in a traingle abc a=5 b=4 and cos(a-b)=31/32
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1
Answer:
We know, tan[(A – B)/2] = [(1 – cos(A – B)) / (1 + cos(A – B))]½
Using this we get, tan[(A – B)/2] = 1/63½
Now, tan[(A – B)/2] = (a – b)cot(C/2) / (a + b)
using this we get, cotC/2 = 3/7½
or, tanC/2 = 7½/3
Now, cosC = (1 – tan2C/2) / (1 + tan2C/2) = 1/8
Now, c2 = a2 + b2 – 2abcosC
Using this we get, c = 6
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Answered by
5
In a △ABC, if a=5,b=4 and cos(A−B)=
32
31
, then side c is
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ANSWER
Given, a=5,b=4,cos(A−B)=
32
31
Therefore, tan(
2
A−B
)=
1+cos(A+B)
1−cos(A−B)
=
63
1
⇒
a+b
a−b
cot
2
C
=
63
1
⇒
5+4
5−4
cot
2
C
=
63
1
⇒tan
2
C
=
3
7
Now, cosC=
1+tan
2
2
C
1−tan
2
2
C
⇒cosC=
1+
9
7
1−
9
7
=
8
1
Now, c
2
=a
2
+b
2
−2abcosC
⇒c
2
=25+16−40×
8
1
=36
⇒c=6
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