The problem states that $\triangle ABC$ is a right-angled triangle at $B$ where $3 \cdot AC = 5 \cdot BC$. We need to find $\tan A$.

Step-by-Step Solution:

1. Assign the lengths: Since $\triangle ABC$ is right-angled at $B$, we can apply the Pythagorean theorem. Let:

   • $AC = h$ (hypotenuse),
   • $BC = b$ (adjacent side to $\angle A$),
   • $AB = a$ (opposite side to $\angle A$).

2. Given Ratio: We are given $3 \cdot AC = 5 \cdot BC$, which simplifies to: $\frac{AC}{BC} = \frac{5}{3}$ This implies that $\frac{h}{b} = \frac{5}{3}$, so we can assume $h = 5k$ and $b = 3k$ for some constant $k$.

3. Using the Pythagorean Theorem: In a right triangle, $AC^2 = AB^2 + BC^2$. Substituting $h = 5k$ and $b = 3k$: $(5k)^2 = a^2 + (3k)^2$ $25k^2 = a^2 + 9k^2$ $a^2 = 25k^2 - 9k^2 = 16k^2$ $a = 4k$

4. Finding $\tan A$: By definition, $\tan A = \frac{AB}{BC} = \frac{a}{b}$. Substituting the values of $a$ and $b$: $\tan A = \frac{4k}{3k} = \frac{4}{3}$

Answer:

The value of $\tan A$ is $\frac{4}{3}$.

Thus, the correct option is: (d) $\frac{4}{3}$

Would you like more details on any of these steps?

Here are some related questions to expand your understanding:

1. How would the answer change if the ratio $3 \cdot AC = 4 \cdot BC$ were given instead?
2. How is the tangent ratio related to the angle in right-angled triangles?
3. What would be the value of $\sin A$ and $\cos A$ for this triangle?
4. How can the Pythagorean Theorem be derived from the properties of similar triangles?
5. How do the trigonometric ratios change as the angle $A$ approaches $90^\circ$ or $0^\circ$?

Tip: When dealing with right triangles, assigning a variable (like $k$) based on the ratio can simplify calculations.