The figure shows a triangle ABC,m< A = 4x, m< C = 3x, m< CBD = 5x, and AB = CD. Find x.

User posted image: User posted image

4x + 3x +5x = 12x
12x = 180 (Degrees)
x = 180/12 = 15 (Degrees)

Correct me if I'm wrong...

Are you in 9th Grade Geometry by any chance? I learned this last year, and may have well forgotten, so...

Oops, I didn't even bother to look at the image, and only viewed the triangle in my mind... this is inaccurate, and I will take a look.

Well, for now We'll wait and see... I'm a little 'rusty', and I keep coming up with 0 = 0 ... This would mean that there are infinite many solutions, but I doubt that there are... For now though, I will go with that answer... Infinite many solutions.

The sine rule makes it in to a fairly simple problem. You only need to write down two equations and then solve for x, no need to work out any relation between side lengths or anything like that.

can you pleas complete the solution

i know its x=10 but how

Do you understand how the sine rule can be applied here?

http://www.gogeometry.com/problem/problem013.htm

Given point D lies between point A and point C.
Given AB = CD.
Given x is an angle.
Given angle BAD = 4x
Given angle CBD = 5x
Given angle BCD = 3x

Since angle ADB is an exterior angle of triangle BCD, then angle ADB = angle CBD + angle BCD. Thus angle ADB = 8x.

Since BD is a side shared by the two triangles, call this length M. Since AB = CD, call this length L. Then by the sine theorem, referenced by AlphaNumeric,

(sine angle BAD) / (BD) = (sine angle ADB)/(AB) and
(sine angle BCD) / (BD) = (sine angle CBD)/(CD) or
(sine 4x)/M = (sine 8x)/L and (sine 3x)/M = (sine 5x)/L or
(sine 4x)/(sine 8x) = M/L and (sine 3x)/(sine 5x) = M/L

without knowing M/L, we set equals against equals in the equation of one variable

(sine 4x)/(sine 8x) = (sine 3x)/(sine 5x) or
(sine 5x) = (sine 3x)(sine 8x)/(sine 4x)

Using the fact that (in radians) sine x = [ e^(ix) - e^(-ix) ] / (2i) we expand and reduce the expression.

[ e^(i5x) - e^(-i5x) ] / (2i) = [ [ e^(i3x) - e^(-i3x) ] / (2i) ] [ e^(i8x) - e^(-i8x) ] / [ e^(i4x) - e^(-i4x) ]
or
[ e^(i5x) - e^(-i5x) ] / (2i) = [ [ e^(i3x) - e^(-i3x) ] / (2i) ] [ e^(i4x) + e^(-i4x) ]
or
e^(i7x) - e^(i5x) - e^(ix) + e^(-ix) + e^(-i5x) - e^(-i7x) = 0 or
(e^(i2x) - 1)(e^(-i7x))(e^(12ix) - e^(6ix) - 1) = 0
Since we know the solution is a finite, acute angle, by convention in the first quadrent, we discard the first two factors.
e^(12ix) - e^(6ix) - 1 = 0 says
e^(6ix) = ½ ± i½√3 which says sine 6x = ±½√3 which says x = 10°

Knowing this, every angle and side length ratio is known.

It should be noted that sine 30° and sine 60° and sine 90° are nice, while sine 10° and 40° and sine 50° are not in terms of simpler expressions. But because of trigonometry, relations like 2 sine 40° sine 50° = sine 100° hold true.

It should be noted that on the original page, the text on the page has little to do with the problem in the diagram. It was incorrectly copied from problem 1
http://www.gogeometry.com/problem/problem001.htm
... as were the hints.

Note: This is my post 2500