My friend gave me this problem today, and it’s been bugging me. I want to figure it out on my own, but I have no idea what to do. See if you can figure it out. I attached the problem and a picture using Word. Hope it works. Oh, and it requires use of the law of sines (sinA/a=sinB/b=sinC/c) and quite a bit of basic algebra. Good luck!
triangle.doc (26 KB)
I found this very messy even though I approached it in what I thought was a simple way. I eliminated C and c (giving up the known c=12) to isolate A, B, a, and b. You know that A+B+C=180 and A-B=2C so you can eliminate C with these two equations. This way you can assign a=9 and b=5 and scale the triangle at the end. I had to use one of the triple angle identities using this approach. I also had to use a calculator to determine the final (scaled) lengths a=17.99998 and b=9.99998. Obviously, the solution is a=18, b=10, c=12 and it can probably be arrived at algebraically. That is, I think it should be possible to do this without calculating sines and arcsines repeatedly.
Edit: Oh, another thing that was useful to determine ahead of time is that A has to be the largest angle and 90<A<120. This comes from reduction to 3A+B=360.
Whew. It can be solved with much algebra.
a=18, b=10, c=12
sinA=sqrt(8/9), sinB=(5/9)*sqrt(8/9), sinC=(2/3)sqrt(8/9)
A=109.47, B=31.59, C=38.94
I had to use a triple angle identity and a double angle identity. A calculator is only required for the value of the angles, not the sines of the angles or the sides. Finding the lengths of the other two sides solves the triangle by SSS.
When I think of Harper solving this I see his eyes rolling like one-armed-bandit wheels and ‘beep-boop’ sounds coming from his head.
Then he opens his mouth and a small piece of paper pops out on his tongue making a ‘ping’ noise with the answer on it.
Maybe a little smoke from his ears too.
I had time to write it out. Hope this is clear enough.
Given: A-B=2C, b=5a/9, c=12
Known: A+B+C=180, sinC/12=sinA/a=sinB/b
1.) A-B-2C=0
2.) A+B+C=180
multiply eq. 2. by 2 and add to eq. 1.
3.) 3A+B=360
4.) B=360-3A (this indicates 120>A>90 for 0<B<90.)
therefore A is obtuse.
if sinA/a=sinB/b and b=5a/9 then
5.) sinB=(5/9)sinA
substitute eq. 4. for B
6.) sin(360-3A)=(5/9)sinA
7.) -sin3A=(5/9)sinA
use triple angle identity -sin3A=-3sinA+4(sinA)^3
8.) -3sinA+4(sinA)^3=(5/9)sinA
9.) 4(sinA)^2-3=(5/9)
10.) 36(sinA)^2-27=5
11.) (sinA)^2=32/36=8/9
12.) sinA=sqrt(8/9)
13.) sinB=(5/9)sqrt(8/9)
again, from eq. 1. and eq. 2. extract
14.) 2A-C=180
15.) C=2A-180
16.) sinC/12-sinA/a so
17.) sin(2A-180)=12sinA/a
18.) asin(2A-180)=12sinA=12sqrt(8/9)
19.) -asin2A=12sqrt(8/9)
double angle identity
20.) -2a*sinAcosA=12sqrt(8/9)
from cos^2+sin^2=1 solve for cosA
21.) cosA= sqrt(1-8/9)=sqrt(1/9)=1/3
therefore
22.) (-2a/3)sqrt(8/9)=12sqrt(8/9)
23.) a=(-3/2)12=-18
only care about the magnitude
24.) a=18, b=5a/9=10, c=12
Edit: Remember that A is obtuse so it is a second quadrant angle. To use a first quadrant sine to determine a second quadrant angle, A=180-arcsin(sqrt(8/9))
Idiot. That’s a typo.
16.) sinC/12=sinA/a
How could anyone follow a mess like that with a typo in it? Where is dear Miss Ayelery when you need her?