Pece NO
Date 121 121
H. W
show that 8cos A t 16 cos B t u cos (=17
a=s
b = 6 c=8
C-8
8 SOSA + 16 cos B + 4 cosc = 17
The sides of a trlangle area a = 5, b = 6,C=8
Answers
Answer:
=
12
c+a
=
13
a+b
=
36
2(a+b+c)
or
11
b+c
=
12
c+a
=
13
a+b
=
18
a+b+c
=k
∴b+c=11k,c+a=12k,a+b=13k,a+b+c=18k
Substituting b+c=11k in a+b+c=18k we get
a+11k=18k or a=7k
Substituting a=7k in c+a=12k we get
c+7k=12k or c=5k
Substituting c=5k in b+c=11k we get
b+5k=11k or b=6k
∴a=7k,b=6k and c=5k
Using cosine rule, we get
cosA=
2bc
b
2
+c
2
−a
2
=
2×6k×5k
(6k)
2
+(5k)
2
−(7k)
2
=
60k
2
36k
2
+25k
2
−49k
2
=
5
1
(On simplification)
cosB=
2ca
c
2
+a
2
−b
2
=
2×5k×7k
(5k)
2
+(7k)
2
−(6k)
2
=
70k
2
25k
2
+49k
2
−36k
2
=
35
19
(On simplification)
cosC=
2ab
a
2
+b
2
−c
2
=
2×7k×6k
(7k)
2
+(6k)
2
−(5k)
2
=
84k
2
49k
2
+36k
2
−25k
2
=
7
5
(On simplification)
cosA:cosB:cosC=
5
1
:
35
19
:
7
5
On simplification, we get
cosA:cosB:cosC=
35
7
:
35
19
:
35
25
=7:19:25
Step-by-step explanation:
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