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tato1982
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Find the unknown angle in the triangle. It is quite an interesting task. I will put a trigonometric solution, Mabra I wonder how it opens geometrically. Help someone
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I am getting the same solutions, i.e ##x=12## and ##y=60## i only used pythagoras theorem, SOHCAHTOA and sine rule in my working...no trigonometric identities like you have used.tato1982 said:Find the unknown angle in the triangle. It is quite an interesting task. I will put a trigonometric solution, Mabra I wonder how it opens geometrically. Help someoneView attachment 312479View attachment 312480
I must have been smarter then, than I am now. Can't remember right now how I showed the similarity of the triangles in post #27 where I told readers to "go figure"Charles Link said:See https://www.physicsforums.com/threads/find-angle-b-in-the-trigonometry-problem.1006011/#post-6536554
for another interesting problem regarding triangles that @chwala posted about a year ago.
Am trying to follow the working here; the trig identities are correct but a bit confusing ...particularly on the line...##2\sin 18^0\cos18^0=\cos^{3}18^0 - 3\sin^{2}18^0 \cos 18^0## ...tato1982 said:Find the unknown angle in the triangle. It is quite an interesting task. I will put a trigonometric solution, Mabra I wonder how it opens geometrically. Help someoneView attachment 312479View attachment 312480
Allow me to ask this ; Consider the highlighted part in red...tato1982 said:Find the unknown angle in the triangle. It is quite an interesting task. I will put a trigonometric solution, Mabra I wonder how it opens geometrically. Help someoneView attachment 312479View attachment 312480
I doubt that you'll get an answer back.chwala said:Allow me to ask this ; Consider the highlighted part in red...
View attachment 322065
...you plugged in ##\sin 18^0## on the rhs and used it to show your justification...i do not think that's the correct approach for justifying a trigonometric identity. In my thinking we need to use only the lhs of the identity to realise/prove the envisaged right hand side.
This was a nice one...almost choked ...i think the intention was for the OP to continue with the hint; but having said that i ended up with;neilparker62 said:You can solve: $$\sin x \sin48=\sin18\sin(x+18)$$ Wolfram Alpha 'choked'on it for some reason but the manual solution is fairly straightforward.
I used the tan rule to get to 12 degrees. But yes - it's a numerical answer which does not exactly prove 12 degrees. Have tried all sort of geometric 'tricks' to no avail. However I have been able to determine a surd expression for tan(12) (degrees) as sin(30-18)/cos(30-18) and then show it is equivalent to the expression I got using the tan rule. Well I will get to post that but it's quite a lot of Latex and not really a "simple 2 or 3 step" process!Charles Link said:I was able to follow what the OP did, and he later justified a couple of steps which he assumed to be correct, because he needed the result to complete another step. The OP asked if anyone has a simpler way of solving this, and I haven't been able to come up with one.
It is very easy to get a numerical result, but proving that it is exactly 12 degrees I found to be much more difficult.
e.g. You can get TV immediately from the law of sines, and then TD from the law of cosines, and from that you can get ##\cos{x} ## or ## \sin{x} ## with the law of cosines or the law of sines. I don't see a simple two or three step process though to get that ## x=12 ## degrees.
Boss! Impressive!neilparker62 said:From post #4 we had in effect: $$\tan(x) = \frac{\sin^2 18^{\circ}}{\sin 48^{\circ}-\sin 18^{\circ} \cos 18^{\circ}} = \frac{\sin^2 18^{\circ}}{\sin (30+18)^{\circ}-\sin 18^{\circ} \cos 18^{\circ}}$$ $$=\frac{\frac{\phi^2}{4}} {\frac{1}{2}\frac{\sqrt{3+\phi}}{2}+\frac{\sqrt{3}}{2} \frac{\phi}{2} - \frac{\phi}{2} \frac{\sqrt{3+\phi}}{2} } = \frac{\phi^2} {\phi^2 \sqrt{3+\phi} + \sqrt{3} \phi }$$ $$=\frac{\phi}{\phi \sqrt{3+\phi} + \sqrt{3} }$$ Since this is the same expression we obtained above for ##\tan 12^{\circ}##, we may reasonably conclude that ##x = 12^{\circ}##.
Looks good - certainly a very good exercise for your trig skills! I've tried all sorts of "tricks" with geometry but as yet cannot find a geometric solution.tato1982 said:Find the unknown angle in the triangle. It is quite an interesting task. I will put a trigonometric solution, Mabra I wonder how it opens geometrically. Help someoneView attachment 312479View attachment 312480
looks brilliant man! but just a question- why did you have to come up with the golden ratio specific relationship 'transforms' i.eneilparker62 said:##18^{\circ}## is a Pentagon related angle along with ##36^{\circ}, 54^{\circ}## and ##72^{\circ}##. All of these are characterised by their connection to the golden ratio ##\phi \approx 0.618 ##.
