In the triangle ABC, f is a point on the edge of BC, and PC = 2PB,

In the triangle ABC, f is a point on the edge of BC, and PC = 2PB,

Let ad ⊥ BC be D, let PD = 1,
(1) B and P are on the same side of D
∵∠APC=60°
∴AD=√3
Easy to get BD = √ 3
∴BP=√3 - 1
∴PC=2√3 - 2
∴CD=PC-PD=2√3 - 3
∴tan∠ACB=AD/CD=2+√3
∴∠ACB=75°

It is known that in the triangle ABC, a = 120 degrees, a = 7, B + C = 8, find B, C and SINB

From the cosine theorem, it is concluded that
a^2=b^2+c^2-2bccosA=7^2
b^2+c^2+bc=49 ---(1) (cosA=cos120°=-1/2)
b+c=8
(b+c)^2=64
b^2+c^2+2bc=64 ---(2)
(2)-(1):
bc=15 ---(3)
b(8-b)=15
8b-b^2=15
b^2-8b+15=0
(b-3)(b-5)=0
b1=3,
b2=5.
∴c1=8-b1=5
c2=8-b2=3.
a/sinA=b/sinB
sinB=bsinA/a
sinB1=b1sinA/a=（3*√3/2）/7=3√3/14
B1=arcsin(3√3/14)=21.79°；
In the same way:
B2=arcsin(5√3/14)=38.21°
∴b1=3；b2=5
c1=5；c2=3
B1=21.79°；
B2=38.21°.

In triangle ABC, angle a = 120 °, a = 7, B + C = 8, C > b, find B, C, SINB

According to the cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bccosa = 7 ^ 2 B ^ 2 + C ^ 2 + BC = 49 -- - (1) (COSA = cos120 ° = - 1 / 2) B + C = 8 (B + C) ^ 2 = 64b ^ 2 + C ^ 2 + 2BC = 64 -- - (2) (2) - (1): BC = 15 -- - (3) B (8-b) = 158b-b ^ 2 = 15b ^ 2-8b + 15 = 0 (B-3) (B-5) = 0b1 = 3, B2 = 5

Given △ ABC, (Tana + 1) (tanb + 1) = 2, ab = 2, find: (1) the degree of angle c; (2) the maximum area of triangle ABC

The opposite sides of corner a, corner B and corner C are a, B and C (1) Tana + Tana + tanb + tanataanb + 1 = 2, that is, Tana + tanb = 1-tanataanb, \\\cornera, corner B and corner C are a, B and C (1) Tana + Tana + tanb + tanb + tanataanb + 1 = 2, namely, a, B, C (1) Tana + Tana + Tana + tanb = 1 + Tana + tanb + tanb + tanb + tanb + tanatab + tanataanb + tanataanb + 1 + Tana + tanb + Tana + Tana + B + B + B + B + B (1) (Tanc [π (0, π), (0, π), (0, π), (0, (0, π), (0,π) \\\\a ≥ 2Ab So s △ ABC = 12absinc = 24ab ≤ 24 (4-22) = 2-1