1. **Problem Statement:** Calculate the length of edge $AD$ in the given triangle-based pyramid, where angles at $D$, $A$, and $B$ are $39^\circ$, $44^\circ$, and $126^\circ$ respectively, and edge $BC = 46$ cm. 2. **Understanding the problem:** The pyramid has a base triangle $ABC$ with a right angle at $B$ (since $\angle B = 126^\circ$ and a right angle symbol is shown, the $126^\circ$ likely refers to an external angle, so the internal angle at $B$ is $54^\circ$). We want to find the length $AD$, where $D$ is a vertex above the base. 3. **Step 1: Find the length of $AB$ and $AC$ in triangle $ABC$ using the Law of Sines.** - Angles in triangle $ABC$ sum to $180^\circ$. - Given $\angle A = 44^\circ$, $\angle B = 54^\circ$ (since $126^\circ$ external angle means internal angle $180^\circ - 126^\circ = 54^\circ$), so $\angle C = 180^\circ - 44^\circ - 54^\circ = 82^\circ$. - Using Law of Sines: $$\frac{BC}{\sin A} = \frac{AB}{\sin C} = \frac{AC}{\sin B}$$ - Given $BC = 46$ cm, so: $$\frac{46}{\sin 44^\circ} = \frac{AB}{\sin 82^\circ} = \frac{AC}{\sin 54^\circ}$$ - Calculate $\sin$ values: $\sin 44^\circ \approx 0.6947$ $\sin 82^\circ \approx 0.9903$ $\sin 54^\circ \approx 0.8090$ - Calculate $\frac{46}{0.6947} \approx 66.23$ - Then: $$AB = 66.23 \times 0.9903 \approx 65.56 \text{ cm}$$ $$AC = 66.23 \times 0.8090 \approx 53.56 \text{ cm}$$ 4. **Step 2: Use the right triangle formed by $D$, the foot of the perpendicular from $D$ to base $ABC$, and $A$ to find $AD$.** - Given $\angle D = 39^\circ$ and $\angle A = 44^\circ$ in the pyramid, and the right angle at $B$, the height from $D$ to base $ABC$ is perpendicular. - Since the problem states the dotted line from $D$ to the base is perpendicular, we can use trigonometry in triangle $ABD$ or $ACD$. - However, without explicit height or other lengths, we assume $AD$ is the hypotenuse of a right triangle with angle $39^\circ$ at $D$. - Using the Law of Cosines in triangle $ABD$ or $ACD$ is complicated without more data. 5. **Step 3: Approximate $AD$ using the Law of Cosines in triangle $ABD$.** - We know $AB \approx 65.56$ cm. - Angle at $D$ is $39^\circ$. - Assume $BD$ is the height (perpendicular), so $AD$ is the hypotenuse. - Using right triangle trigonometry: $$AD = \frac{AB}{\cos 39^\circ}$$ - Calculate $\cos 39^\circ \approx 0.7771$ - Then: $$AD = \frac{65.56}{0.7771} \approx 84.36 \text{ cm}$$ 6. **Final answer:** The length of edge $AD$ is approximately $84.36$ cm to 2 decimal places. --- **Answer:** $\boxed{84.36}$ cm