✔ 最佳答案
1)In triangle ABC,(a+b):(b+c):(c+a )=5:6:7,find the angles of triangle ABC correct to the nearest 0.1 degree.
Let
a+b=5k...(1)
b+c=6k...(2)
c+a=7k...(3)
(3)-(1):c-b=2k...(4)
(2)+(4):2c=8k
c=4k,b=2k,a=3k
So
cosA=(b^2+c^2-a^2)/2bc
A=46.6
cosB=(a^2+c^2-b^2)/2ac
B=29.0
C=180-A-B=104.4
2)The three sides of triangle ABC are a=x^2-1,b=2x+1,c=x^2+x+1.
a)From a>0 and b>0, find the range of values of x.
b)Hence,prove that c>a and a>b
c)Find the largest angle of triangle ABC.
(a)
Since
a+b>c; x^2+2x>x^2+x+1;x>1
b+c>a; x^2+3x+2>x^2-1;3x>-3;x>-1
a+c>b; 2x^2+x>2x+1;2x^2-x-1>0;(2x+1)(x-1)>0;x<-1/2 or x>1
So the range of values of x is x>1
(b)
c-a=x+2>0
c-b=x^2-x=x(x-1)>0
(c)
The largest angle of triangle ABC is angle ACB (because c is the longest side)
