In a triangle,if b=20, c=21 and sin A=frac{3} | vedclass

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(B) We have,$b=20, c=21$ and $\sin A=\frac{3}{5}$.Since $\cos^2 A = 1 - \sin^2 A$,we have $\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{1

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$13$

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(B) We have,$b=20, c=21$ and $\sin A=\frac{3}{5}$.Since $\cos^2 A = 1 - \sin^2 A$,we have $\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.Thus,$\cos A = \frac{4}{5}$ (assuming $A$ is acute).Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.Substituting the values: $\frac{4}{5} = \frac{20^2+21^2-a^2}{2 \cdot 20 \cdot 21}$.$\frac{4}{5} = \frac{400+441-a^2}{840}$.$840 \cdot \frac{4}{5} = 841 - a^2$.$168 \cdot 4 = 841 - a^2$.$672 = 841 - a^2$.$a^2 = 841 - 672 = 169$.Therefore,$a = \sqrt{169} = 13$.

In a triangle,if $b=20, c=21$ and $\sin A=\frac{3}{5}$,then $a$ is equal to :

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