2024-03-06 14:59:12 +0200 | received badge | ● Popular Question (source) |
2018-02-07 16:58:20 +0200 | asked a question | How do you find all three angles ?Are there any proof of how to claim all three? For a triangle of $\angle A B C$ the sides of $ a,b,c$ are written in a way of $a=\frac{\sin A}{\sin C}$, $b=\frac{\sin B}{\sin C}$, $c=\frac{\sin C}{\sin C}$ and the heights $h_a,h_b,h_c$ are written in a form $\frac {h_c}{h_a}=$,$\frac{h_c}{h_b}=$,$\frac{h_c}{h_c}$ to give us $ a$ and $b$. and the base $c=1$. If I have a triangle with sides $5,5,4 $ and their altitudes are $\sqrt {21},\sqrt {13.44},\sqrt {13.44}$ , why do they simplify to give us a special kind of triangle where $a=1.25,b=1.25,c=1$
Angles $3=0.16+0.16+0.68^2+0.84+0.84+(1-0.68^2)$ Laws of Cosine ,when we have all $3$ lengths are:
Here we have sides $1.25,1.25 and 1$,a simplest version of the triangle measuring $5,5,4$. And for three sides of a triangle $a,b,c$,and $\angle ABC$. The legs of the heights $h_a,h_b,h_c$ are situated on three sides of the triangle. For one side $a$ I have $\frac{b-\cos(A)}{\cos(C)}=a$ and for the second side $b$ I have $\frac{a-\cos(B)}{\cos(C)}=b$ and the third side which is $c$ as the base of the triangle equals to $1$.
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