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tan. (H 10)

sin. L tan. H2 = sin. L x tan. 30o =

✓3 which gives

H2 = H (10) = m by supposition.

Again, let P2 = x, P 10 = y. then

sin. x : sin. m :: sin. H : sin. 0
sin. m : sin. y :: sin. (-a) : sin. H
sin. x sin. (-a)
sin.
y

sin. A Also

COS. m cos. X cos. L cos. 6 =

sin, x sin. L.

(1)

(2)

COS. m cos. (-a)= cos. y cos. L

(3) sin. y. sin. L And cos. H= cos. X-cos. m. cos. L

cos. y -cos. m. cos. L sin, m. sin. L

sin. m. sin. L Hence

cos. x + cos. y = 2 cos. m. cos. L Whence by the arithmetic of sines

X, y and o

may be found ; and these being known, it is easy to find

90° PH2 which will give the dip required.

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Again,

tan. (H) = sin. L x tan. 6 15

tan. H4 = sin. L x tan. 4 X 15 :: tan. (4, 6) = Stan. H6 – tan. I 4

1+tan, H6 x tan. H4

sin. L X (00-73) 1+ 3. X00 x sin. L

1

sin. L XN 3 Hence, by the question

1 sin. L

sin. L X 7 3

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840. Let a be the angular distance of the shadow of the gnomon from the meridian at a second past 4 o'clock, and B that at 4 o'clock; then, by 833, we have tan. a = sin. L. tan. (4 x 15° +

24
tan. B = sin. L. tan. (4 x 15°)
tan. a –

tan. 8
:. tan. (@-B) =
1 + tan. a. tan. ß

1 tan. (60

tan. 60°

240 = sin. LX

1 1 +sin. 'L x tan. 60 x tan. 60°

240 But by the question the angle due to a second at noon is

B

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a

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1
1-sin. "L x tan. O

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841. Let ZHZ (Fig. 110,) be the meridian of the place, PP' the earth's axis, QQ' the equator, SS' the path of the sun, and HH' the horizon, whose intersection with SS' is s. Then sPH measures the time of sunrise, from midnight. Let PH the latitude be denoted by L, Ps the co-declination of the sun by D, and by the rules of trigonometry, we have

tan. L cos. (sPH) =

tan. D which gives sPH or the time required.

842.

By the preceding problem

tan. L COS. SPH

tan. De

tan. L COS. 45°

=N 2 tan. L

which gives the co-declination of the star, and therefore the declination.

843. We will premise the solution of this problem, with a general investigation of the nature of the curve traced by the shadow of a point, elevated above the horizon, upon the horizontal plane.

Let A (Fig. 111, a, b) be the given point, AC I horizon, CB the shadow at noon, and CP for any azimuth Q. From P

let fall PM I CB, and make CM = x, PM = y, CA = h, ZP = co-lat. of the place = L, PS = co-declination of the sun = D, and ZS = co-altitude of the sun = a; then by trig.

cos. D - cos. a x cos. L COS. P =

sin. a X sin. L But

CP

(.ro + y2)
sin. a =
AP

(h* + x2 + y)
CA

h
AP v (k" + x + y)

COS. a =

and cos. O

✓ (x + y) Hence we get

cos. Dv (ha + x2 + y) = h cos. L + x sin. L....(a) which gives y cos.' D + x* (cos.' D – sin.' L) - xh sin. 2L + h (cos.'Dcos.' L) = 0.... (1) the equation to a conic section, which is a parabola, an ellipse, or hyperbola, according as cos. D – cos. L is zero positive or negative. The transverse axis 2a is found by making y = 0; for then h sin. 2 L

cos.' D - cos.' L

h?. - sin.' L

cos. D - sin.'L

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cos.2 D

and x =

h

x (sin. 2D + sin. 2L) 2 cos.2 D-sin. L) and 2a = sum of these values of x. h sin. 2D

(2) cos.. D - sin.'L Also the other axis is twice the maximum value of y, which being found by putting the differential of the value of y2 cos. D equa! to o, thence getting r and substituting in equat. (1), gives 2h sin. D

(3) V (cos. D – sin.2 L)

2b =

The curve traced on the horizon being thus determined, let us proceed to deduce from it the curve traced on any plane bpC, which

is I to the meridian ACB, and inclined to the horizon at the < bCB = L PCP = B. Let pm (which is I Co) - =ý

cm = X. And x = x x

sin. (B + M)

sin. M
= x' x {sin. B. X cot. M + cos. B}
x {x sin. B + h cos.

B}

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hy

a tan. 8 +
y
a'h cos. B

x tan. 8 +
h - a sin.B

hy :: y =

(5) h - sin. B These values of x and y being substituted in equation (1) give

y" cos.' D + 2'? {cos.' L - sin.' B - sin.' D + 2 sin. BX sin. L (1 + sin. B sin. L) } ht' {cos. ß sin. 2L + 2 sin. BX (cos.' D - cos.' L) } + ho (cos.' D - cos.' L) = 0,.. .. (6) the equation to a conic section, which is a parabola, an ellipse, or an hyperbola, according as the co-efficient of z" is zero, positive or negative.

The semiaxes of this curve may be found in the same way as those of the horizontal trace were determined.

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