January 1st, 2019, 07:20 PM  #1 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0  Detailed section of all triangles which have unexplained properties
Facts for all triangles when the base of the triangle is one. Are there any explication? have three measures of three altitudes $h_0,h_1,h_2$ where sides $a,b,c$ are the constant lengths of all triangles and base $c=1$ and $h_0$ is situated on base $c$. $\frac{h_0}{h_1}=b$,$\frac{h_0}{h_2}=a$ and $ h_1=\sin A$ as well $h_2=\sin B$. $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ d = diameter $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ if I multiply by d, the diameter, it will give an answer of 1, and 1 is the base. And if I multiply the base times 4, all of 3 sides are multiple of 4 including the altitudes. d = diameter $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ jet.jpg Last edited by skipjack; January 2nd, 2019 at 03:44 AM. 
January 1st, 2019, 08:05 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907  
January 1st, 2019, 09:19 PM  #3 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 
I have three measures of three altitudes $h_0,h_1,h_2$, where sides $a,b,c$ are the constant lengths of all triangles and base $c=1$ and $h_0$ is situated on base $c$. $\frac{h_0}{h_1}=b$,$\frac{h_0}{h_2}=a$ and $ h_1=\sin A$ as well $h_2=\sin B$. $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ d = diameter $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ if I multiply by d, the diameter, it will give an answer of 1, and 1 is the base. And if I multiply the base times 4, all of 3 sides are multiple of 4 including the altitudes. d = diameter $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ Sides 5,5,4 where 4 is the base and by dividing 5÷4=1.25, 1.25, 4÷4=1 and the base equals 1. Last edited by skipjack; January 2nd, 2019 at 03:45 AM. 
January 1st, 2019, 09:43 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
I don't know what you mean by "base" but if you take the following equation: $\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ and multiply all sides by "d" you get $\displaystyle d \cdot \dfrac{\sin A}{a}= d \cdot \dfrac{\sin B}{b}= d \cdot \frac{\sin C}{c}=d \cdot \dfrac{1}{d}$ and you get $\displaystyle \dfrac{d~\sin A}{a}= \dfrac{d~\sin B}{b}= \dfrac{d~\sin C}{c}= 1$ If you multiply both all sides by 4d you get $\displaystyle \dfrac{4d~\sin A}{a}= \dfrac{4d~\sin B}{b}= \dfrac{4d~\sin C}{c}= 4$ There's no magic here. Geometrically speaking all you are doing is rescaling your triangle into a similar one. Dan Addendum That is if I'm interpreting the problem correctly. Last edited by skipjack; January 2nd, 2019 at 03:41 AM.  
January 10th, 2019, 07:24 AM  #5 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 
According to Pythagorean theorem, these are unexplained properties for all triangles. $$ \cos A + (b  \cos A)=b $$ $$ \cos B + (a  \cos B)=a $$ $$ (\sin B)^2+(a \cos B)^2=b^2 $$ $$ (\sin A)^2+(b \cos A)^2=a^2 $$ $$ (\sin B)^2+(\cos B)^2=c^2 $$ $$ (\sin A)^2+(\cos A)^2=c^2 $$ $$ (b\times\cos C)=(a \cos B) $$ $$ (a\times\cos C)=(b \cos A) $$ $$ (a\times\cos B)+(b \times\cos A)=c $$ Last edited by skipjack; January 10th, 2019 at 12:56 PM. 
January 10th, 2019, 12:53 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
None of those are "unexplained". Some of them are true only when c = 1.

January 11th, 2019, 01:23 PM  #7 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 
These are new to me and it describes the schematic shown above, but using a simplified version of the triangles when base equals 1, the lengths of altitudes or sides are measures of angles. $(c\times\sin B)^2+(a\times\cos C)^2=b^2$ $(c\times\sin A)^2+(b\times\cos C)^2=a^2$ $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ $(b\times\cos C)+(c\times\cos B)=a$ $(a\times\cos C)+(c\times\cos A)=b$ $(a\times\cos B)+(b \times\cos A)=c$ $(c\times\cos B)+(b \times\cos C)=a$ $(c\times\cos A)+(a\times\cos C)=b$ and to find one of the altitudes: $h_1=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)a^4b^4c^4}}{2a}$ Last edited by skipjack; January 11th, 2019 at 05:16 PM. 
January 11th, 2019, 05:15 PM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; January 11th, 2019 at 05:18 PM.  
January 11th, 2019, 05:39 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
The first two were evidently mistyped and should have been as follows: $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ $(c\times\sin A)^2+(a\times\cos C)^2=a^2$ 
January 11th, 2019, 06:08 PM  #10 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 
It was my mistake. The first 2 lines were found to be with errors and I just corrected them and now it is verified and as long as sides $a+b>c. a+c>b. b+c>a $ I don't know why one triangle is different than others if they hold the same properties. The only difference is that one triangle has its three angles and and its six properties $\sin A,\sin B,\sin C,\cos A,\cos B,\cos C$. $f^1)$ $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ $f^2)$ $(c\times\sin A)^2+(a\times\cos C)^2=a^2$ $(\sin B)^2+(\cos B)^2=c^2$ $(\sin A)^2+(\cos A)^2=c^2$ $(b\times\cos C)+(c\times\cos B)=a$ $(a\times\cos C)+(c\times\cos A)=b$ $(a\times\cos B)+(b \times\cos A)=c$ $(c\times\cos B)+(b \times\cos C)=a$ $(c\times\cos A)+(a\times\cos C)=b$ and to find one of the altitude: $h_1=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)a^4b^4c^4}}{2a}$ Thanks. Last edited by skipjack; January 11th, 2019 at 06:45 PM. 

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