sum of angles SQT and STQ is 40.
Since the measure of angles a plus b plus c plus SQT and STQ is 180, then a + b + c = 140.
anon--
I'm guessing you meant QR, not QT.so I replaced T by R throughout your explanation. Yes, your method works and is one I've seen some students like.
Another high-level explanation I've seen is to use the fact that the sum of the 6 angles in the 2 triangles is of course 360. The 2 unmarked angles, RTS and PTQ, have a sum of 180. Therefore,
a+b+c= 360 - (40+180) = 140.
However, neither of these nice solutions uses the exterior angle theorem - the method I requested.
I'll get you started:
Denote the measure of angle RTS by x. Then 140 = c+x by the Exterior Angle Theorem. Now use the theorem again!
Using exterior angles:
Extend segments QP, PR, RS, and SQ to get rays QP, PR, RS, and SQ. The rays form the four sides of the quadrilateral QPRS. The sum of the exterior angles of any polygon is equal to 360 degrees. That means exterior angle P (formed by ray QP and ray PR) + exterior angle R + exterior angle S (note that this exterior angle is -40 degrees, since it is concave)+ exterior angle Q = 360 degrees. Here's the equation, then, substituting the exterior angles with 180 (a line) minus angles a, b, and c, and using the vertical angle of the angle marked 40 degrees:
(180-a)+(180-b)+(180-c)+(-40)=360
Move a, b, and c to the one side, and the constants to the other:
180+180+180-40-360=a+b+c
Therefore, a+b+c=140.
Nice, Melissa!
I was looking for exterior angles of a triangle but your approach works! Your idea to use -40 for one of the exterior angles was inspired. It makes sense since one normally uses 180 - (interior) and 180-220 = -40. It's a bit harder to use the usual definition of exterior angle (adjacent and supplementary to interior angle) in the non-convex case - a negative angle makes sense here.
You might want to try the usual exterior angle theorem for triangles. I started you off in my last comment to anaonymous.
Oh!!! I see it now.
Yes, mark angle STR as x degrees. That means x+c=140, using the Exterior Angle Theorem. Then mark angle x's vertical angle as x degrees as well. That angle is equal to a+b. So now x+c=140 and x=a+b. Using substitution, I arrived at a+b+c=140.
This took me a while to see, but finally I realized this is a very useful method.
Oh, by the way, cotton blossom is me, Melissa. (Hi!)
I read the directions, but I still like interior angles here. Why ignore a 1-step approach?
Blogger is not letting me post as me, only as my old blogger id. Yecch.
Here's me: jonathan
"Jonathan": Perhaps the reason Mr. Marain posted this problem was to see how many different ways there are to solving it. The way he initially did it was (i think) interior angles of a quadrilateral as well. I guess this could be seen as a rather simple problem, but you wouldn't believe some other ways i've seen--things like labeling angles STR and STP (lol--STP) with DOTS and then subtracting their sum, 180 from...I don't even remember. But it was yet another way.
The most obvious method to me is:
QTP = c + 40
a + b + QTP = 180
a + b + c + 40 = 180
Therefore, a + b + c = 140
This reminds me of the type of homework puzzles we have in the geometry section of our Singapore math book.
Oops...
I just realized x itself was an exterior angle itself--no need to label the vertical angle.
denise--
Your solution is equivalent to mine but here's what I did using exterior angles twice:
Let x = the measure of ∠RTS.
(1) ∠RSQ is an exterior angle of triangle RTS so:
140 = x+c
(2) But ∠RTS is an exterior angle of triangle PQT so:
x = a+b
Now substitute (2) into (1).
jonathan--
Melissa expressed my reasons for this post better than I! While some students become confused when seeing a variety of solution paths, I believe the benefits outweigh the disadvantages. In many cases, these different approaches only occur to me because students came up with them at one time or another. As you know well (because you do it in your classroom!), this is what happens when we take the time to enable these kinds of dialogues in class. I certainly like the efficiency of the quadrilateral method but I also believe there's something elegant about exterior angles.
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