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in triangle ABC, if a=5,b=4 and cos(A-B)=31/32 then c is equal to? A 9 B 8 C 7 D 6

in triangle ABC, if a=5,b=4 and cos(A-B)=31/32 then c is equal to?
 
A  9
B  8
C  7
D  6

Grade:11

2 Answers

jagdish singh singh
173 Points
8 years ago
\hspace{-0.45 cm}$Using $\cos\left(\frac{A-B}{2}\right) = \frac{1-\tan^2\frac{A-B}{2}}{1+\tan^2\frac{A-B}{2}} = \frac{31}{32}.\Rightarrow \tan\frac{A-B}{2} = \frac{1}{\sqrt{63}}$\\\\\\Now Using $\tan \frac{A-B}{2} = \frac{a-b}{a+b}\cot\frac{C}{2}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{63}}{9}$\\\\\\Now Using $\cos C = \frac{1-\tan^2 \frac{C}{2}}{1+\tan^2 \frac{C}{2}}\Rightarrow \cos C = \frac{1}{8}$\\\\ Now Using $c^2=a^2+b^2-2ab\cos C=36\Rightarrow c=6$
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
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
 
Hope it helps.
Thanks and regards,
Kushagra

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