What is x?

2. ## Re: What's the angle?

<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">30

3. ## Re: What's the angle?

OK, I must be running on 3 cylinders on this lazy Sunday morning. I can only get an indeterminate value unless I make some extra assumptions.

Alan

4. ## Re: What's the angle?

Well done, it took me quite a while to solve this one.

5. ## Re: What's the angle?

I found that using just the angles was not enough, but you had to take into account the lengths of sides. I won't show all the detail, but an overview:

<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold"> Apex is Pt A, Lower left B, Lower Rt is C, Pt M is the intersection pt on AC (containing angle x), Pt N is the intersection pt on AB.
I assumed the base (BC) was 1 "arbitrary unit".

Given the base size and the angle of the "peak" (Angle BAC) you can calculate the length of the 2 other sides of the triangle (AB = BC) in "units".

Triangle AMB is isoceles since angle mab = mba. Since we "know" AB, we can calc AM = MB. Then we can calc MC = AC-AM. Knowing MA gives us 1 side of the triangle MNC.

If you "drop the perpendicular" from N to AB at pt Q, you can calc the BQ and QC with some trig functions (get the relationship of NQ for each, and then set them equal).
Once you have QC, and you have angle NQC you can calc the length of NC.

With NC and MC and and the angle MCN, you can calculate the length of NM using "Law of Cosines"

Given NM, MC, CN you can calculatea the angle NMC using the "law of cosines". Once you you have angle NMC, you can "easily" calc angle "X"

An alternative approach, is to contruct the diagram, (it is completely defined) and then measure the angle x <img src=/S/grin.gif border=0 alt=grin width=15 height=15></span hide>

Steve

6. ## Re: What's the angle?

The updated graphic shows some additional angles that are easily calculated.

<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">BC / BE = sin 80

7. ## Re: What's the angle?

I actually thought it was going to be a simple "jump out and hit you" - type problem, judging from Tony's description*! I see there's a bit more to it than that - nice descriptive solution Steve, but shouldn't that read:
"drop the perpendicular" from N to BC at pt Q
rather than
"drop the perpendicular" from N to AB at pt Q,

Alan

*Sorry Tony, but your wording of the problem reminds me of that Monty Python sketch of the TV program "How To Do It":

"...but first here's Jackie to tell you how to rid the world of all known diseases."

Jackie: "Well first of all, become a doctor and discover a marvellous cure for something and then, when the medical world really starts to take notice of you, you can jolly well tell them what to do and make sure they get everything right, so that there'll never be diseases anymore."

8. ## Re: What's the angle?

I see it all! But I seem to get reciprocal values of what you get initially in your analysis, using the sine rule:
a/sin(A) = b/sin([img]/forums/images/smilies/cool.gif[/img] ???

Alan

9. ## Re: What's the angle?

Yes, you are correct. I was starting to mix up my variables. I kept redrawing my drawings, figuring I was missing something simple.

I started to try (like Tony did) to get similar triangles, but never found it, so I went thru the brute force method. The technique I used (if made more general), should work even if you do not have isosceles triangles or similar triangles

Steve

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