The angle at the base of Mycerinus, if built to a ratio of 4 to 5 (half-base to altitude), and not to the more practical but nearly perfect ratio of 32 to 20 (apothem to half-base) would be the complement of angle ADC, thus—
| 4 | = ·8 = Tan. < ADC | = 38° 39′ 35 | 165″ |
| 5 | 477 | ||
| ∴ < DAC | = 51° 20′ 24 | 312″ | |
| 477 |
but as it is probable that the pyramid was built to the ratio of 32 to 20, I have shown its base angle in Figure 19, as 51° 19′ 4″.
Figure 72 shows how the slopes of Cephren were arrived at.
(Cephren) Fig. 72.
LHNM represents the base of the pyramid. On the half-base AC, I described a 3, 4, 5 triangle ABC. I then projected the line CF (ratio 21 to BC 20), thus erecting the 20, 21, 29 triangle BCF. From this datum, I arrived at the triangles BEA, ADC, and GKH; GK, EA and AD each representing apothem; CF and CD, altitude; and HK, edge. The lengths of the lines AD, HK and CH being got at as in the pyramid Mycerinus. These measures reduced to cubits, calling AC = ratio 16 = 210 cubits (half-base of pyramid) give the following result.
| R. B. CUBITS. | BRITISH FEET. | |
| Half-base | 210·00 | 353·85 = LA |
| Apothem | 346·50 | 583·85 = EA |
| Edge | 405·16 | 682·69 = HK |
| Altitude | 275·625 | 464·43 = CD |
| Half-diag. of base | 296·985 | 500·42 = CH |
thus I get the ratios of—Apothem : Half-Base :: 33 : 20, &c. The planes in the diagram are placed in their correct positions, as directed for Figure 71.
The angle at the base of Cephren, if built to the ratio of 16 to 21 (half-base to altitude), and not to the practical ratio of 33 to 20 (apothem to half-base), would be the complement of < ADC, thus—