Question

In a triangle abc a=5 b=4 and cos(a-b)=31/32 solution of triangles

Answer 1

Given,

a = 5

b = 4

We know that,

$\cos(A-B) = \frac{1 - tan^{2}\frac{(A-B)}{2}}{1 + tan^{2}\frac{(A-B)}{2}} \\ \\ \frac{31}{32} = \frac{1 - tan^{2}\frac{(A-B)}{2}}{1 + tan^{2}\frac{(A-B)}{2}} \\ \\ 31 + 31*tan^{2}\frac{(A-B)}{2} = 32 - 32*tan^{2}\frac{(A-B)}{2} \\ \\ tan^{2}\frac{(A-B)}{2} = \frac{1}{63} \\ \\ tan \frac{(A-B)}{2} = \frac{1}{\sqrt{63}} \text{Again, We know that,} \\ \\ tan \frac{(A-B)}{2} = \frac{a-b}{a+b} \cot \frac{C}{2} \\ \\ \frac{1}{\sqrt{63}} = \frac{5-4}{5+4} \cot \frac{C}{2} \\ \\ \tan \frac{C}{2} = \frac{\sqrt{63}}{9} \\ \\ \cos C= \frac{1 - tan^{2}\frac{C}{2}}{1 + tan^{2}\frac{C}{2}} \\ \\ \cos C = \frac{1 - \frac{63}{81}}{1 + \frac{63}{81}} \\ \\ \cos C = \frac{81-63}{81+63} \\ \\ \cos C = \frac{18}{144} \\ \\ \cos C = \frac{1}{8} \\ \\ Again, \\ \\ c^{2} = a^{2} + b^{2} - 2ab \cos C \\ \\ \\\ = 5^{2} + 4^{2} - 2*5*4*\frac{1}{8} \\ \\ = 25 + 16 - 5 \\ \\ = 36 \\ \\ \textbf{c = 6}$

Answer 2

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Asked on October 15, 2019 by

Novin Jaykumar

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