1. **Problem statement:** We need to find the length of edge $AD$ in a triangular pyramid with given angles and side $BC = 37$ cm. 2. **Given data:** - Angle at $A = 54^\circ$ - Angle at $D = 33^\circ$ - Angle at $C = 125^\circ$ - Side $BC = 37$ cm - Right angle at $B$ between edges $BD$ and $BC$ 3. **Step 1: Understand the shape and angles.** The triangle $BCD$ has a right angle at $B$, so $\angle B = 90^\circ$. 4. **Step 2: Use triangle angle sum in $BCD$.** $$\angle B + \angle C + \angle D = 180^\circ$$ $$90^\circ + 125^\circ + 33^\circ = 248^\circ$$ This sum is more than $180^\circ$, so the angles $C$ and $D$ given are likely angles in different triangles or the pyramid's faces, not all in triangle $BCD$. 5. **Step 3: Use right triangle $BCD$ to find $BD$ and $CD$.** Since $\angle B = 90^\circ$, and $BC = 37$ cm, we can use trigonometry if we know one more side or angle in triangle $BCD$. But we don't have that directly. 6. **Step 4: Use Law of Cosines in triangle $ADC$.** We know $\angle A = 54^\circ$, $\angle D = 33^\circ$, and $\angle C = 125^\circ$ are angles in the pyramid, so consider triangle $ADC$. 7. **Step 5: Calculate length $AD$ using Law of Cosines in triangle $ADC$.** We need side $AC$ or $DC$ to use Law of Cosines. Since $BC = 37$ cm and $\angle B$ is right, we can find $AC$ using triangle $ABC$ if possible. 8. **Step 6: Use triangle $ABC$ with $\angle A = 54^\circ$ and right angle at $B$.** Assuming $\angle B$ is right in triangle $ABC$, then: $$AC = \frac{BC}{\cos 54^\circ} = \frac{37}{\cos 54^\circ}$$ Calculate $\cos 54^\circ \approx 0.5878$: $$AC = \frac{37}{0.5878} \approx 62.93 \text{ cm}$$ 9. **Step 7: Use Law of Cosines in triangle $ADC$ with $\angle C = 125^\circ$, sides $AC = 62.93$ cm, and $DC$ unknown.** We need $DC$ to find $AD$, but $DC$ is unknown. 10. **Step 8: Use triangle $BCD$ to find $DC$.** Since $\angle B = 90^\circ$ and $BC = 37$ cm, if we find $BD$, then: $$DC = \sqrt{BD^2 + BC^2}$$ But $BD$ is unknown. 11. **Step 9: Use angle $D = 33^\circ$ in triangle $BCD$ to find $BD$.** In right triangle $BCD$, $\tan 33^\circ = \frac{BD}{BC}$ $$BD = BC \times \tan 33^\circ = 37 \times 0.6494 = 24.03 \text{ cm}$$ 12. **Step 10: Calculate $DC$ using Pythagoras:** $$DC = \sqrt{BD^2 + BC^2} = \sqrt{24.03^2 + 37^2} = \sqrt{577.44 + 1369} = \sqrt{1946.44} \approx 44.11 \text{ cm}$$ 13. **Step 11: Use Law of Cosines in triangle $ADC$ to find $AD$:** $$AD^2 = AC^2 + DC^2 - 2 \times AC \times DC \times \cos 125^\circ$$ Calculate $\cos 125^\circ \approx -0.5736$: $$AD^2 = 62.93^2 + 44.11^2 - 2 \times 62.93 \times 44.11 \times (-0.5736)$$ $$= 3960.5 + 1945.7 + 3187.3 = 9093.5$$ 14. **Step 12: Calculate $AD$:** $$AD = \sqrt{9093.5} \approx 95.38 \text{ cm}$$ **Final answer:** $$\boxed{AD \approx 95.38 \text{ cm}}$$