三角一題~~~

a) ( a + b ) : ( b + c ) = 5 : 6
( a + b ) / ( b + c ) = 5 / 6
5b + 5c = 6a + 6b
b = 5c – 6a --- ( 1 )
( b + c ) / ( c + a ) = 6 / 7
6c + 6a = 7b + 7c
c = 6a – 7b --- ( 2 )
Put ( 2 ) into ( 1 ).
b = 5 ( 6a – 7b ) – 6a
= 30 a – 35 b – 6a
36 b = 24 a
a = 3b / 2 --- ( 3 )
Put ( 3 ) into ( 2 ).
c = 6 ( 3b / 2 ) – 7b
c = 2b
So a : b : c = 3b / 2 : b : 2b
= 3 : 2 : 4
b) Let a = 3k , b = 2k and c = 4k, where k is a constant
By cosine rule, cos C = - 1 / 4
c) By cosine rule again,
cos A = 11 / 16, cos B = 7 / 8
Therefore cos A + cos B + cos C
= 11 / 16 + 7 / 8 – 1 / 4
= 21 / 16

a) ( a + b ) : ( b + c ) = 5 : 6

( a + b ) / ( b + c ) = 5 / 6

5b + 5c = 6a + 6b

b = 5c – 6a --- ( 1 )

( b + c ) / ( c + a ) = 6 / 7

6c + 6a = 7b + 7c

c = 6a – 7b --- ( 2 )

Put ( 2 ) into ( 1 ).

b = 5 ( 6a – 7b ) – 6a

= 30 a – 35 b – 6a

36 b = 24 a

a = 3b / 2 --- ( 3 )

Put ( 3 ) into ( 2 ).

c = 6 ( 3b / 2 ) – 7b

c = 2b

So a : b : c = 3b / 2 : b : 2b

= 3 : 2 : 4

b) Let a = 3k , b = 2k and c = 4k, where k is a constant

By cosine rule, cos C = - 1 / 4

c) By cosine rule again,

cos A = 11 / 16, cos B = 7 / 8

Therefore cos A + cos B + cos C

= 11 / 16 + 7 / 8 – 1 / 4

= 21 / 16

(a+b)：(b+c)：(c+a)=5：6：7
a+b=5k -1
b+c=6k -2
c+a=7k -3
2-1
c-a=k
2a=6k
a=3k
b=2k
c=4k
a:b:c=3:2:4
a=3k b=2k c=4k
cos C = 9k^2+4k^2-16k^2 / 2(3k)(2k)
=-1/4

cos B=9k^2+16k^2-4k^2 /2(3k)(4k)
=7/8
cos A=4k^2+16k^2-9k^2/ 2(2k)(4k)
=11/16
cos A+ cosB+cosC=21/16
cosC=-1/4
C= COS-1 -1/4