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a

a

B

cos 2C

a

tan b

(38.) Solution of triangles by series. [1] Given the sides a, b, and the angle C of a plane triangle:

62

73 B sin C+

sin 2C+ sin 3C + &c.
2 a

33
b
72

63
log?c=log, a
cos C

cos 3C- &c. 2a?

3a3 [2] Given the sides a, b, and the angle C of a spherical triangle:

tan lb sin 2C 1(A + B)=90°— C+ sin C cotia cot

2 tan

tan b) sin 20 (A B)=90° — C sin C

tanza

tan logsin dc=log: (sinla.cos lb)

tansa

&c.

2 log, cosc=log (cosa.cos lb) + tanja.tan lb.cos C (tanka.tan16)

2

+ &c. v

(

cotia

&c. 8

tanza

2

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cos 20

E

cos 20

+ &c.

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in sin mc

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* B=s. (1.)"
Blog.c=log, a–S ()
? }(A+B)=90°– ļC+Šm(–1)---(tan)". sin m C
Į (A B)=90°– ļC-Sm (ta

.. (tan
• log, sin dc=log, (sin La.cos l_b) – 5 m (tan 11

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log,cos c=log (cosa.cos}6) +S,(-1)=\(tanja.tanb)m

m

(39.) Correction for a compound base. If the base be composed of two straight lines a and b, inclined at an angle 180° - 2", the correction is

ab x? ?

0,00000000001175.

a + b

(40.) Reduction of an oblique to a horizontal angle. Let a, ß, be the angles of elevation of two objects;

72 their angular distance;

C, the horizontal angle: then [1] C-y=${(a +B)' tanıy-la-B) cotx}.

sinį (a +a–B).sin}(a-a-B), [2] (sinC)=

cos a.cos ß log SinļC=}{log Sin} (a +a-B) + log Sin (a-a-B) log Cos a -- log Cos ß} + log r.

cos observed _ For a single object, cos reduced 2=

cos Z of elevation (41.) Reduction of a spherical to a plane triangle. Let C be the spherical angle,

C-x the corresponding plane angle; then

1

[1] {(a+b)* tanļC-(ab)*cot {C}.

16 [2] Legendre's Theorem : area of triangle

= }(A+B+C–180°). 372

The value of x in seconds is

area of triangle

37% < 0,000004848 If the area be found in feet, and a degree on the Earth's surface=365155 feet, then log10 divisor=9,803894.

Reduction of a plane to a spherical triangle.
Let y be the plane angle contained by the sides a, ß;
C the corresponding spherical angle; then

cosy-4aß

(1–1a*)(1–133) (L. 293_313; C. ch. xx; W. ch. xii; Enc. Met. V. 1. p. 698,9.)

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sin 15°=}(1+)*- }(1 – })=8,;
sin ? 15°=} (1+8;)! – } (1 – 8,)}=8,;
sin 15°=} (1+8,)} – } (1 – 8,)}=83;
&c. = &c.

15°
By continuing this process we obtain sin

210 sin 1' by taking it proportional to the arc. 15° 41"

= 52" 44" 3"" 45'. 210

and thence

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a.

V3+1

sin 3o =

872 (V5-1)

412V5-V6.

Let 2 sin 6° = x, then

25-503 +5x-1=0, which equation may be solved by approximation. sin (a + b)=sin a + {sin a — sin (a−b)} - 4 sin a.

- } – (sin { b): Putting b=lo, we obtain Delambre’s formula,

sin (a +19)=sin a+{sin a-sin (a -1°)} - 4 sin a.(sin 30%).
sin (60° +a)=sin a + sin (60° – a).
tan (45° + a)=2 tan 2a + tan (45o - a).
cosec a=cot a + tanda

(W. ch. iv; C. ch. vi.) (43.) Sines of arcs expressed by surds.

73-1

75+75. 8

V3 sin 6°=-(V5+1) +

1 sin 9° =

4V2)-115–
V3

1 sin 12° =

V5 – 1 8

1 sin 15o =

2V2 sin 18°=£(V5 – 1). 73-1

V3+1 sin 21°=

)

5-15. 872

8 V3

1 sin 24° - -+)

8

1 sin 27°

472 sin 30°= $ V3+1

V3-1 sin 33° (V5-1) +

5 + 15. 872

8 1 sin 36° =

5-V5

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212V5-V5.

T

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)

(V5–1)+aVaV5+

V3+1

V3-1 sin 39°= (V5+1) 5-V5. 8V2

8

V3 sin 42° - ; (V5-1) + 75+V5.

472
1
sin 45°

V 2
V3

1 sin 48° =

V5. 8

412 V 3-1

V3+1 sin 51°=

1

15-V5. 2

8 sin 54°=} (V 5+1). V3-1

V3+1 sin 57°

(V5-1) + 5+15. 8V 2

8 sin 60°= {V3.

1 sin 63° = (V5-1)+15+V5. 4V 2

V3 sin 66°= ; (V5+1).+

75-75.

472 V3+1

V3-1 sin 69°

(V5+1)+ -V5-V5. 8V 2

8 1

75+75. 2V2

1
sin 75o = (V3+1).
272

V3
V5 – 1

+

sin 72°

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