If in Delta ABC, a = 6, b = 3 and cos(A - B) | vedclass

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(C) Given: $a = 6, b = 3, \cos(A - B) = \frac{4}{5}.$Using the formula $\cos(A - B) = \frac{1 - 	an^2(\frac{A - B}{2})}{1 + 	an^2(\frac{A - B}{2})},

Vedclass · Accepted Answer

$9$

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(C) Given: $a = 6, b = 3, \cos(A - B) = \frac{4}{5}.$Using the formula $\cos(A - B) = \frac{1 - 	an^2(\frac{A - B}{2})}{1 + 	an^2(\frac{A - B}{2})},$$\frac{4}{5} = \frac{1 - t^2}{1 + t^2} \implies 4 + 4t^2 = 5 - 5t^2 \implies 9t^2 = 1 \implies t = \frac{1}{3}.$So,$	an(\frac{A - B}{2}) = \frac{1}{3}.$Using Napier's Analogy: $	an(\frac{A - B}{2}) = \frac{a - b}{a + b} \cot(\frac{C}{2}).$$\frac{1}{3} = \frac{6 - 3}{6 + 3} \cot(\frac{C}{2}) = \frac{3}{9} \cot(\frac{C}{2}) = \frac{1}{3} \cot(\frac{C}{2}).$$\cot(\frac{C}{2}) = 1 \implies \frac{C}{2} = 45^\circ \implies C = 90^\circ.$The area of $\Delta ABC = \frac{1}{2} ab \sin C = \frac{1}{2} 	imes 6 	imes 3 	imes \sin 90^\circ = 9 	imes 1 = 9 \, square \, unit.$

If in $\Delta ABC,$ $a = 6, b = 3$ and $\cos(A - B) = \frac{4}{5},$ then its area will be ..... $square \, unit.$

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