Trig problem

A

apple2357

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I am working on this problem. Trying to find alpha ( lets call it 'x' instead) for this isosceles triangle.

WhatsApp Image 2021-03-05 at 10.43.06.jpeg

I have been using the sine rule on the two middle triangles and managed to get to this point:

sin(30) sin (90+x) = sin50sin(150-x)

By 'guesswork and some graphical work' i have established the answer is x= 10. I am pretty sure that is correct.

However, I can't see how this is going to fall out easily.

I can do a little more, so the next line is

(0.5) cos x = sin50( 0.5 cosx+ (root3)/2 sinx)

But I can't seem to see a way around the sin50.

There might be a better way to solve the original problem that I am missing too, so any thoughts welcome!
 
1614955840239.png
Here's I got the answer: Call AB = 1; find AD in triangle ABD; find AC in triangle ABC; find CD in triangle ACD; and then find angle ACD.

I've seen a number of somewhat similar problems that can in principle be solved exactly, without trig; I haven't attempted that, since you are using trig.
 
Dr.Peterson said:
View attachment 25557
Here's I got the answer: Call AB = 1; find AD in triangle ABD; find AC in triangle ABC; find CD in triangle ACD; and then find angle ACD.

I've seen a number of somewhat similar problems that can in principle be solved exactly, without trig; I haven't attempted that, since you are using trig.
Click to expand...

When you say without trig?? I will pursue my line but would also be curious to have a hint that doesn't rely on trig!
 
apple2357 said:
When you say without trig?? I will pursue my line but would also be curious to have a hint that doesn't rely on trig!
Click to expand...
I don't actually have a hint (and haven't looked for one, as I said). But since the answer looks like an exact integer, there should be a proof that it is exact; and I've seen other problems where the same issue arises. Here is an example:

World's Hardest Easy Geometry Problem

World's Hardest Easy Geometry Problem
thinkzone.wlonk.com thinkzone.wlonk.com
 
Which triangle is isosceles? And which of its sides are equal?
 
apple2357 said:
I can do a little more, so the next line is

(0.5) cos x = sin50( 0.5 cosx+ (root3)/2 sinx)

But I can't seem to see a way around the sin50.
Click to expand...


What you have done is perfect - one step left to do.

You have: Cos [MATH]\alpha[/MATH] = Sin 50 (Cos [MATH]\alpha[/MATH] + [MATH]\surd[/MATH]3 Sin[MATH]\alpha[/MATH]) (multiplying by 2)
Since Cos [MATH]\alpha[/MATH] is clearly non-zero, you can divide across by this:

1615140146726.png

This question could actually be done using 'simple' trig - i.e. using right-angled triangles. (See attached file).

However, as Dr.Peterson suggests, it is an interesting question to see if it is possible to prove that [MATH]\alpha[/MATH] is exactly 10 (we have used a calculator), with or without trigonometry! I too would be very interested to see a proof.
 

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apple2357 said:
I am working on this problem. Trying to find alpha ( lets call it 'x' instead) for this isosceles triangle.

View attachment 25545

I have been using the sine rule on the two middle triangles and managed to get to this point:

sin(30) sin (90+x) = sin50sin(150-x)

By 'guesswork and some graphical work' i have established the answer is x= 10. I am pretty sure that is correct.

However, I can't see how this is going to fall out easily.

I can do a little more, so the next line is

(0.5) cos x = sin50( 0.5 cosx+ (root3)/2 sinx)

But I can't seem to see a way around the sin50.

There might be a better way to solve the original problem that I am missing too, so any thoughts welcome!
Click to expand...
A terribly worded problem.

"Trying to find alpha for this isosceles triangle" but it is not clear that alpha is even in the isosceles triangle, which is not identified.

No points are labelled. There are a bunch of measurements in degrees, but it is not clear which angles correspond to which angles.

If triangle ABC is supposed to isosceles, why do the base angles appear to be unequal?

I might work on this problem if I had a clue what it was.

We ask in the guidelines for a complete and exact statement of the problem for a reason.
 
JeffM said:
If triangle ABC is supposed to isosceles, why do the base angles appear to be unequal?
Click to expand...
The base angles are both 70 degrees.

Yes, the wording given would be poor for an actual exercise, but I think it's hard to interpret apple2357's paraphrase otherwise than "find angle alpha in this picture, given the indicated angles". The mention of the triangle being isosceles, and the false implication that alpha is an angle of the isosceles triangle, was unnecessary.
 
Dr.Peterson said:
The base angles are both 70 degrees.

Yes, the wording given would be poor for an actual exercise, but I think it's hard to interpret apple2357's paraphrase otherwise than "find angle alpha in this picture, given the indicated angles". The mention of the triangle being isosceles, and the false implication that alpha is an angle of the isosceles triangle, was unnecessary.
Click to expand...
Oh. I really was confused. I thought it was a three-dimensional prism.
 
lex said:
What you have done is perfect - one step left to do.

You have: Cos [MATH]\alpha[/MATH] = Sin 50 (Cos [MATH]\alpha[/MATH] + [MATH]\surd[/MATH]3 Sin[MATH]\alpha[/MATH]) (multiplying by 2)
Since Cos [MATH]\alpha[/MATH] is clearly non-zero, you can divide across by this:

View attachment 25594

This question could actually be done using 'simple' trig - i.e. using right-angled triangles. (See attached file).

However, as Dr.Peterson suggests, it is an interesting question to see if it is possible to prove that [MATH]\alpha[/MATH] is exactly 10 (we have used a calculator), with or without trigonometry! I too would be very interested to see a proof.
Click to expand...


Very impressed with your method here. I did spot that there was a potential right-angled triangle within it but couldn't see what to do next. Thanks!
 
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