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Trigonometry> in triangle ABC, if a=5,b=4 and cos(A-B)=...

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in triangle ABC, if a=5,b=4 and cos(A-B)=31/32 then c is equal to? A9B8C7D6

Amogh Dasture , 8 Years ago

Grade 11

2 Answers

jagdish singh singh

Last Activity: 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

Last Activity: 4 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] = 1/63^½

Now, tan[(A – B)/2] = (a – b)cot(C/2) / (a + b)

using this we get, cotC/2 = 3/7^½

or, tanC/2 = 7^½/3

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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