In the triangle ABC, if C 2B and 2a b c, find c b

Because a 2=b(b+c), s (sina) 2=(sinb) 2+sinbsin(a+b).

So (sina+sinb)(sina-sinb)=sinbsin(a+b).

So 4sin[(a+b) 2]*cos[(a-b) 2]*cos[(a+b) 2]*sin[(a-b) Zheng Qingsen2]=sinbsin(a+b).

This sentence uses the formula of the sum and differential product: shouting acres.

sina+sinb=2sin[(a+b)/2]*cos[(a-b)/2]

sina-sinb=2cos[(a+b)/2]*sin[(a-b)/2]

So sin(a+b)sin(a-b)=sinbsin(a+b).

So sin(a-b)=sinb

So a=2b

by the cosine theorem.

cosa = (b 2 + c 2-a 2) 2bc = bc 2bc = 1 2a = 60° by the sinusoidal theorem.

a/sina=b/sinb

sina/sinb=a/b=√3

3/2)/sinb=√3

sinb=1/2

b = 30° or 150° (round).

c=180°-60°-30°=90°

Answer: c=90°

Since b+c=2a, sinb+sinc=2sina is known by the sinusoidal theorem

sinb+sinc=2sin(b+c) 2cos(b-c) 2, so there is sin(b+c) 2cos(b-c) 2=sina The inscription gives c=2b, and the triangle has (b+c) 2=90°-A 2 will c=2b, (b+c) 2=90°-a 2 generations of tombs into sin(b+c) 2cos(b-c) 2=cos(a 2)cos(b 2)=2sin(a 2) group stool cos(a 2).

So cos(b 2) = 2sin(a 2) = 2cos(b + c) 2 = 2cos(3b wang ruler 2).

By cos(b2) = 2cos(3b2).

Solve the equation cos(b 2) = 2cos(3b 2) to get sin(b 2) = (root number 2) 4

b=c=2b=2*

a=180°

Therefore, the triangle is an acute triangle.

Right angle. The equation can be formed into a 2+c 2=b 2 satisfies the Pythagorean theorem, so it is a right triangle.

I wrote it myself, don't mind.

Because a 2=b(b+c), s (sina) 2=(sinb) 2+sinbsin(a+b).

So (sina+sinb)(sina-sinb)=sinbsin(a+b).

So 4sin[(a+b) 2]*cos[(a-b) 2]*cos[(a+b) 2]*sin[(a-b) 2]=sinbsin(a+b).

Here we use the formula for the sum and difference product:

sina+sinb=2sin[(a+b)/2]*cos[(a-b)/2]

sina-sinb=2cos[(a+b)/2]*sin[(a-b)/2]

So sin(a+b)sin(a-b)=sinbsin(a+b).

So sin(a-b)=sinb

So a=2b