On the opposite sides of the triangle ABC, ABC is known as a = 2 times root, 3C = 21 + Tana / tanb = 2C / B, and the area s of the triangle ABC is calculated

On the opposite sides of the triangle ABC, ABC is known as a = 2 times root, 3C = 21 + Tana / tanb = 2C / B, and the area s of the triangle ABC is calculated

√11
From 1 + Tana / tanb = 2C / b
By using sine theorem
b=4+2√3cosB/cosA
Using cosine theorem
2b^2＝24
b＝2√3
So far, a = 2 √ 3, B = 2 √ 3, C = 2
Solution of triangle
s＝√11
(root 11)

Prove ACOS a + bcos B + CCOS C = 2asin B sin C in triangle ABC

The sine theorem is equivalent to proving sinacosa + sinbcosb + sinccosc = 2sinasinbsin (a + b) = 2Sin ^ 2asinbcosb + 2Sin ^ 2bsinacosa
The formula of double angle is used to shift the term
It is equivalent to cos2a * sin2b + cos2b * sin2a + sin2c = 0
It is equivalent to sin (2a + 2b) + sin2c = 0
sin(360-2c)+sin2c=0
Obviously
Immediate proof

Ad is the middle line of triangle ABC, e is the point on AC, AE = 1 / 2ec, be is the intersection of AD and f; AF = FD

Certification:
Take the midpoint of CE as G and connect DG
∵ D is the midpoint of BC
The DG is the median of △ BCE
∴DG∥BE
∵AE=1/2EC
∴AE=EG
∵BE∥DG
Ψ EF is the median of △ ADG
∴AF=FD

It is known that in △ ABC, ad is the midline on the edge of BC, e is the point on ad, and be = AC, extend be to AC to F, and prove that AF = EF

Prove: as shown in the figure, extend ad to point G, so that ad = DG, connect BG. ∵ ad is the midline (known) on the edge of BC, ∵ DC = dB, in △ ADC and △ GDB, ad = DG ≌ ADC ≌ GDB (SAS), DC = DB ≌ ADC ≌ GDB (SAS), ≌ CAD = g, BG = AC and ∵ be = AC, ≌ be = BG, ≌