If in a triangle ABC,AB=5 units,angle B=cos ^ | vedclass

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(A) Given $\cos B = \frac{3}{5}$,then $\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$.Using the sine rule,$\frac{b}{\sin B} = 2

Vedclass · Accepted Answer

$6+8 \sqrt{3}$

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(A) Given $\cos B = \frac{3}{5}$,then $\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$.Using the sine rule,$\frac{b}{\sin B} = 2R$,where $R=5$ is the circumradius.$b = 2R \sin B = 2(5)\left(\frac{4}{5}ight) = 8$.Using the cosine rule for $\angle B$:$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$\frac{3}{5} = \frac{a^2 + 5^2 - 8^2}{2(a)(5)} = \frac{a^2 + 25 - 64}{10a} = \frac{a^2 - 39}{10a}$.$6a = a^2 - 39 \Rightarrow a^2 - 6a - 39 = 0$.Solving for $a$ using the quadratic formula: $a = \frac{6 \pm \sqrt{36 - 4(1)(-39)}}{2} = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3}$.Since $a > 0$,we take $a = 3 + 4\sqrt{3}$.The area of $	riangle ABC$ is given by $\Delta = \frac{1}{2}ac \sin B$.$\Delta = \frac{1}{2}(3 + 4\sqrt{3})(5)\left(\frac{4}{5}ight) = 2(3 + 4\sqrt{3}) = 6 + 8\sqrt{3}$.

If in a triangle $ABC$,$AB=5$ units,$\angle B=\cos ^{-1}\left(\frac{3}{5}\right)$ and the radius of the circumcircle of $\triangle ABC$ is $5$ units,then the area (in sq. units) of $\triangle ABC$ is:

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