Trigonometry: A regular octagon is inscribed in a circle of
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Re: Trigonometry
Hello, watson542!
What a dirty trick . . . They actually expect you to think about it!
\(\displaystyle O\) is the center of the circle.
\(\displaystyle AB\) is the side of the octagon.
Note that the central angle \(\displaystyle AOB\,=\,45^o.\)
\(\displaystyle \Delta AOB\) is an isosceles triangle with vertex angle \(\displaystyle 45^o\) and equal sides \(\displaystyle r\,=\,15.8\)
Using the Law of Cosines: \(\displaystyle \:AB^2\:=\:15.8^2\,+\,15.8^2\,-\,2(15.8)(15.8)\cos45^o\)
. . . . . \(\displaystyle AB^2\:=\:2(15.8)^2\,-\,2(15.8)^2\cos45^o \:=\:2(15.8)^2(1\,-\,\cos45^o)\)
. . . . . \(\displaystyle AB^2\:=\:4(15.8)^2\cos22.5^o\)
Hence: \(\displaystyle \,AB \:=\:2(15.8)\cos22.5^o\)
Can you finish it now?
Hello, watson542!
What a dirty trick . . . They actually expect you to think about it!
Code:
A
* * *
* | \ *
* | *B
* r| / *
|45°/r
* | / *
* - - - - * - - - - *
O\(\displaystyle O\) is the center of the circle.
\(\displaystyle AB\) is the side of the octagon.
Note that the central angle \(\displaystyle AOB\,=\,45^o.\)
\(\displaystyle \Delta AOB\) is an isosceles triangle with vertex angle \(\displaystyle 45^o\) and equal sides \(\displaystyle r\,=\,15.8\)
Using the Law of Cosines: \(\displaystyle \:AB^2\:=\:15.8^2\,+\,15.8^2\,-\,2(15.8)(15.8)\cos45^o\)
. . . . . \(\displaystyle AB^2\:=\:2(15.8)^2\,-\,2(15.8)^2\cos45^o \:=\:2(15.8)^2(1\,-\,\cos45^o)\)
. . . . . \(\displaystyle AB^2\:=\:4(15.8)^2\cos22.5^o\)
Hence: \(\displaystyle \,AB \:=\:2(15.8)\cos22.5^o\)
Can you finish it now?
Soroban's cosine is the easiest way to go about this sort of thing. But suppose you couldn't use trig. Archimedes didn't have trig as we know it, but he managed to estimate pi very closely by inscribing polygons in a circle. Starting with a square, then a octagon, then a 16-gon, 32-gon, etc.
My looking that the diagram we can see:
a)\(\displaystyle \L\\(\frac{79}{5}-x)^{2}+(\frac{s}{2})^{2}=(\frac{79}{5})^{2}\)
b) \(\displaystyle \L\\x^{2}+(\frac{s}{2})^{2}=t^{2}\)
Squaring the binomial in a, we have:
\(\displaystyle \L\\x^{2}-\frac{158}{5}x+\frac{s^{2}}{4}=0\)....[c]
Using b in c and solving for x:
\(\displaystyle \L\\x=\frac{5t^{2}}{158}\)....[d]
Now, using a:
\(\displaystyle \L\\(\frac{79}{5}-x)^{2}=(\frac{79}{5})^{2}-(\frac{s}{2})^{2}\)
Square both sides:
\(\displaystyle \L\\\frac{79}{5}-x=\frac{\sqrt{24964-25s^{2}}}{10}\)
Sub in [d]:
\(\displaystyle \L\\\frac{79}{5}-\frac{\sqrt{24964-25s^{2}}}{10}=\frac{5t^{2}}{158}\)
Solve for t:
\(\displaystyle \L\\t=\frac{\sqrt{79}\sqrt{158-\sqrt{24964-25s^{2}}}}{5}\)
Now, though it appears to be a monster, we can use this to find the perimeter of any subsequent n-gons, given the preceeding n-gon.
Starting with the square, which we know already has side length \(\displaystyle \frac{79\sqrt{2}}{5}\). Call it \(\displaystyle s_{1}\)
Subbing that into the formula above, we find the the side length of the octagon is \(\displaystyle \L\\\frac{79\sqrt{2-\sqrt{2}}}{5}\approx{12.09}\)
Multiply by 8 and we see the perimeter is \(\displaystyle \H\\\frac{632\sqrt{2-\sqrt{2}}}{5}\approx{96.74}\).
Which agrees with the cosine method.
