Find the value of angle X ?

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nehilm

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What will be the Value of X?
Attached is the image of question.
 

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I suspect you may have to solve some triangles using trigonometry. How much have you learned of that?
 
I looked at this and found a few additional facts, but no solution. Here's a diagram I drew on Geogebra. Geogebra won't label the angle in question as x, so I labeled it a.


Calculate Angle Trig Problem.png

I got four equations out of this:
\(\displaystyle \begin{array}{l} a + b = 60\\ a + c = 165\\ b + d = 175\\ c + d = 280 \end{array}\)
My first thought was to put these into a matrix and row reduce. They do look like 4 linearly independent equations in 4 variables, so I thought I'd get a unique solution.
\(\displaystyle \left[ {\begin{array}{*{20}{c}} 1&1&0&0&{60}\\ 1&0&1&0&{165}\\ 0&1&0&1&{175}\\ 0&0&1&1&{280} \end{array}} \right]\)
But when I row-reduced, I got this:
\(\displaystyle \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 1}&{ - 115}\\ 0&1&0&1&{175}\\ 0&0&1&1&{280}\\ 0&0&0&0&0 \end{array}} \right]\)
Yes, rank of 3. No solution.

I thought of the sine and cosine laws. We're not given any lengths for the sides. I suppose I could assign an arbitrary length to \(\displaystyle \overline {AB} \), and then gradually work out the lengths of the other sides in terms of that. Once I knew the sides of \(\displaystyle \triangle{CDB} \), I could then work out the angle in question.

Somehow I feel that's not the solution we're meant to find here.
 
(You can make GeoGebra use any label you want by using the Caption feature for the label.)

I expected that working with the angles alone would not be sufficient, even though clearly the angles determine the answer, because I've seen enough problems like this. I would take your idea of letting AB = 1 (or 100, or whatever) and solving some of the triangles. I haven't carried that out yet to confirm my thoughts, but will do so as I have time.
 
bZNyQ7C2 said:
I looked at this and found a few additional facts, but no solution. Here's a diagram I drew on Geogebra. Geogebra won't label the angle in question as x, so I labeled it a.


View attachment 13943

I got four equations out of this:
\(\displaystyle \begin{array}{l} a + b = 60\\ a + c = 165\\ b + d = 175\\ c + d = 280 \end{array}\)
My first thought was to put these into a matrix and row reduce. They do look like 4 linearly independent equations in 4 variables, so I thought I'd get a unique solution.
\(\displaystyle \left[ {\begin{array}{*{20}{c}} 1&1&0&0&{60}\\ 1&0&1&0&{165}\\ 0&1&0&1&{175}\\ 0&0&1&1&{280} \end{array}} \right]\)
But when I row-reduced, I got this:
\(\displaystyle \left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 1}&{ - 115}\\ 0&1&0&1&{175}\\ 0&0&1&1&{280}\\ 0&0&0&0&0 \end{array}} \right]\)
Yes, rank of 3. No solution.
Click to expand...
No no! Not no solution, but infinitely many solutions! Given what you did with the matrix I would now look at the general solution (which will be in terms of d) and see if this information will help. Let us know where this takes you.

You also have at least one mistake in your final matrix. Please look at the 1st row.
 
@Jomo ... ? I know what you mean, I have enough LA for that. But look at the original problem. The information we're given is enough to guarantee that every unknown angle, including x, has just one possible value. A unique solution does exist. Infinitely many solutions is not an option. I admit there is something going on here that I don't see yet. It's as if there's another equation that I haven't found.

I'm afraid I don't see what mistake I made. The first four columns represent a, b, c and d in that order. The first equation is \(\displaystyle a + b = 60\)
or, written differently, \(\displaystyle a + b + 0 \times c + 0 \times d = 60\). Where is the mistake? Is it in the 5th column? Well, \(\displaystyle \left( {65 + 15} \right) + \left( {5 + 35} \right) = 120 = 180 - 60\). The angle at C measures 60.

Maybe you're referring to the RREF. There was a problem with the translation of other of the Latex. The (1,4) entry is -1, not 1. If you look closely, you'll see there is a minus sign there, it just isn't aligned quite right with the 1. Otherwise, we wouldn't have a negative number in (1,5).

I did actually think of writing out the general solution with d as the parameter (or free variable it's sometimes called), but that didn't lead me anywhere. I will take another look. Maybe I missed something.

Meanwhile, I will see what I can do with the law of sines.
 
I solved the problem using trig as I outlined, and found that the answer is very simple, as I expected.

You would think there would be a simple, purely geometric method, but I suspect this problem is related to some standard trick problems in which the numbers are very special. These typically can be solved geometrically, but by considerably more elaborate methods than one expects, such as introducing auxiliary lines and congruent triangles.

Here are a couple references to the problems I'm thinking of, which may be related to this one:
I haven't determined whether any of the methods used for these fit this problem.
 
I did work it out. Assigning to AB the value 100, the first step was

\(\displaystyle \frac{{100}}{{\sin {{80}^ \circ }}} = \frac{{\overline {BD} }}{{\sin {{35}^ \circ }}}\)

leading to \(\displaystyle \overline {BD} \approx 58.242\). After several more steps, I got \(\displaystyle x \approx 39.66\). I believe the exact answer would be 40 degrees.

I can show more details if anyone is interested.

As I said before, I strongly suspect this is not the solution we're meant to find.
 
@Dr.Peterson ... that sounds right. Did you follow the same procedure I did? If so, I expect you found it tedious, albeit straightforward. If not, please let me know what you did. I used the sine law three times and the cosine law once.

I will look at the links you provided.

BTW, did you find any errors in my matrix? If there are any, I'd like to know where they are.
 
I found BC and BD by the law of sines (which is very easy with AB=1); then found CD by the law of cosines, and finally x by the law of sines. None was particularly hard (and my result was 39.999), but with enough searching we can probably find a purely geometric, and exact, solution. Possibly knowing the solution will help in recognizing some triangles of interest.

Your reduced matrix looks fine to me.
 
Last edited:
I did just the same thing you did, except I used AB = 100. If you got more digits in the result than I did, I assume you must've carried more digits in the intermediate calculation.

Thanks for the help.
 
Yes, your rref (last matrix) did not have an easy - sign to see where there should be a -1. So in the end it looks all good.
 
@Jomo .. Thanks.

@Dr.Peterson ... I looked up the links you provided. Interesting. I have a vague memory of seeing this triangle in another context. I seem to remember it was one of the web pages from Jim Wilson (U of Georgia), but that doesn't help me find it, he has a lot of stuff online. I think it was described as a tool for solving certain problems, and I think it was called a "mystic triangle" or something like that. I haven't been able to find it again.

I'll have to take it on faith that these problems can be solved. I don't know if I'm up for that challenge right now. But it strengthens my impression that the problem posed by this thread does have an elementary solution, though probably not an easy one.
 
Did you ever tell us where you got the problem?

But, yes, "elementary" doesn't mean "easy", despite common usage.

I've found no direct connection between this and those I referred to, despite the presence of some of the same angles; but it is still my impression that there is something similar going on.
 
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