Can you solve this right triangle puzzle with a unique angle ratio?

[Wilmer]

B
    b-a
            D
a
                   c-b+a

C                b                  A

Right triangle ABC, with the standard a, b, c side lengths.
Angle BAC = 2u degrees.

Point D is on the hypotenuse AB, such that:
BD = b-a, angle BCD = 3u degrees.

Calculate u.

 

[MarkFL]

Is this a problem with which you are asking for help, or is it a problem that you are posting as a challenge?

 

[Wilmer]

Is this a problem with which you are asking for help, or is it a problem that you are posting as a challenge?

Hey Mark!
It's a problem I found.
My solution to it is that it HAS to be a 30-60-90 triangle,
else the "b-a" length is impossible (combined with 2u and 3u)..
Posted it to sort of "see" if anyone agrees.

Anyhoo...if I've done something sinful, I apologize :)

 

[MarkFL]

Hey Mark!
It's a problem I found.
My solution to it is that it HAS to be a 30-60-90 triangle,
else the "b-a" length is impossible (combined with 2u and 3u)..
Posted it to sort of "see" if anyone agrees.

Anyhoo...if I've done something sinful, I apologize :)

Since you aren't sure if you are correct and are wanting to see if people agree with your solution, I would say you posted in the correct place. (Yes)

 

[MarkFL]

Let's use some coordinate geometry here and WLOG, let $a=1$...let the hypotenuse lie along the line:

\(\displaystyle y=1-\frac{x}{b}\)

And line segment $\overline{CD}$ lie along the line:

\(\displaystyle y=\cot(3u)x\)

And so we find $D$ is at:

\(\displaystyle (x,y)=\left(\frac{b}{b\cot(3u)+1},\frac{b\cot(3u)}{b\cot(3u)+1}\right)\)

Now, we require:

\(\displaystyle (b-1)^2=\left(\frac{b}{b\cot(3u)+1}\right)^2+\left(\frac{1}{b\cot(3u)+1}\right)^2\)

From this, we determine:

\(\displaystyle \cot(3u)=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}\)

And using a triple-angle identity for cotangent, we have:

\(\displaystyle \frac{3\cot(u)-\cot^3(u)}{1-3\cot^2(u)}=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}\)

We also know:

\(\displaystyle \cot(2u)=b\)

and using a double-angle identity for cotangent, we have:

\(\displaystyle \frac{\cot^2(u)-1}{2\cot(u)}=b\)

Form this, we determine:

\(\displaystyle \cot(u)=b+\sqrt{b^2+1}\)

And so we have:

\(\displaystyle \frac{\left(b+\sqrt{b^2+1}\right)\left(b^2+b\sqrt{b^2+1}-1\right)}{3b^2+3b\sqrt{b^2+1}+1}=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}\)

Taking the positive real root, we find:

\(\displaystyle b=\sqrt{3}\)

And this in turn implies:

\(\displaystyle \cot(3u)=1\)

or:

\(\displaystyle 3u=45^{\circ}\)

\(\displaystyle u=15^{\circ}\)

 

[Wilmer]

Thanks Mark. I'm lazy, so I went at it this way:

Assume ABC is a 30-60-90 triangle.

Let a = 1: then c = 2 and b = sqrt(3)

Since angleBAC = 2u and angleBCD = 3u,
then angleBCD = 45 and angleBDC = 75

BD = SIN(45) / SIN(75) = sqrt(3) - 1
also:
BD = a - b = sqrt(3) - 1

That's it...u=15 ! Only case where above works...

Anything wrong with that?
I'd sure go that way on a multiple choice timed test!

 

[Wilmer]

Interestingly(?) enough, seems to be only one pytagoreaner
that has these angles in a mu:nu ratio (m and n = integers).

Happens to be the 3-4-5er (and its multiples, of course).
BD : DA = 1 : 4

AngleBAC = 2u and angleBCD = u

Checked 'em all where a < 10000.

File 13 stuff!