How do you find all three angles ?Are there any proof of how to claim all three?

asked 2018-02-07 15:45:18 +0100

Larrousse gravatar image

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 $a=1.25,b=1.25,c=1$

$\sqrt\frac {21}{13.44}=1.25$

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:

$a^2=b^2+c^2-2bc\cos(A)$

$b^2=a^2+c^2-2ac\cos(B)$

$c^2=a^2+b^2-2ab\cos(C)$

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 $1$.

$\sqrt{\sin^2(B)+(a-\cos(B))^2}=b$

$\sqrt{\sin^2(A)+(b-\cos(A))^2}=a$

$\sqrt{\sin^2(A)+(\cos(A))^2}=c$

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$\frac{a}{c}=\cos(C)$ $(1-\frac{a}{c})\times\sqrt\frac{(a+c)}{(c-a)}=\sin(C)$

$\sqrt\frac{(c-b)}{c}=\cos(B)$ $\sqrt\frac{b}{c}=\sin(B)$

Find

$\cos(A)$ & $\sin(A)$ which we already know.

Question is: why $\frac{h_c}{h_b}$; $\frac{h_c}{h_a}$;$\frac{h_c}{h_c}$ and. $\frac{\sin A}{\sin C}$; $\frac{\sin B}{\sin C}$ $\frac{\sin C}{\sin C}$equal to the sides $a,b,c$?

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Comments

This is not the right place, this site is supporting the computer algebra system named Sage and/or SageMath. However, since the answer to the only question can be quickly given...

Above: $\sin C/\sin C$ is obviously $1$, which is not always equal to $c$. So the question claims a false statement. What is on the path of the truth is as follows:

Let $R$ be the side of the circumscribed circle. Then $a=2R\sin A$ and the other relations, so we get the "sine law" $a/\sin A=b/\sin B=c/\sin C=2R$.

The other claim is also false, but something can be simply traced back from $2S =ah_a=bh_b=ch_c$, where $S$ is the surface of $\Delta ABC$.

So the proportion $a:b:c$ corresponds to $(1/h_a);(1/h_b):(1/h_c)$.

dan_fulea gravatar imagedan_fulea ( 2018-02-07 19:21:37 +0100 )edit

Please take a look at the sagemath tour .

For instance, let $A,B,C$ be the angles in a triangle. Then we may ask sage for the following...

sage: var( 'A,B,C' );
sage: E = 4*sin(A)*sin(B)*sin(C)
sage: E.reduce_trig()
-sin(A + B + C) + sin(A - B + C) + sin(-A + B + C) - sin(-A - B + C)

Now, the term $\sin(A+B+C)=\sin180^\circ$ vanishes. And for the other terms, we may write simpler expressions, e.g. $$ \sin(-A + B + C)=\sin((A+B+C)-2A=\sin(180^\circ-2A)=\sin(2A) \ . $$

We can check the equality:

sage: var( 'A,B' );
sage: C = pi-A-B
sage: bool( ( 4*sin(A)*sin(B)*sin(C) == sin(2*A)+sin(2*B)+sin(2*C) ).reduce_trig() )
True
dan_fulea gravatar imagedan_fulea ( 2018-02-07 19:29:57 +0100 )edit
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