By inspection they're similar triangles, therefore the two unknown (ie, not 90deg) angles are the same. Therefore only need to solve one unknown (I've gone for the smaller of the angles).
150 = 50cosX + 300tanX
3 = cosX + 6tanX
Which looks like it should be solvable for X but I can't for the life of me remember my trig identities right now.
...and to make it fully constrained you need a few (not visible) assumptions: perpendicular, coincident... :-)
Thanks everyone for the replies.
Yes I had tried that approach, one which I found to be the least complicated. But unfortunately it was not successful:
3 = cosx + 6tanx
I apllied the standard trig function equivalents which are:
sin^2x + cos^2x = 1
tanx = sinx/cosx
and these produced
cosx = 1/ (square root of (1+ tan^2x)).
Putting this into our simple equation produces an equation with only one trig function but it is not pretty:
-8/9 = 4tan^4x - 4tan^3x + 5tan^2x - 4tanx
which I cannot solve for tanx. Perhaps you can.
Fredrik, John, and Anna,
Sorrry that I did not explain myself clearly. I am just practising my mathematic skills (or lack of). SW required only the three measurements plus the two right angles in the sketch to make it "defined", but so far I have found that three measurements are not enough to obtain the same result using only pen and paper.
Yes my question is a mathematical question and probably should not be posted in a SW forum. But there are a lot of clever people in this forum and I thought I would try here for a method.
I may have access to a mathematics teacher (for 17 yo's). If so, I will let you know the response.
Without resorting to the impressively lengthy general quartic solution I can't think of another way of approaching it at the point you've got it to. At least you know it's got at least one real root heading in!
Thanks Dave for the pointers to the solution of quartic equations (I had never heard of quartic eqns so didn't google).
As you say, it is an impressive, non-trivial, lengthy process with many detours depending on intermediate results. And yes, knowing there is a real solution, I could crank the handle for the answer, but really all I wanted to know is the method.
The original real-life design issue I had was to find the max width of a thin flat component that could be vertically inserted into a wide, shallow Tee slot shaped recess (blind at both ends), by lowering one edge into the recess first. My sketch in the OP above is a simplified end view, dimensioned to make the small triangle easier to see.
The act of graphically constructing the sketch defines the sides and the angles. SW Measure tool will tell you what they all are.
What is there to solve?
You have geometric constraints in your sketch. Horizontal, Vertical, Perpendicular which help to fully define your sketch.
In SolidWorks go to View and toggle on Sketch Relations to see those geometric constraints.
Both Geometric and Dimensional Constraints work together to fully define your sketches.
There are certainly high school trigonometry methods of solving these types of problems, but normal computer programs cannot handle these methods.
Theses sketches are setting up a set of equations.
If the equations are linear, there is a simple matrix inversion that gives an analytical solution within machine precision. If the system turns out to be non-linear, the computer will iterate to the answer to produce a numerical solution within the arbitrarily selected precision. Normal computer programs will not solve even the simplest non-linear equations analytically. Only specialized programs, like Mathematica, or custom equation solvers do this. The The simplest and most commonly taught method for this is Newton Raphson (http://www.math.ohiou.edu/courses/math3600/lecture13.pdf).
It would be very interesting if someone at SolidWorks could give some high-level insight into the algorithm that sets up the system of equations.
As I see it, the graphical construction (by mouse clicks) defines the coordinates of the verticies.
SW Fully Define checks that there are just sufficient constraints to prevent morphing, rotation and translation. This does not need to know values of lengths or angles, merely that they are defined.
e.g. A polygon of n sides is constrained against morphing by n+(n-3) constraints (linear or angular dimensions). One fixed vertex constrains against translation and a second fixed vertex constrains against rotation.
Measuring the undefined lengths and angles is high school maths.
If each straight line is assumed to be a hypotenuse relative to the H & V axes, then, knowing the offsets of each vertex, Pythagoras theorem will give the length of the line.
The angle of the line relative to H & V axes can be calculated by arccos(opposite/adjacent).
(In high school I had a book of tables to tell me arccos values, and later a slide rule, and later an electronic computer. There must be a formula behind all this.)
Angles between lines can be calculated by subtraction.
We are probably saying the same thing.
I'm not trying to be belligerent with the tags--just trying to separate what I know from what I'm pretty sure about. You are much much smarter than the SW sketch solver. You can spot shortcuts and elegant solutions. While I don't know for certain, I doubt that the sketch solver can even spot a right triangle. Pythagoras' theorem is certainly part of the algorithm that sets up the equations. I'm just saying that trig identities and the quadratic equation are almost definitely not. This stuf is getting dumped into a matrix of equations in a very general manner, and solved with an iterative method.