##\sin18^{\circ}=\frac{\phi}{2}##
##\sin36^{\circ}=\frac{\sqrt{2-\phi}}{2}##
##\sin54^{\circ}=\frac{1}{2\phi}##
##\sin72^{\circ}=\frac{\sqrt{3+\phi}}{2}##
For this application we'll need ##\sin18^{\circ}## and ##\sin72^{\circ}=\cos18^{\circ}##. And we'll need the golden ratio specific relationship: ##1-\phi^2=\phi## or alternatively ##1 - \phi = \phi^2##.
##1-\phi^2=\phi## or alternatively ##1 - \phi = \phi^2##.
Apart from "Uncle Google" , you may also refer to Euclid XIII 9 !Charles Link said:The following I found in a google: (to get the expression in post 14)
Let 18 degrees=## a ##.
Then
## 2a=90-3a ##.
## \sin(2a)=\sin(90-3a)=\cos(3a) ##.
Next:
##2 \sin(a) \cos(a)=4 \cos^3(a)-3 \cos(a)=\cos(a)(4-4 \sin^2(a)-3) ##.
So that:
## \cos(a)(4 \sin^2(a)+2 \sin(a)-1)=0 ##.
Can then solve for ## \sin(a) ##.
Yes - that's one instance it was needed. I also used it in the derivation of the expression I gave for ##\tan 12^\circ##. And in showing this was equivalent to the other expression I obtained in post #4. You'll also need it if you pick up the challenge above to show ##\frac{TD}{TO}=\sqrt 3##.chwala said:looks brilliant man! but just a question- why did you have to come up with the golden ratio specific relationship 'transforms' i.eFor convenience/aesthetics i suppose...of course i can also see that you used the property ##\cos^2 ∅ + \sin^2∅=1## to find ##\cos 18^0=\dfrac{\sqrt {4-∅^2}}{2}=\dfrac{\sqrt {4-(1-∅)}}{2}=\dfrac{\sqrt {3+∅)}}{2}## according to your transform.
Can't say I was too happy with this in the OP's post because he basically assumed answers for x and also for DT. If you show ##DT=2 \sin 48^\circ## it's game over straight away. I tried very hard to do that and failed.Charles Link said:The way I have is a little simpler, but still slightly complex:
We need to show that ## 2 \sin(48) \sin(12)=\sin(18) ##, (that will make the one triangle a 30-60-90, and the angle x to be 12 degrees ), so that
## 2 \sin(30+18) \sin(30-18)=\sin(18) ##, and expanding with trig identity
##2 ((1/4)\cos^2(18)-(3/4) \sin^2(18))= \sin(18) ## so that
## 4 \sin^2(18)+2 \sin(18)-1=0 ##.
We get that ## \sin(18) ## must satisfy the post 14 expression, and the identity is shown to be correct.
neilparker62 said:We can also find expressions for TO ##( \sin 48^\circ)## and TD ##( \frac{\sin 48^\circ \cos 18^\circ}{\sin 18^\circ}-1)##. Then we should have ##\frac{TD}{TO}=\cot 18^{\circ} - \csc 48^{\circ} = \sqrt{3}## and I will leave proof of this identity to the reader.
Have fun!
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. To find the unknown angle, you can use the formula: sin(A)/a = sin(B)/b = sin(C)/c, where A, B, and C are the angles and a, b, and c are the corresponding sides. Simply plug in the known values and solve for the unknown angle.
Yes, the Law of Sines can be used to find the unknown angle in any triangle, as long as you have enough information to set up the equation. This means you must know at least one angle and its corresponding side length, or two angles and one side length.
The Law of Sines is used to find the unknown angle in a triangle, while the Law of Cosines is used to find the unknown side length in a triangle. The Law of Cosines also takes into account the lengths of all three sides, while the Law of Sines only uses one side and its corresponding angle.
Yes, the Law of Sines can be used to find the unknown angle in a right triangle. However, it is usually easier to use the trigonometric functions of sine, cosine, and tangent to find the unknown angle in a right triangle.
If you get a negative angle when using the Law of Sines, it means that you have used the incorrect angle in the equation. Make sure to use the angle that is opposite the side you are trying to find, and if necessary, draw a diagram to help you visualize the triangle and its angles.