To find the distance $DC$ in the given triangle, we can use the Law of Sines. In the triangle $\triangle ABC$, we are given:

• $AB = 91 \, \text{m}$
• $\angle BAC = 54^\circ$
• $\angle ABC = 48^\circ$

Since the angles in a triangle add up to $180^\circ$, we can find the angle $\angle ACB$:

$\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 54^\circ - 48^\circ = 78^\circ$

Now, using the Law of Sines in $\triangle ABC$, we have:

$\frac{AB}{\sin \angle ACB} = \frac{AC}{\sin \angle ABC} = \frac{BC}{\sin \angle BAC}$

Thus,

$AC = \frac{AB \cdot \sin \angle ABC}{\sin \angle ACB} = \frac{91 \cdot \sin 48^\circ}{\sin 78^\circ}$

Calculating $AC$:

1. $\sin 48^\circ \approx 0.7431$
2. $\sin 78^\circ \approx 0.9781$

$AC = \frac{91 \times 0.7431}{0.9781} \approx 69.1 \, \text{m}$

Now that we have $AC$, we need more information or additional steps to find $DC$, as it seems to involve additional relationships or right-angle triangles with points $D$ and $C$. Please confirm if there’s any further data on angles or if $D$ is a particular type of point like the height from $C$ (forming a right triangle), so I can provide the final calculation for $DC$.

Would you like additional help with these calculations or have specific questions?

Here are five related questions to extend this problem:

1. What is the significance of point $D$ in the triangle?
2. Can we find $DC$ if we consider it a line perpendicular to $AB$?
3. How does knowing $\angle DCA$ or $\angle DCB$ help find $DC$?
4. If $D$ represents a height from $C$ to $AB$, what is the formula for calculating this distance?
5. How would using the cosine rule change our approach in this scenario?

Tip: For non-right triangles, the Law of Sines and the Law of Cosines are powerful tools to relate sides and angles.