How do you describe the fact that all triangles with base one are measure of theta(an angle)? The only reason I can think of is because the base is one.

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 mutiply 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 a multiple of 4 including the altitudes. d=diameter

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 mutiply 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 a multiple of 4 including the altitudes. d=diameter

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