. Reports upon the survey of the boundary between the territory of the United States and the possessions of Great Britain from the Lake of the woods to the summit of the Rocky mountains . - ,multiplied by distance in miles of point A Irom point ofdeparture. 381. 65.•. angle F A M = 90° + 59. 75 x —gij— = 90° + 4 45 = 90° 4 45The perpendicular at A, i. e. A P, makes with A K anangle = 90° - M A K = C0° - (90° 4 45 - 17° 12 01. 5.)= 17° 07 16. 5K P = A H sin 17° 7 16. 5. -,, . ^ sin A B KAK = AB,. = sin A K. B i sin 94^ 48: sin 23 51 : 271. 49 ch. ..KP = ? sin 94° 4851.
. Reports upon the survey of the boundary between the territory of the United States and the possessions of Great Britain from the Lake of the woods to the summit of the Rocky mountains . - ,multiplied by distance in miles of point A Irom point ofdeparture. 381. 65.•. angle F A M = 90° + 59. 75 x —gij— = 90° + 4 45 = 90° 4 45The perpendicular at A, i. e. A P, makes with A K anangle = 90° - M A K = C0° - (90° 4 45 - 17° 12 01. 5.)= 17° 07 16. 5K P = A H sin 17° 7 16. 5. -,, . ^ sin A B KAK = AB,. = sin A K. B i sin 94^ 48: sin 23 51 : 271. 49 ch. ..KP = ? sin 94° 4851. 3 sin 15° 2351AP = A Kcos 17° 7 10 sin 17° 7 16. 5 = eh. 5 = 259. 4(i eh. P p = otfset to parallel due to distance from A consid*ered as initial point = 8. 05 It. K is north ot A 79. 92.) ch. X 66 - 8. 05 ft. = 5,275. 05 8. 05 ft. = 5,267 find position of A with respect to 49° parallel: Feet. Initial point of tangent north of 49° parallel Ofiset from A due to distance, A north of 49° parallel. .65 .•. K is norlli of English of 49°. 5,267 ft. 4- .05 ft. = 5, of K from initial point along is ch. -|-259. 40 ch. = ch 250 EEPOET OP THE CHIEF ASTEONOMEE. 251 British Tangent No. 20— STATION EEKOR. By Captain Featberstonhanghs triangnlation K is Dorth of Ms determination of 49° pnrallel = 5 ^3S c5 Byliientenant Galwey S^ianorth of 49= parallel , .jseries Astron
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