If in a triangle ABC, AB = 5 units, ∠ B = ^ - 1 ((3)/(5)) and...

As,

$cosB = \dfrac{3}{5} ⇒ sinB = \dfrac{4}{5}$

Given:

$R = 5$

By sine rule, we have

$\dfrac{c}{sinC} = 2R$

$⇒ \dfrac{5}{10} = sinC$

$⇒ C = 30 °$
Now,

$\dfrac{b}{sinB} = 2R$

$⇒ b = \dfrac{4}{5} \times 10 = 8$

Now, by cosine formula

$cosB = \dfrac{{a}^{2} + {c}^{2} - {b}^{2}}{2ac}$

$⇒ \dfrac{3}{5} = \dfrac{{a}^{2} + 25 - 64}{2\left(5\right)a}$

$⇒ {a}^{2} - 6a - 39 = 0$

$⇒ a = \dfrac{6 \pm \sqrt{192}}{2} = \dfrac{6 \pm 8\sqrt{3}}{2}$

$⇒ a = \left(3 + 4\sqrt{3}\right)$

(Reject $a = 3 - 4\sqrt{3} < 0$)

Now, $\Delta = \dfrac{abc}{4R} = \dfrac{\left(3 + 4\sqrt{3}\right) \cdot 8 \cdot 5}{4 \cdot 5} = 2\left(3 + 4\sqrt{3}\right)$

$⇒ \Delta = \left(6 + 8\sqrt{3}\right)$ sq. units.