The largest angle A in triangle ABC is 2 times the minimum angle C, AB 7AC 8 BC

Let the three stools of this triangle be rough and thick, and burn 2, and 4

So" 2 a>

a=4 so the three sides are 4, 5, 6

acb=2∠b

c+∠b+∠a=180

3∠b+∠a=180

ACB=2 B angle corresponds to AB-AC>0(1)3 B+ A=180 for thick grip

Iwawangqing b = 60 - a 3

b-∠a=60-∠a/3(2)

A0 corner corresponds to ab-bc>0

1) + (2), got.

ab-ac+ab-bc>0

2ab>ac+bc

a+b+c)(b+c-a)=bc, i.e., (b+c) 2-a2=bc, then there is no dust b 2+c 2-a 2=-bc

cosa=(b 2+c 2-a 2) 2bc=-1 and the loss of the god 2 is a=120 degrees.

Let the three sides be a=n 1, b=n, c=n 1, and the opposite angles are a, b, and c respectively, then c=2a. From the sinusoidal theorem: a sina = c sinc, i.e. (n 1) sina = (n 1) sinc=(n 1) [2sinacosa], so, cosa = (n 1) (2n 2).

Because cosa=[n (n 1) n 1) ] [2n(n 1)]=(n 1) (2n 2),n 4n) [2n(n 1)]=(n 1) (2n 2),(n 4) (n 1)=(n 1) (n 1).

The solution is n=5, i.e., the three sides are .

sin 2α=2sin α×cos α

Therefore, the formulas of sin 2c=2sin c cos csin2c sinc=2sin c cos c sinc=2cosc are not difficult to prove, the key is to remember them as 1+1=2.

Sine theorem: a sina = b sinb sinb = c sinc double angle formula: sin(2c) = 2*sinc*cosc As long as the above sine formula is deformed, you can get a c = sina sinc because the maximum angle a is 2 times the minimum angle c.

So a c=sina sinc=sin2c sinc=2*sinc*cosc sinc=2cosc

You take ab as the base, do cd perpendicular to ab, we know that the edge corresponding to angle a is a, and the edge corresponding to angle c is c, so sina = cd c, sinc = cd a, so there is a c = sinc sina, and the latter one is simplified according to the formula sin2x=2sinxcosx!

According to the sinusoidal theorem a sina = c sinc sinc exchange to get a c = sina sinc and because a = 2c so sin2c sinc, according to sin2c = 2 sinc * cosc, sinc and sinc are eliminated, get =2 cosc

sin2c=2sinccosc I learned this formula in high school.

Set the remaining angle of the silver tan cx°

2x+2x+x=180

5x=180

x=362x=72

a= b=72°

c=36°

On the one hand, from the sine theorem, we get: ab sinc bc sina 7 sinc bc (2sinccosc) and thus there is: cosc bc 14

On the other hand, cosc (AC2BC2AB2) (2·AC·BC) (BC215) (16BC).

BC 14 (BC 2 15) (16BC) Solution: BC 105

Because a is the largest intersection, bc is the longest side.

bc*bc=7*7+8*8=113

bc = root number 113