How to find the altitudes of all triangles?

Question

For Cosine:

For a triangle with angles$\angle ABC$ there exist lengths $a,b,c$ opposite to the angles. By using the trigonometric function for cosines, I multiplied the length of $AC\times\cos\angle A$ for example:

5+5+4=14 where $\cos\angle B=0.4,\cos\angle A=0.4,\cos\angle C=0.68$ and there are $3$ $\theta$,

$AC\times\cos\angle A=2$ or $BC\times\cos\angle B=2$ which is the place of the altitude and $2+2=4$,

$AB\times\cos\angle B=1.6$ $BC\times\cos\angle C=3.4$ $1.6+3.4=5$ Next step is all I have to use is the Pythagorean theorem. Can anyone explain if this is true or false?

for Sine:

$AC\times\sin\angle B=$

$BC\times\sin\angle A=$

And to find Sine I subtract 1 from all three $\cos ^2 A,\cos^2 B,\cos^2 C$

And in general the sum of all angles is:

$\sin^2 A+\sin^2 B+\sin^2 C+\cos^2 A+\cos^2 B+\cos^2 C=3$

Answer 1

You can find the length of an altitude by multiplying length of a side by sine of an angle. For example, to find altitude AD to side BC, you multiple AB by sine of angle B or AC by sine of angle C.

Answer 2

Let $a$, $b$ and $c$ be sides-lengths of the triangle $ABC$, $h_a$, $h_b$ and $h_c$ be altitudes of our triangle.

Thus, $$h_a=\frac{2S_{\Delta ABC}}{a}=\frac{\sqrt{2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4}}{2a}$$