Hello All and especially Ray....welcome aboard!
My guess is they're using Ho-229 to solve the problem. Look at these pages from Pub. 229 Volume 2 page 339 to get Hc d and Z. For interpolation of "d" see Pub. 229 Volume 2 page 2. The pages are attached below.
As you'll see quickly that 229 is a bit closer to an accurate figure than Pub. 249. You may find some difficulty in correcting "d" but look at the interpolation table and it'll be easy.
Very quickly I got an Hc of 13° 54.9' which when compared to Ho of 13° 46.5' produced an Intercept of 8.4 Away. The trick in getting that figure was correct use of 229's Interpolation Table. I might be off by 0.1' as I'm not exactly sure how to use the "double second correction".
The Z figure (which of course will equal Zn) is 122.7° (equal to the Zn in the problem)
Get the entire Pub. No. 229 here- Pub. No. 229
Ho-229 Vol. 2 page 339.pdf (Size: 42.12 KB / Downloads: 338)
Ho-229 Vol. 2 page 2 Interpolation.pdf (Size: 30.29 KB / Downloads: 329)
I hope that clears things up for you.
Louis
My guess is they're using Ho-229 to solve the problem. Look at these pages from Pub. 229 Volume 2 page 339 to get Hc d and Z. For interpolation of "d" see Pub. 229 Volume 2 page 2. The pages are attached below.
As you'll see quickly that 229 is a bit closer to an accurate figure than Pub. 249. You may find some difficulty in correcting "d" but look at the interpolation table and it'll be easy.
Very quickly I got an Hc of 13° 54.9' which when compared to Ho of 13° 46.5' produced an Intercept of 8.4 Away. The trick in getting that figure was correct use of 229's Interpolation Table. I might be off by 0.1' as I'm not exactly sure how to use the "double second correction".
The Z figure (which of course will equal Zn) is 122.7° (equal to the Zn in the problem)
Get the entire Pub. No. 229 here- Pub. No. 229
Ho-229 Vol. 2 page 339.pdf (Size: 42.12 KB / Downloads: 338)
Ho-229 Vol. 2 page 2 Interpolation.pdf (Size: 30.29 KB / Downloads: 329)
I hope that clears things up for you.
Louis