If $\sec 6A=\csc (A-29^{\circ })$, where $2A$ is an acute angle, then find the value of $\angle A$.

Given the trigonometric equation:
$$\sec 6A=\csc (A-29^{\circ })$$

We know the complementary angle identity:

$$\sec \theta =\csc (90^{\circ }-\theta )$$

Therefore, we can equate:

$$6A=90^{\circ }-(A-29^{\circ })$$

$$6A=90^{\circ }-A+29^{\circ }$$

$$6A+A=119^{\circ }$$

$$7A=119^{\circ }$$

$$A=\frac{119^{\circ }}{7}=17^{\circ }$$

Let's verify the condition: $2A=2\times 17^{\circ }=34^{\circ }$, which is an acute angle ($<90^{\circ }$). Hence, the condition is satisfied.