In Delta ABC,if a = 3, b = 4, c = 5,then sin | vedclass

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(C) Given sides are $a = 3, b = 4, c = 5$. Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting the values: $\cos B = \frac{

Vedclass · Accepted Answer

$24/25$

Vedclass · Answer

(C) Given sides are $a = 3, b = 4, c = 5$. Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting the values: $\cos B = \frac{3^2 + 5^2 - 4^2}{2 	imes 3 	imes 5} = \frac{9 + 25 - 16}{30} = \frac{18}{30} = \frac{3}{5}$. Since $\sin^2 B + \cos^2 B = 1$,we have $\sin B = \sqrt{1 - (3/5)^2} = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5$. Using the double angle formula: $\sin 2B = 2 \sin B \cos B$. Therefore,$\sin 2B = 2 	imes (4/5) 	imes (3/5) = 24/25$.

In $\Delta ABC$,if $a = 3, b = 4, c = 5$,then $\sin 2B = $

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