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        	IN A TRIANGLE ABC given that a=5   b=4   c=?????? If COS(A-B)=31/32
one year ago

Arun
23045 Points
							Dear student I have attahed image in which solution is given. In case of any difficulty please feel free to ask $\hspace{-0.45 cm}Using \cos\left(\frac{A-B}{2}\right) = \frac{1-\tan^2\frac{A-B}{2}}{1+\tan^2\frac{A-B}{2}} = \frac{31}{32}.\Rightarrow \tan\frac{A-B}{2} = \frac{1}{\sqrt{63}}\\\\\\Now Using \tan \frac{A-B}{2} = \frac{a-b}{a+b}\cot\frac{C}{2}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{63}}{9}\\\\\\Now Using \cos C = \frac{1-\tan^2 \frac{C}{2}}{1+\tan^2 \frac{C}{2}}\Rightarrow \cos C = \frac{1}{8}\\\\ Now Using c^2=a^2+b^2-2ab\cos C=36\Rightarrow c=6$

one year ago
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• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions