|
Steve Metke
Posts: 1
Joined: 2/10/2009
|
May Problem Corner |
Flag »
Reply »
|
May 4th (6:20 pm)
Hello Ashley,
Here are my solutions to Problems 184 and 185.
---
Problem #184: Length of AB is 705.52 (705.51938)
When bisecting an angle within a triangle, the bisecting line cuts the opposite side into two smaller lengths which, when compared, are the same ratio as comparing the two sides adjacent the bisected angle. So, in this example, AQ is to QB as AC is to CB.
To solve this problem I used my handy dandy TI83+ graphing calculator and stored these stats:
E= 530.822
D= 587.532
I= 433.293
then, in the graphing function I typed(but did not display):
Y1=sq root of [Isq + Esq-2IEcos(X/2)]
Y2=sq root of [Isq + Dsq-2IDcos(X/2)]
next, I typed and displayed:
Y3=Y1/Y2
Y4=E/D
At the intersection of these Y3 and Y4 you find angle X (78.046447 degrees) which makes this possible.
Lastly, I plugged 78.046447 back into equations Y1 and Y2 to find that
AQ = 334.8645
QB = 370.65488
so, AB = 705.52
----
Problem 185
Nice try but this kind of fuzzy math doesn't hold up when a=b as you'll be dividing by zero which causes the algebra to zoom towards infinity and beyond.
-----
Thanks
Steve Metke
City of Portland (Oregon)
Survey Department
503-807-6179
scmetke@gmail.com
|
|
| |
Monday, May 04, 2009 at 9:23:11 PM |