You know the circumference of the circle is \(\displaystyle 2{\pi}(\frac{79}{5})=\frac{158{\pi}}{5}\approx{99.27}\)
The octagon perimeter must be smaller than that. As you increase the number of sides on the n-gon, the perimeter of the n-gon approaches the circumference of the circle. That's intuitive. That's how Archimedes estimated pi.
My looking that the diagram we can see:
a)\(\displaystyle \L\\(\frac{79}{5}-x)^{2}+(\frac{s}{2})^{2}=(\frac{79}{5})^{2}\)
b) \(\displaystyle \L\\x^{2}+(\frac{s}{2})^{2}=t^{2}\)
Squaring the binomial in a, we have:
\(\displaystyle \L\\x^{2}-\frac{158}{5}x+\frac{s^{2}}{4}=0\)....[c]
Using b in c and solving for x:
\(\displaystyle \L\\x=\frac{5t^{2}}{158}\)....[d]
Now, using a:
\(\displaystyle \L\\(\frac{79}{5}-x)^{2}=(\frac{79}{5})^{2}-(\frac{s}{2})^{2}\)
Square both sides:
\(\displaystyle \L\\\frac{79}{5}-x=\frac{\sqrt{24964-25s^{2}}}{10}\)
Sub in [d]:
\(\displaystyle \L\\\frac{79}{5}-\frac{\sqrt{24964-25s^{2}}}{10}=\frac{5t^{2}}{158}\)
Solve for t:
\(\displaystyle \L\\t=\frac{\sqrt{79}\sqrt{158-\sqrt{24964-25s^{2}}}}{5}\)
Now, though it appears to be a monster, we can use this to find the perimeter of any subsequent n-gons, given the preceeding n-gon.
Starting with the square, which we know already has side length \(\displaystyle \frac{79\sqrt{2}}{5}\). Call it \(\displaystyle s_{1}\)
Subbing that into the formula above, we find the the side length of the octagon is \(\displaystyle \L\\\frac{79\sqrt{2-\sqrt{2}}}{5}\approx{12.09}\)
Multiply by 8 and we see the perimeter is \(\displaystyle \H\\\frac{632\sqrt{2-\sqrt{2}}}{5}\approx{96.74}\).
Which agrees with the cosine method.
You know the circumference of the circle is \(\displaystyle 2{\pi}(\frac{79}{5})=\frac{158{\pi}}{5}\approx{99.27}\)
The octagon perimeter must be smaller than that. As you increase the number of sides on the n-gon, the perimeter of the n-gon approaches the circumference of the circle. That's intuitive. That's how Archimedes estimated pi.
Thanks galactus; very interesting.
I think Soroban typoed and means AB = 2(15.8)sin(22.5)
I don't think the Cosine Law is the easiesy way:
dropping a perpendicular from O to AB (Soroban's diagram)
gives you 2 identical right triangles, so right away you get
AB/2 = 15.8sin(22.5).
Note: also much simpler than my "area" suggestion :shock:
I think Soroban typoed and means AB = 2(15.8)sin(22.5)
I don't think the Cosine Law is the easiesy way:
dropping a perpendicular from O to AB (Soroban's diagram)
gives you 2 identical right triangles, so right away you get
AB/2 = 15.8sin(22.5).
Note: also much simpler than my "area" suggestion :shock:
Trigonometry
Ok, I have read all responses and I have tried to figure them all out, but if Denis is right and Soroban made an error by using Cos instead of sin, then the outcomes are different. cos= 29.19 and sin = 12.09. Which is correct? Please help.
Ok, I have read all responses and I have tried to figure them all out, but if Denis is right and Soroban made an error by using Cos instead of sin, then the outcomes are different. cos= 29.19 and sin = 12.09. Which is correct? Please help.
Did you Look at my post at all?. It appears all that typing was for naught. What did I arrive at?. Think about it. 8(29.19)=233.52. That's larger than the circumference of the circle.
What did Denis get: \(\displaystyle \L\\2(15.8)sin(22.5)=12.09\)
Law of cosines: \(\displaystyle \L\\\sqrt{(15.8)^{2}+(15.8)^{2}-2(15.8)(15.8)cos(45)}=12.09\)
What did Denis get: \(\displaystyle \L\\2(15.8)sin(22.5)=12.09\)
Law of cosines: \(\displaystyle \L\\\sqrt{(15.8)^{2}+(15.8)^{2}-2(15.8)(15.8)cos(45)}=12.09\)