SchematicFor a $\triangle ABC$ there exist three sides, $a, b, c$ and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa. At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.
The following equations represent a schematic of all triangles to obtain the result needed to find 3 sides and three angles of triangles.We can use the Law of Cosine for all three to get the three angles also.
$(c\times\sin B)^2+( b\times\cos C)^2=b^2$
$(c\times\sin A)^2+(a\times\cos C)^2=a^2$
$(c\times\sin B)^2+(c\times\cos B)^2=c^2$
$(c\times\sin A)^2+(c\times\cos A)^2=c^2$
$(b\times\cos C)+(c\times\cos B)=a$
$(a\times\cos C)+(c\times\cos A)=b$
$(a\times\cos B)+(b \times\cos A)=c$
and the formula to find the altitudes for all triangles:
$h_a=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa without the Law of cosine even though I have three sides without three angles.
$(\sin A)^2+(\sin B)^2+(\sin C)^2+(\cos A)^2+(\cos B)^2+(\cos C)^2=3$ they apply for all triangles.
And the law of sine states when one of the base of the triangle is one then:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}=$ where the reciprocal of the diameter is $\angle C$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ where d is the diameter
Let $\triangle ABC$ be sides $a$, $b$, and $c$. I am using that the length of the side opposite vertex $A$ is called $a$, $B$ for $b$ and $C$ for $c$.
If $\theta=\angle C$. Then the Cosine says that
$c^2=a^2+b^2-2ab\cos \theta$. Since we know $a$, $b$, and $c$, we can use the formula for all three altitudes or above formula to calculate $\cos\theta$.
We can use the Law of Cosine and the formula of three altitudes to get all six angles.