In ABC, if a 3 b 3 c 3 a b c c 2, and tanatanb 3, try to determine the shape of abc

A 3+B 3+C 3) (A+B+C)=C 2> (A 3+B 3+C 3)=(A+B+C)*C 2> A 2+B 2-ab=C 2 and C 2=A 2+B 2-2ABCOSC

c=60tan(a+b)=tan120=3^(1/2)=(tana+tanb)/(1-tanatanb) tanatanb=3

tana+tanb=-2*3^(1/2)> tana=tanb=-3^(1/2)

So it's an equilateral triangle.

First agree with the method of the first floor above, he is the traditional method. Look at my approach. The triangle abc is an equilateral triangle.

It's okay to keep it simple. Move (a b c) to the right, make it a 3 b 3 c 3 = a*c 2 b*c 2 c 3, and then simplify it to a(a 2-c 2) b( b 2-c 2) = 0 Because a, b, and c must be greater than 0 (the side of the triangle), so a = c = b, so your tana*tanb= 3 condition is useless, hehe. This question is simple, how to say it, mathematics is a wonderful thing, do mathematics to learn to be clever (such as substituting special values), like doing some fill-in-the-blank multiple-choice questions, but also to do more questions to have a feeling.

Hypothesis. a-c=-2n,=> a=c-2n,(1)a+b=7n,(2)

c-b=n,=>b=c-n,(3)

Bring (1), (2) into (3) to get the boy closed dress, c-2n+c-n=7n2c=10n

c=5n, then the state rises a=3n, b=4n

A2+B2=C2

So the triangle is a right-angled triangle.

The shape of abc is generally judged by the cosine theorem.

The angle opposite the largest side of the filial piety circle.

Let the angle of the pair of sides of a be a

So. A2=B2+C2-2BCCOSA {The square of the tassel orange of A is equal to the square of B plus the square of C-2BCCOSA}

7^2=5^2+3^2-30cosa

So. 30cosa=49-25-9=15

So cosa=

A angle = 120°

So the Qiaojia collapse is an abc obtuse triangle.

(b+c)²-a²=3bc

b²+c²-a²=bc

So cosa=(b +c -a b +c -a) 2bc=bc 2bc=1 2

a=60 degrees.

sina=2sinbcosc

a/sina=b/sinb

So there is a=2bcosc

a=2b*(a²+b²-c²)/2ab

a²=a²+b²-c²

b²-c²=0

b=c is an isosceles triangle and a=60 degrees.

So it's an equilateral triangle.

(A 3+B 3-C 3) (A+B-C)=C 2 A 3+B 3-C 3=AC 2+BC 2-C 3 A 3+B 3=C 2(A+B) (A+B)(A 2-AB+B 2)=C 2(A+B) A 2+B 2-AB=C 2 From the cosine theorem a 2+b 2-c 2=2ABCOSC: A 2+B 2 C 2+2ABCOSC A 2+B 2=C 2+AB Cosc=1 2 C 60° A+B=2 3sinasinb=sinasin[(2 3)-a] =sina(sin2 3cosa-cos2 3sina) = root number 3 4sin2a + 1 4-1 4cos2a = 3 4 sin(2a- 6) = 1

again - 6<2a- 6<11 6, 2a- 6= 2, a= 3The triangle is a regular triangle, i.e., an acute triangle Answer Supplement.

a +b -c ) (a+b-c)=c -a +b -c =ac +bc -c -a +b =ac +bc -a+b)(a -ab+b )=(a+b)c -a -ab+b =c --a +b -c =ab --cosc=(a +b -c ) (2ab)=1 2---c=60°--a+b=120 sinasinb=(1 2)[cos(a-b)-cos(a+b)]=3 4 --cos(a-b)=3 2+cos120°=1---a=b=60°--abc is an equilateral triangle.

Is it that the title is not given? It's considered to be an equilateral triangle, but I can't really calculate it according to the existing topic, and the level is limited, so I'm sorry.

Solution: Let c+a=x, then b+c=7x, a+b=2x, so 2x+7x+x=(a+b)+(b+c)+(c+a)=2(a+b+c)=48

Therefore 10x=48

So x=so a+c=, a+b=

So b-c=

and b+c=7*

So b = (c = so a = < 0

So this can't make a triangle.

a, b, and c will produce negative numbers, so they cannot form a triangle.