Proof of plane geometry:?

At any delta △ABC?

Do AD⊥BC.

The opposite of C, the opposite of B and the opposite of A are all C?

Have BD=cosB*c, AD=sinB*c, DC=BC-BD=a-cosB*c?

According to Pythagorean theorem, we can get:?

AC^2=AD^2+DC^2？

b^2=(sinB*c)^2+(a-cosB*c)^2？

b^2=sin^2b*c^2+a^2+cos^2b*c^2-2ac*cosb？

b^2=(sin^2b+cos^2b)*c^2-2ac*cosb+a^2？

b^2=c^2+a^2-2ac*cosB？

cosB=(c^2+a^2-b^2)/2ac

Vector proof: (The graph is a vector) (References:

Vector? 4 cosine theorem is a compulsory course for high school people's education textbook? High school people's education textbook compulsory 5)

Prove:

∵ As shown in the figure, there is a→+b→=c→

∴c c=(a+b) (a+b)

∴c^2=a a+2a b+bb∴c^2=a^2+b^2+2|a||b|cos(π-θ)

Get c 2 = a 2+b 2-2 | a || b | cos θ (note: the formula of trigonometric function is used here).

Then disassemble it to get C 2 = A 2+B 2-2 * A * B * COSC.

The same principle can prove to others that the following CosC = (c 2-b 2-a 2)/2ab means to move CosC to the left.