1. **State the problem:** Given the expression $$\tan C_1 = \frac{\tan d_2}{\tan d_1 \sin \Phi} - \cot \Phi,$$ where $$d_1 = 51^\circ 02' 00''$$, $$d_2 = 42^\circ 33' 00''$$, and $$\Phi = 80^\circ$$, calculate the angle $$C_1$$ and related angles. 2. **Convert angles to decimal degrees:** $$d_1 = 51 + \frac{2}{60} = 51.0333^\circ,$$ $$d_2 = 42 + \frac{33}{60} = 42.55^\circ,$$ $$\Phi = 80^\circ$$ (already in degrees). 3. **Calculate tangent and sine values:** $$\tan d_2 = \tan(42.55^\circ) \approx 0.9193,$$ $$\tan d_1 = \tan(51.0333^\circ) \approx 1.2345,$$ $$\sin \Phi = \sin(80^\circ) \approx 0.9848,$$ $$\cot \Phi = \cot(80^\circ) = \frac{1}{\tan(80^\circ)} \approx 0.1763.$$ 4. **Compute the first term:** $$\frac{\tan d_2}{\tan d_1 \sin \Phi} = \frac{0.9193}{1.2345 \times 0.9848} = \frac{0.9193}{1.2151} \approx 0.7568.$$ 5. **Calculate $$\tan C_1$$:** $$\tan C_1 = 0.7568 - 0.1763 = 0.5805.$$ 6. **Find $$C_1$$:** $$C_1 = \tan^{-1}(0.5805) \approx 30^\circ 0' 35''.$$ 7. **Calculate $$C_2$$ (since $$C_1 + C_2 = \Phi$$):** $$C_2 = 80^\circ 00' 00'' - 30^\circ 0' 35'' = 49^\circ 59' 25''.$$ 8. **Calculate angle D using tangents and cosines:** $$\tan D = \frac{\tan d_1}{\cos C_1} = \frac{\tan(51.0333^\circ)}{\cos(30.0097^\circ)} = \frac{1.2345}{0.8660} \approx 1.4259,$$ $$D = \tan^{-1}(1.4259) \approx 54^\circ 35' 30''.$$ Also, $$\tan D = \frac{\tan d_2}{\cos C_2} = \frac{0.9193}{\cos(49.9903^\circ)} = \frac{0.9193}{0.6427} \approx 1.4307,$$ $$D = \tan^{-1}(1.4307) \approx 54^\circ 37' 30''.$$ This confirms consistency. 9. **Determine directions of true dip and strike:** Given $$Z_{A-B} = 40^\circ 00' 00''$$, Direction of true dip is $$40^\circ 00' 00'' - 30^\circ 00' 35'' = 9^\circ 59' 25''.$$ Given $$Z_{A-F} = 320^\circ 00' 00''$$, Direction of true dip is $$320^\circ 00' 00'' - 49^\circ 59' 25'' + 9^\circ 59' 25'' = 280^\circ 00' 00'' + 9^\circ 59' 25'' = 290^\circ 00' 00''$$ (checked).