Note. In finding the height of objects, to the observed height must be added, that of the observer's eye above the plane of the horizon. Let AB be the obelisk, CB the distance of the observer, and BE the height of his eye; then AE is the part required to be found. In the tri E angle ACE, we have given CE=113, the angle ACE=40°, consequently CAE=(90-40=) 50°, and the angle CEA a right angle; to find AE. Now (Art. 67.) CE: EA :: sin A: sin ACE, ·· EA= CE. sin ACE sin A and log. EA=log. CE+log. sin ACE-log. sin A =2.0530784+9.SOS0675-9.8842540=1.9768919, the natural number, corresponding to which is 94.8182=AE, AE+EB= 94.8182+5=99.8182 feet 99 feet 9 inches the height required. Prob. 2. The angular altitude of a spire, known to be 137 feet high, is 51°; now supposing the height of the observer's eye to be 5 feet, how far is he distant from the foot of the spire? Note. In questions of this kind, the height of the eye must be subtracted from the given height, previous to the operation. Here are given AB=137, EB 5, AE=137-54= 131.5, AED a right angle, and angle ADE=51°, . ang. DAE= (90°-51o) 39°·.· (Art. 67.) DE : EA: sin DAE: sin ADE ·· EA. sin DAE DE= sin ADE 131.5 × sin 390 sin 51° ...log DE-log D 131.5+log. sin 39-log. sin 51° 2.1189258+9.79887189.8905026=2.0272950 .・ DE=106.487 feet=106 feet 5 inches Prob. 3. Wanting to calculate the perpendicular height of a cliff, I took its angular altitude 12° 30', but after measuring 950 yards in a direct line towards its base, I was unexpectedly stopped by a river; here however I again took its altitude 69° 30'; required the height of the cliff, and my distance from the centre of its base? Let A be the first station, B the second, C the summit of the cliff, and D its base; then AB=950, the angle A=12° 30′, angle ABC (180° −69° 30′=) 110° B C 30′ ·.· ang ACB=(180—12° 30′+110° 30′=180° — 123′=) 57°; in the triangle ABC we have the side AB and the three angles given, to find BC. Now (Art. 67.) AB: BC :: sin AB. sin A ACB: sin A·: BC=" sin ACB and log BC=log AB+log. sin A-log. sin ACB= (log 950+log. sin 12° 30'-log. sin 57°) 2.9777236+9.3353368-9.9235914=2.3894690, ·.· BC =245.171; having found BC, there is given in the triangle BCD the right angle BDC, the angle CBD=69° 30′, the angle BCD (90°-69° 30′) 20° 30′ and the side BC=245.171, BC.sin BCD (Art. 63.) BC: BD :: rad: sin BCD, ·. BD= rad 85.8608 yards. Also (Art. 63.) BC: CD :: rad: sin CBD; BC.sin CBD · CD=. rad =229.645 yards. Prob. 4. Two persons, situated at A and B, distant 24 miles, observed a bright spot in a thunder cloud at the same instant; its altitude at A was 46°, and at B 63° 30′; required its perpendicular height from the earth? First. Angle ACB=(180°—46°+63° 30′) 70° 30′, then (Art. 67.) AB: BC:: sin ACB: sin BAC, BC=2 AB. sin BAC sin ACB B Ac 2.13612 miles. Wherefore in the right angled triangle BCD, BC: CD rad: sin CBD (Art. 63.), ··· CD= = BC. sin CBD rad Prob. 5. Two towns, A and B, are invisible and inaccessible to each other, by reason of an impassible mountain, situated between them; but both of them are visible and accessible from the point C, viz. A bears NE from C distance 3 miles, and B bears Nb W from C distance 5 miles; required the bearings and distance of A and B from each other? (61° 52′ 30′′+27° 1′ 57′′=) 88° 54′ 27", and angle B (61° 52′ 30′′27° 1′ 57′′) 34° 50′ 33′′; next, (Art. 67.) CA: AB:: sin B: sin C, C.4. sin C :AB= sin B =4.36606 miles. Lastly, through the centre C draw ab parallel to AB, and measure the circumference Na, and it will be found to contain 46° 6', which, by referring to the table (Art. 87.), will be found to answer to the N W point nearly; that is, B bears from A NW 1o 6' W distance (4.36606 miles) 4 miles 3 furlongs nearly. VOL. II. Ff Prob. 6. A general arriving with his army on the bank of a river is desirous of crossing it, but there are two of the enemy's fortresses, A and B, on the opposite shore, and he wishes to know their bearings and distance from each other; for this purpose two stations C and D are chosen close to the river side, C being directly east, from D at mile distance; at C the angles are as follow, viz. ACB=68°, BCD=32°; at D the angles are ADB=62°, ADC=64°. Now suppose he crosses directly from the point D, required the bearings and distance of A and B from each other; the width of the river at the point of crossing, and the distance of the point where he proposes to land from A and B? First. In the triangle DAC, there are given DC= mile= .75, the angle ADC=64°, DCA=(32°+68°=) 100, and DAC =(180-164) 16°; to find DA. By Art. 67. DC: DA :: DCX sin DCA .75 X sin 100° sin DAC: sin DCA, . DA= 2.67963 miles. Secondly. In the triangle BDC, there are given DC=.75, BDC=(62°+64°=) 126°, DCB=32°, and DBC= (180°— 126°+32°=) 22°, to find BD. By Art. 67. DC: BD :: sin DCX sin DCB .75 x sin 32° Thirdly. In the triangle BDA there are given D.4= 2.67963, BD=1.06095, and the included angle ADB=62°; to DBA+BAD 2 find the angles DBA, BAD, and the side BA. Now 180°-62° 2 =59°=half the sum of the angles DBA, BAD at the base; also AD+DB=2.67963+1.06095=3.74058=sum of the sides, and AD-DB=2.67963-1.06095=1.61868= diff. of the sides. But (Art. 72.) AD+DB: AD-BD :: tan DBA+BAD DBA-BAD ; that is, 3.74058: 1.61868 :: 2 1.61868 x tan 59° 2 tan 59° : : tan 3.74058 =tan 35° 42′ 5′′-half the difference of the angles DBA, BAD at the base. (Art. 69.) { Also (Art. 67.) BDxsin BDA sin BAD = 59°+35° 42′ 5′′=94° 42′ 5′′ the angle DBA. 59°-35° 42′ 5′′=23° 17′ 55′′"the angle BAD. BD : BA :: sin BAD : sin BDA, ·: BA= 1.06095 x sin 62° sin 23° 17′ 55′′ =2.36842 miles. Fourthly. In the triangle DBE there are given the angle E a right angle DBE=(180°-DBA=180°—94° 42′ 5′′=) 85° 17′ 55′′, the angle BDE=(90°—DBE=90°—85° 17′ 55′′=) 4° 42′ 5′′, and the side BD=1.06095; to find the sides BE and DE. By Art. 63. DB: BE:: rad: sin BDE, BE= |