Chap. VIII.

GEOMETRY.

859

If you desire to measure the acute angle A C B, it will only be necessary to subtract 40°fVom 90°, the remainder, 50°, is the angle. To have the length of the hypotenuse, or sideB C, follow the rule given for the second triangle by proportion, placing in the first term10000, in the second the value of the side A B, 1191seconds, and in the third thesecant of the angle C B A, 40°, which is 130540: the quotient will give 15552 lines forthe distance B C, which, reduced into feet, will give 108 feet.

It is required to find the distance A B, having formed an acute-angled triangle, ABC,the angle A C B is supposed to be 79°, the angle C B A, 37°, and the length of the

side C B, 58 feet. To find the angle C A B, add a

together the two known angles, 79° and 37°, and subtracttheir sum, 116°, from 180°, the remainder, 64°, will be thevalue of the angle CAB. To obtain the length of theside A B by proportion, place in the first term the sineof the angle C A B, 64°, opposite the known side, A B,

58 feet; the sine will be 89879 ; in the second term thelength of the side C B 58 feet; in the third term the sine

of the angle A C B, 79°, opposite the required side; this c

sine will be 98162 ; the quotient will give 63 feet for the .side A B, and since there remains 31019, reduce theminto inches and seconds. If the side A C is required, place for the first term the sine ofthe angle C A B, 64°, viz. 89879, for the second term the value of the side C B, 58,and in the third term the sine of the angle C B A, 37°, which is 60181 ; the quotient willgive 38 feet for the length of the side A C: what remains must be reduced into inches andseconds.

Fig. 1086.

Required the distance from A to B, the side A Chaving 46, C B 66, and the angle CAB, 81°. First findthe angle C B A, opposite the known side A C, by pro-portion : place in the first term the length of the otherside, C B, 66, opposite the known angle, C A B, 81°.

In the second term the sine of the angle, C A B, 81 whichis 98768, and in the third term the length of the knownside, A C, 46, opposite the angle sought, C B A ; thequotient, 68838, will be the sine: seek the nearest numberin the table, 68835, which will give 43° 30' for the angleC B A. Add the sums of the two triangles C A B, 81°,and C B A, 43° S<y, their sum, 124° 30', subtracted from180°, leaves 55° SO 7 for the angle A C B. Then find thelength of the side A B, by the rule before given. Place in the first term the sine, 98768,of the angle C A B, 81°, opposite the side C B, 66; in the second term, the value of theknown side, CB, 66, and in the third term, the sine, 82412, of the angle A C B, 55° SO',opposite the required side, AB; the quotient will give 55 for the length of A B, with aremainder.

Measuring Lengths by the calculation of Sines , forming an acute-angled triangle, of whichtwo sides and the comprised angle are known. Required the distance from A to B, byforming an acute-angled triangle ABC, the side of which A C is 63 chains, B C 30 chains,and the angle A C B comprised by these two sides 69°. To find the two angles CAB andCB A, add together the two sides, AC 60 chains, andB C 30 chains; their sum will be 90°; subtract theshort side, B C, 30, from the long side, A C, 60; thereremains 30 : subtract from 180° the known angle A CB69°; the remainder, 111°, will be the value of the twounknown angles, CAB and CBA, of which take thehalf, 55° 30'. Place in the first term the sum of the twosides, AC 60 and BC 30 = 90; in the second, the dif-ference of the two sides 30, and in the third term thetangent of half the two unknown angles 55° 30', which

is 145500; the quotient will give another tangent, 48500, the nearest number to whichin the tables, 48413, will give 25° 50', to which add half the value of the unknownangle 55° 30', and it will give 81° 20/ for the greatest angle of the two, viz. C B A, but ifyou subtract the 25° 50' from 55° 30', it will give 29° 40' for the angle CAB. To findthe side A B by proportion, place in the first term the sine of the angle CAB 29° 40'opposite the side B C, 30 chains; this sine will be 49495; in the second term the sideB C 30 chains, and in the third term the sum of the angle A C B 69° opposite the siderequired ; this sine will be 93358 ; the quotient will give 56 chains for A B. If youwould reduce the remainder 29020 chains into feet, you will have 14 feet more, whichgives 54 chains 14 feet for the distance from A to B.

Measuring Lengths by forming an Obtuse-Angled Triangle , of which two angles and the