178 
PRESTON. 
Second term. 
-To find b B = -—cos a tan 8 X B (see plate 9). 
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The cosine of a is the line u v = — 0.368 x 20. The 
tangent of 8 is cd=og== 1.33. Project v to v' and draw 
v' o. The intersection of this line with the vertical through 
d gives the point s'. We then have g s' — cos « tan 8 = 
— 0.49. Draw ox so that j x — x j o. The intersec¬ 
tion of a horizontal line through s' with the line x o gives 
0.33. The value 
the point s" and s" y =-^9 cos a tan 8 = 
of B on June 9 is — 8.35. This distance laid off on the 
line p n, or, which is the same thing, the ordinate for June 9 
being projected to the vertical at a distance of 10 units from 
the origin, gives the point Z. The intersection of a vertical 
line through s" with the line Zo gives the point Z' and the 
distance 
Z' Z" =~K cos cl tan 8 x B = -f- 0.27. 
10 
The object in laying off j x — j o is to secure one 
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more decimal place in the value of b. The correct value in 
the final result is obtained in the multiplication by B, since 
the construction gives us of cos a tan 8. 
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The line x o is used in the construction of c G and d D as 
well as b B. 
Introducing the factor ^ serves the double purpose of 
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giving one more decimal place, thus increasing the accuracy, 
and also of restoring the final result to the correct scale after 
multiplying by B. 
Third term .—-To find c C — — cos a sec 8 X C (see plate 9). 
The secant of the declination is the line A B = sec 53° 7' 
= + 1.67. For verification see plate 7. The intersection of 
