Difficult exam question

You are suppose to be able to get exact values, and also to be able to do using only the pythagoras theorem, but my friend and I have been working on it for a while and get see how to do it.

That can be done, but not with only the pythagorean theorem. You would have to make use of some other theorems, and I believe a few of the special right-triangle proportions. I can't remember what they are though; I left my math book at school.

Assume that a=1. Find the length of DB. Now find the length of AB. Now use trigonometry to find angles A and B.

Using Pythagoras' theorem in the triangles ABC, ADC, BCD gives you three quadratic equations involving a, AC, CD, DB. From that, you can get the CD/DB ratio, which is the tangent of the angle at B.

Quote:

Original post by Anonymous Poster Quote: Original post by Sneftel Assume that a=1. Find the length of DB. Now find the length of AB. Now use trigonometry to find angles A and B. That would be in terms of a. There are no exact values given, making it impossible to find exact values.

The question asks for angles. The scale of the triangle thus does not matter. This is basic geometry.

Trivial.
let h=DC
let alpha=CBD
let beta=CAD=90-alpha
then h=sin(alpha)*a=tan(beta)*a=tan(90-alpha)*a
we have:
sin(alpha)*a=tan(90-alpha)*a
There, a cancels out. Har har har.
sin(alpha)=tan(90-alpha)
sin(alpha)=cos(alpha)/sin(alpha) // that's because sin(90-alpha)=cos(alpha) and cos(90-alpha)=sin(alpha)

sin(alpha)^2=cos(alpha)
sin(alpha)^2+cos(alpha)^2=1=cos(alpha)+cos(alpha)^2
cos(alpha)^2+cos(alpha)-1=0
it have roots
-1/2 [+/-] sqrt(1/4 + 1)
so
cos(alpha) = sqrt(1.25)-0.5

alpha = 0.904556894302381364 radians = 51.8272923729877525 degrees. You can easily derive all other angles from there.
Note: it's late night there.

i know police on nongamemath, but anyway.

As exercies, you can try solving as others proposed.

Oh dear, this problem looks just like one I had on a math test on friday. I didn't want to see it again so soon. The one on my test were two similiar triangles in themselves, but this one would need one more piece of information to be similiar triangles (either AC=DC, or angle B=angle A). If I were you I would draw tri ACD and tri BAC (those are corresponding angles) seperatly, then it might trigger something.

~zix~

BTW: the astute reader will recognize an interesting congruence between this problem, and my user icon.

Quote:

Original post by Sneftel BTW: the astute reader will recognize an interesting congruence between this problem, and my user icon.

Indeed.[grin]

How many degrees are in every triangle? You don't need to prove anything, you are given one 90 degree angle and one 45 degree angle for both triangles. Whats left, 45 degrees.

Quote:

Original post by jeff33991 How many degrees are in every triangle? You don't need to prove anything, you are given one 90 degree angle and one 45 degree angle for both triangles. Whats left, 45 degrees.

Where can you see the 45 degree angle?

I actually had this exact question in my matricular exam few weeks ago... It is easily solved if you recognize that there are two similar triangles (as they both have two similar corners). If you mark BD as b, then you get this:

a/(a+b) = b/a

You get two answers, and only other is valid (because the lenght cannot be negative):

b = .5(-1+sqrt(5))a

Then you can solve the angle B (DBC) with cosine, as you can cancel out a. Angles are 38, 52 and 90. Sorry for my lack of proper english math terms, I only know what they are for sure in finnish :P