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I will demonstrate another Bygrave sight reduction on the E-6B. The
correct values are:

LHA = 295 51.2
lat = -36 29.6
dec = -17 54.1,

az = 90 01.3
alt = 31 05.5

It's no accident azimuth is almost 90. Let's see if the solution is
inaccurate at that angle.

First convert sexagesimal angles to decimal degrees:

lat = -36 29.6 = -36.49
LHA = 295 51.2 = 295.85
dec. = -17 54.1 = -17.90

t = 360 - LHA = 64.15

The Bygrave formula requires the cosine or secant of t. Either will do,
since the cosine and secant are simply reciprocals of each other.
However, the E-6B has only a tangent scale, so to obtain a secant we use
the trigonometric identity sec² t = 1 + tan² t. This relationship 
between tangent and secant will be used several times. For angle t:

tan t = 2.076
tan² t = 4.33
sec² t = 5.33
sec t = 2.304

Now that sec t is known, solve for y:

y = arctan(tan dec * sec t) = arctan(.742) = 36.6

The cosine of y is needed later, so compute it from tan y:
sec² y = 1 + tan² y = 1.550
sec y = 1.245
cos y = .803

Compute upper case Y:
Y = 90 - lat + y = 91.1

With Y known, compute its secant, needed later. Start with the tangent.
Unfortunately tan 91.1° is not on the scale. But it's equal to 1 / tan
1.1°. To practical accuracy, tan 1.1° is simply 1.1° converted to
radians. So opposite 1.1° on the outer scale set 57.3 on the inner. Read
tangent at the inner index. But we really wanted its reciprocal, which
is at the outer index: 52.0.

With tan Y known, find its secant via the secant and tanget identity 
again: sec² Y = 1 + tan² Y. This time it's easy. The tangent of Y is so 
big, adding 1 to its square makes hardly any difference. To E-6B 
accuracy, tan Y = sec Y = 52.0.

Next solve for az = arctan(tan t * cos y * sec Y) = arctan(2.076 * .803
* 52.0) = arctan(86.6). The arctangent of 86.6 is way outside the range
of the scale. However, arctan 86.6 = 90 - arctan(1/86.6) = 90° -
arctan(.01155). Again use the fact that the tan of a small angle is
practically equal to the angle itself, expressed in radians. Therefore,
arctan(.01155) = .01155 * 57.3 = .662°, and 90° - .662° = az 89.34°.

According to the Bygrave rules, that value must be subtracted from 180
because Y is greater than 90. So az = 90.66°.

We're almost done. To compute altitude we need the cosine or secant of
azimuth. Again the tangent is large, so to practical accuracy sec az =
tan az = 86.6. (Formally, the value is negative, but the sign is ignored.)

Finally Hc = arctan(cos az * tan Y) = arctan(tan Y / sec az) =
arctan(52.0 / 86.6). At 52.0 on the outer scale, set 86.6 on the inner.
Read arctan = 31.0° on the drift scale.

The correct values are az = 90 01.3 (vs. 90.66), Hc = 31 05.5 (vs.
31.0). Not bad, especially the azimuth was intentionally chosen to make
the solution difficult. And it was difficult! Trig on any slide rule
demands knowledge of identities. (As a schoolboy I disliked trig 
identities, but have become more comfortable with them over the years.) 
The minimum facilities for trig on the E-6B, plus the parameters of this 
problem, made the solution even harder.

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