862 



THE IRRIGATION AGE. 



THE PRIMER OF HYDRAULICS* 



By FREDERICK A. SMITH, C. E. 



7. Solution of Triangles. 



In Figure 39 let ABC be any triangle the length of the 

 sides being o, b and c respectively and the angles being a , $ 

 and -y respectively the following relations obtain : 



a : b : c sin a : sin B : sin y', this 

 means that in any triangle any two sides 

 are to each other as the sines of the 

 opposite angles. This rule is used to 

 compute triangles when one side and 

 two angles are given or when two sides 

 and one opposite angle are given. 



If two sides and the enclosed angle 

 are given then the following formula is 

 used : 



a + B a B 



a. + b : a & = tg : tg . 



2 2 



This means: the sum of two sides in a triangle is to its 

 difference as the tangent of YI the sum of the opposite angles 

 is to the tangent of half the difference of the opposite angles. 

 This is the second general formula. 



The third general formula is expressed thus : 

 c' + b" a" 



2 be 



which means the cosinus of any angle is found by adding the 

 squares of the two enclosing sides, subtract the square of the 

 opposite side and divide by twice the product of the enclosing 

 sides. 



Problem 1. In Figure 39 let o = 8 ft., b = 11 ft and g 

 - 70. Find the side c and angles a and y. 



Solution, a : b = sin a : sin B ; 



b sin a = a sin @ and 



sin a = a sin $/b 



Substitute the given quantities : 



Sin a = 8 X .9511 -H 11 = .6917. 



The angle corresponding to the sine .6917 in the tables 

 is 43 46'. 



The angle y = 180 a B = 180 -- 72 - - 43 46' 

 = 64 14'. 



The side c is found as follows : 



c : b = sin -y : sin Q. 



c = b sin y/ sin /J; substitute known quantities: 



c = 11 X .9006 -^ .9511 = 10.416 ft. 



Problem 2. In triangle ABC, Fig. 39, side a = 8 ft, 6 

 = 11 ft. and angle y = 64 14'; find a, B and c. 



a + B a B 



Solution, a +- b : a b = tg : tg 



a-B 



the unknown term is tg - ; 

 means and extremes : 2 



a-B 

 (a + b) (tg - ) = ( &) (tg 



a B 2 a B 



form the product of the 



B 



-) ; hence : 



tg (- -) = (a b) (tg 



-) / (a + b); 



Substitute the known quantities and remember if y is 

 subtracted from 180 it leaves a + B ' 



180 64 14' = 115 46' = a + /J; then divide 115 



46' by 2 = 57 53' 



, also since b = 11 and = 8, 



angle B w >" ' )e larger than angle a and the formula should 

 be modified as follows: 



a B ~ a 



tg (- -) = (b a) (tg- -)/(b + a) 



Hence: tg ( ) = (11 8) (1.5931) / (11 + 8) 

 2 



B-a 



tg ( ) = 3 X 1.5931 -5- 19 = .2515. 



2 



The angle corresponding to this tangent is found in 

 the table of tangents to be: 14 7', thus: 

 B + a 



= 57 53'; then B + a = 115 46' 



= 14 



- a = 28 14' 



Add the two values together : 



20= 114 or B = 72 ; 

 then subtract B a from B + a : 



2 a = 87 32' or a = 43 46'. 



It will be seen if the 3 angles are added together the 

 result is exactly 180, which proves that the work is correct. 



Problem 3. The 3 sides in a triangle are 28, 31 and 42 ft., 

 find the 3 angles. 



Solution. In Fig. 40 let a = 

 28 ft., b = 31 and c = 42 ft, then by 

 using the third general formula we 

 have : 



cos a = 31x31 + 42x42 28x 

 28 -f- 2x31x42 



Cos a = 1641 -^- 2604 = .63031 

 look in cosine table which gives the 

 corresponding angles 50 56'. 

 to find angle B use same rule : 



cos B = 28x28 + 42x42 31x 

 31 -H 2x28x42 



cos B = 2548 -=- 4704 = .5417 

 look for angle B H1 cosine table, 57 12' 



then angle y = 180 (50 56' 4- 57 12') = 180 108 8' 

 =; 71 52'. 



Fig. 40. 



Article V. Mensuration of Plane Figures. 



Under mensuration is meant the determining of the 

 areas of surfaces and the solid contents of bodies occupying 

 space. In preceding articles reference has been made repeat- 

 edly to methods of finding the area of triangles, rectangles, 

 etc. In the following are shown rules and application for 

 the computing of areas of the more important geometrical 

 surfaces. 



1. Triangles. The area of a triangle is found by multi- 

 plying base by height and divide by 2. (See Art. 3.) When 

 the 3 sides of a triangle are given, when a, b and c are the 

 3 sides and .$* = ^ their sum, the area is found as follows : 



A = V J (* a ) ( J k) (s c). To apply this 

 formula let a = 16 ft., b = 17 ft., c = 13 ft., then 5" = 

 16 + 11 + 13 46 

 = = 23. Substitute values in formula : 



Area = v / 2:i < ~ :! lfi ) ( 23 17 ) ( 23 13 ) 



= V '23x7x6xl0 = V 9660 = 9 / 828 SQ- ft - 



2. Quadrilaterals. The area 

 of a rectangle, square, rhombus 

 and parallelogram is found by 

 multiplying base by height. The 

 area of a trapez is found by cut- 

 ting it into 2 triangles, finding 

 the area of each of the tri- 

 angles and adding them. Then 

 in Fig. 41 the area of trapez 

 ABCD is found by drawing AC, 

 which divides the figure into 

 two triangles and serves as a 

 common base for both triangles ; 



draw BE perpendicular to AC and DF perpendicular to AC. 



AC X BE 



then area of triangle ABC = - , and area of tri- 



angle A CD = 

 AC X BE 



ACXDE 



hence area of ABCD = 



ACXDE 



AC 



(BE + FD.) 



"Copyright by D. H. Anderson. 



3. Polygons. This principle may be extended to any poly- 

 gon ; thus for instance the 7-sided figure shown in Fig. 42 



