Given: \(\sin A = \frac{4}{5}\)
First, we'll find \(\cos A\) using the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \\ \left(\frac{4}{5}\right)^2 + \cos^2 A = 1 \\ \frac{16}{25} + \cos^2 A = 1 \\ \cos^2 A = 1 - \frac{16}{25} = \frac{9}{25} \\ \cos A = \frac{3}{5} \quad (\text{We take the positive value}) \]
Next, we find \(\tan A\):
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]
Now, we'll evaluate the expression \((3 - \tan A)(2 + \cos A)\):
\[ (3 - \tan A)(2 + \cos A) = \left(3 - \frac{4}{3}\right)\left(2 + \frac{3}{5}\right) \\ = \left(\frac{9}{3} - \frac{4}{3}\right)\left(\frac{10}{5} + \frac{3}{5}\right) \\ = \left(\frac{5}{3}\right)\left(\frac{13}{5}\right) \\ = \frac{5 \times 13}{3 \times 5} = \frac{13}{3} \]
The correct answer is option (2) \(\frac{13}{3}\).
Using $\sin A = \frac{4}{5}$,
we calculate $\tan A = \frac{4}{3}$ and $\cos A = \frac{3}{5}$.
Substituting into the expression gives $\left(3 − \frac{4}{3}\right)\left(2 + \frac{3}{5}\right) = \left(\frac{5}{3}\right) \times \left(\frac{13}{5}\right) = \frac{13}{3}$.
The given graph illustrates: