FIGURE 7. SUPPLEMENTARY STATION.

Solution. Let A be the position of the station mark observed from B and C, and S the position of the instrument, or supplementary station, then if A, B, C be the

three angles, a, b, c, the three sides corresponding, and B and C be observed, one value of A is obtained = 180°-B+C, and if a is known, b and c can be calculated on this assumption. To verify the value of A, measure AS=m a very short distance, also the angles BSC=S, ASB=B, ASC=y; and suppose the very small angles ABS, ACS to be represented by x and y.

Then as A=BDC-x-S-x+y,

and as the values of x and y in circular measure are

=

Example. Let BC a 2000 feet; and let the observed angles ABC=B and ACB=C, be respectively 60° and 50°; let the distance of the satellite station S from A be 10 feet; and the angles observed at S be ASB=90°, BSC=70° 14′ 45′′; required the angle BAC.

Now using the value of BAC=180°-B+C=70°
the side c=sin 50° cosec 70° x 2000= 1630°42
the side b=sin 60° cosec 70° x 2000=1843°*21
sin 90° sin 19° 45′ 15′′

And A=S-10 cosec 1" (16304

the logarithmic calculation of which is

1843.2

Hence A=70° 14′ 45′′ — 1265′′ +378"

=69° 59′ 58′′, thus checking the above value 70°

The distant triangle. (See page 53.)

Solution.-Using the terms given with the formulæ.

cotg 0.sin 0+p. –cotg 0 cos 0 sin b−sin 0 sin b

Example. Let the given triangle ABC have its three angles, C=70°, A=60°, B=50°, and its sides c=2000 feet, b=163042, a=184321; and the angles observed at P be respectively, a (subtended by a) =11° 9′ 10′′ and ẞ (subtended by b)=23° 40′ o'

Then 0+0=360° — 70° — 34° 49′ 10′′=255° 10′ 50′′.

The calculation with the formula for the angle CAP is

−cotg 8o°, or +cotg Ioo° I'246 3156

And as the last term will be negative, involving a cosine of an angle greater than 180°, and less than 270; and as the whole expression also becomes negative, the angle 0.will be 100°, as calculated and shown by sign,

and PC b.sin 100°. cosec 23° 40′ = 4000 feet. PA and PB may hence be easily obtained.

The calculation of latitudes, longitudes, and azimuths. (See page 54.)

Example. From the English G. T. Survey, ii. p. 88.
In the triangle ABC,

Given, L the observed latitude of A = 50° 37′ 7.3 North
the observed longitude of B=1° 11′ 36′′
M, the azimuth of B, as seen from A=84° 54′ 52.5 NW.
and d, or the arc AB=314 397'5 feet.

Required the latitude of B, the longitude of A, and the azimuth of A as seen from B.

In this case R, the mean radius of curvature at A, is assumed to be=20 963 000, and 1+ Q2 cos2 L=1'002 57 and the logarithmic calculation for difference of latitude is

log (1+ Q2 cos2 L)=0·001 1600

log d

=5'497-3548

log cos M

log R

7.321 4534

log sin I"

L-L'=-274′′·S7+28"·10=—4′ 6′′·77

L'=50° 37' 7'3+4′ 6′′·77=50° 41′ 14′′·07.

and the logarithmic calculation for difference of longitude is

d sin M sec L'

R sin I"