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Originally posted by GinoJfrom C drop a perpendicular to AB such that you have a right angled triangle with side lengths 2, 1, and AC of length sqrt(5), then just calculate asin(1/sqrt(5))
http://i160.photobucket.com/albums/t166/ginoj_2007/Geometry.jpg
This is the question. We need to find the A angle.
(For ppl who cannot read: The B and C angles are equal; [BC]=[AD]=1 ; [DC]=Square Root 2.)
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Originally posted by strokem1Ok. then why is the other length 2?
A^2+B^2=C^2
Does that help you see why its sqrt(5)?
I'm sorry, but unless there is some rule that tells you what the length of the perpendicular is in a right angled triangle constructed in the manner described then the solution to the problem isn't as straightforward as is being suggested.
Nowhere in the diagram does it suggest that the angles or are right angles, and in any case the length DC is sqrt(2) and not 2. How do you drop a perpendicular from C to AB and prove that the lengths are those given - they may well be but I'd like to know why.
I think that the key to this problem is to do with the fact that ADC + ADB = Pi so that you can use the sine formulae to relate all the lengths to one another, and to the angles and solve for A. Which I will attempt once I've had some lunch.
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Here is a partial solution:
The length BC = 1, the length AB = AC = x and is unknown.
Applying the cosine formula (http://en.wikipedia.org/wiki/Law_of_cosines) to the large triangle gives:
1 = 2*x^2 - 2 x^2 Cos(A), (1)
Applying it to the triangle ADC gives:
2 = 1 + x^2 - 2*x Cos(A), (2)
Eliminating Cos(A) we get.
x^3 - 2x^2 - x + 1 = 0 (3)
Solve equation 3 using a symbolic manipulation program (too much of a pain to do by hand) and substitute the largest root into 1 or 2 to get the answer.
We know that x = AB is greater than AD = 1. The cubic equation y = x^3 - 2x^2 - x + 1 has turning points at x = (2 +/- sqrt(7))/3 so has three real roots. y = 1 for x = 0, for x large the cubic term dominates so y is negative for large negative x and positive for large positive x. For x = 1, y = 1 - 2 - 1 +1 = -1 so the roots of equation 3 are negative, between 0 and 1 and greater than 1. We want the root that is larger than 1.
Edit: Solving (3) numerically gives AB = 2.24698 and the angle A as 0.448799 radians or 25.714 degrees.
Edit 2: Note that 2.24698 != sqrt(5) = 2.236
Originally posted by DeepThoughtImppressive!
Here is a partial solution:
The length BC = 1, the length AB = AC = x and is unknown.
Applying the cosine formula (http://en.wikipedia.org/wiki/Law_of_cosines) to the large triangle gives:
1 = 2*x^2 - 2 x^2 Cos(A), (1)
Applying it to the triangle ADC gives:
2 = 1 + x^2 - 2*x Cos(A), (2)
Eliminating Cos(A) we get.
x^3 - 2x^2 - x + 1 = ...[text shortened]... e angle A as 0.448799 radians or 25.714 degrees.
Edit 2: Note that 2.24698 != sqrt(5) = 2.236
Yes, the answer is 180/7 degrees.
Originally posted by DeepThoughtCould you walk me through how you got from equation (2) to equation(3). I can't follow how you eliminated cos(A) or how you end up with a cubic equation.
Here is a partial solution:
The length BC = 1, the length AB = AC = x and is unknown.
Applying the cosine formula (http://en.wikipedia.org/wiki/Law_of_cosines) to the large triangle gives:
1 = 2*x^2 - 2 x^2 Cos(A), (1)
Applying it to the triangle ADC gives:
2 = 1 + x^2 - 2*x Cos(A), (2)
Eliminating Cos(A) we get.
x^3 - 2x^2 - x + 1 = ...[text shortened]... e angle A as 0.448799 radians or 25.714 degrees.
Edit 2: Note that 2.24698 != sqrt(5) = 2.236
Originally posted by luskinRearrange equation (1) to give:
Could you walk me through how you got from equation (2) to equation(3). I can't follow how you eliminated cos(A) or how you end up with a cubic equation.
2 x^2 Cos(A) = 2*x^2 - 1, (A)
do the same with equation (2) to give:
2 x Cos(A) = x^2 - 1,
therefore (this is probably the step you missed - leaving out the power of two wasn't a typo)
2 x^2 Cos(A) = x^3 - x (B)
the left hand sides of A and B are equal and so:
x^2 - 1 = x^3 - x
Which rearranges to give equation 3 from my post above.
I thought I'd got it wrong at first because the numbers didn't come out to anything neat, which they normally do with this type of problem. I didn't like the look of what comes from substituting into the general equation for the roots of a cubic equation so went for Newton Rapheson instead. I'd be interested to know if there's a way of getting to the answer analytically.
Originally posted by DeepThoughtThanks for that. I see it all now. Yes it would be interesting to see any other way of getting to it. I thought from the look of the diagram that it probably involved the golden ratio somehow. But that line kept leading to dead ends and wrong answers.
Rearrange equation (1) to give:
2 x^2 Cos(A) = 2*x^2 - 1, (A)
do the same with equation (2) to give:
2 x Cos(A) = x^2 - 1,
therefore (this is probably the step you missed - leaving out the power of two wasn't a typo)
2 x^2 Cos(A) = x^3 - x (B)
the left hand sides of A and B are equal and so:
x^2 - 1 = x^3 - x
Which rearranges to ...[text shortened]... nstead. I'd be interested to know if there's a way of getting to the answer analytically.
Originally posted by DeepThoughtApologies folks...about half an hour after submitting that post of mine I realised I'd made the tragic mistake of jumping to a wrong conclusion from my diagram and accepting something without justifying it!...I couldn't get back to edit the post neither 🙁
Here is a partial solution:
The length BC = 1, the length AB = AC = x and is unknown.
Applying the cosine formula (http://en.wikipedia.org/wiki/Law_of_cosines) to the large triangle gives:
1 = 2*x^2 - 2 x^2 Cos(A), (1)
Applying it to the triangle ADC gives:
2 = 1 + x^2 - 2*x Cos(A), (2)
Eliminating Cos(A) we get.
x^3 - 2x^2 - x + 1 = ...[text shortened]... e angle A as 0.448799 radians or 25.714 degrees.
Edit 2: Note that 2.24698 != sqrt(5) = 2.236
Originally posted by AgergThat can't be correct. The length of the perpendicular from C on AB has to be less than 1. Let the foot of the perpendicular from C on AB be P;
from C drop a perpendicular to AB such that you have a right angled triangle with side lengths 2, 1, and AC of length sqrt(5), then just calculate asin(1/sqrt(5))
Then obviously CP < BC and BC =1. Hence your assumption about DP being equal to 1 and the right angled triangle APC having sides of length 2, 1 and AC is wrong. Hence AC cannot be sqrt(5).