simple geometry problem

RoundTower

but is there a simple answer?



|AC| = |AB|
CD bisects <ACB
|AD| + |CD| = |BC|
Find θ

MathsManiac

Would 90 degrees be simple enough?

EDIT: ignore this post (see below!)

Tau

I tried this last night, eventually using coordinate geometry and got a huge big equation is cos(theta) which had to have a root somewhere between 60 and 180 degrees. Which you could solve using a computer if you wanted to.

MathManiac, I've been trying to verify your answer, but I can't - could you show how you prove thats it?

Thomas_S_Hunterson

Two similar problems if anyone's interested:


Solve each for X. The latter is tougher (that's a hint).
from: http://thinkzone.wlonk.com/MathFun/Triangle.htm

MathsManiac

Tau wrote: »

I tried this last night, eventually using coordinate geometry and got a huge big equation is cos(theta) which had to have a root somewhere between 60 and 180 degrees. Which you could solve using a computer if you wanted to.

MathManiac, I've been trying to verify your answer, but I can't - could you show how you prove thats it?

Sorry, just realised I misread the question (thought it said |AD| + |CA| = |BC|).

I'll try again, (but this time I reckon it's about 100 degrees!)

RoundTower

Sean_K wrote: »

Two similar problems if anyone's interested:


Solve each for X. The latter is tougher (that's a hint).
from: http://thinkzone.wlonk.com/MathFun/Triangle.htm

this one looks really easy but since it wasn't really easy, it is probably pretty hard.

MathsManiac

The answer (to the original problem) is 100 degrees alright. It's not too bad with trigonometry, but I'm sure there should be a way of doing it without.

Anyway, here's a trig solution:
Let <ACD = t (i.e., the dot).
Then <CBA = 2t, so <CDA=3t and <CAD=180-4t

Taking |AC| to be 1 unit, the sine rule in triangle ACD gives:
|AD| = sin(t) / sin(3t) and |CD| = sin(4t) / sin (3t)

Also, taking E as the midpoint of [CB], triangle ACE gives:
|CE| = cost(2t), so |CB| = 2cos(2t)

We're given that |AD| + |CD| = |BC|, which we can now write as:
[sin(t) / sin (3t)] + [sin(4t) / sin(3t)] = 2cos(2t)

Multiply across by sin(3t) to get:
sin(t) + sin(4t) = 2cos(2t)sin(3t)

Apply a formula to the right-hand side to get:
sin(t) + sin(4t) = sin(5t) + sin (t)

Hence, sin(4t)=sin(5t)
4t + 5t = 180
t = 20

theta = 180 - 4t = 100.

Anyone find a non-trig solution?