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Suppose the three sides of a triangular plåt of ground are to be 50, 60 and 70 yards, under what angle must the first two be laid out.
cos required angle = R X
2500 + 3600
2 × 50 × 60
if we make R = 1; hence angle required. This angle will be found from the table of natural sines and cosines, to be 78° 27' 47".*
or 20000 is the nat. cosine of the
This case might also be solved with the table of log. sines, &c. by subtracting the log. of the whole denominator from that of the whole numerator, and adding 10 the log. of R, or adding at once the arith. comp. of the denominator, the logs. of the numerator and R, rejecting 10 from the sum; in either case the result would be log. cosine of the angle required. The solution is left as an exercise for the student.
69. We proceed now to demonstrate some formulæ which express relations between the different trigonometrical lines of the same arc, and between the trigonometrical lines of two different arcs. They are introduced here because necessary for the solution of the few cases of plane triangles which remain. We shall first derive formulæ by means of which, when the sines and cosines of two arcs are known, the sine and cosine of their sum or difference may be found. Thus if the sine and cosine of 30° and also those of 20° be given, those of 50° 30° + 20° or of 10° 30° -20° may be found.
*The seconds are found as follows: Take the difference between the two cosines next greater and next less than yours, and also the difference between yours and the next greater, multiply the latter difference by 60 and divide the product by the former difference; the quotient will be the seconds sought. The reason appears from the following proportion.
Let A Ma in the diagram, be one of the given arcs, and B M = b, be the other. Then M P = sin α, BI= sin b, since it is the perpendicular let fall from one extremity of the arc b upon the radius which
passes through the other
extremity. CP COS
E KP A
a and c I the distance from the foot of the sine to the centre cos b. A B = AM + M B = a + b BE = sin (a + b) and E c = cos (a+b).
In the similar triangles C P м and с к I we have the proportion
The triangle I B L is similar to C M P, because the two have their sides respectively perpendicular, * hence
CM CP BIBL
R: cos a: sin b: B L
also in the same triangles
CM MP; BI: I L
R sina sin b: IL
Multiplying the means and dividing by the first term of each of these four proportions, we have from the first proportion
* The perpendicular sides are homologous (Geom. B. 4, Prop. 21, Schol.)
Adding the eqations (1) and (3), we have from the addition of the first members
IK BL which is =EL+BL= BE=Sin AB = sin (a+b); adding also the second members, the resulting equation is sin a cos b+ sin b cos a
read thus the sine of the sum of any two arcs is equal to the sine of the first into the cosine of the second plus the sine of the second into the cosine of the first, divided by radius.
If R1 the denominator of the second member disappears.
Subtracting equation (4) from (2) we have for the difference of the first members
ск IL CK EKCE = COS AB = cos (a+b) performing the subtraction also upon the second members there results
read thus the cosine of the sum of any two arcs is equal to the rectangle of their cosines minus the rectangle of their sines, divided by radius.
In the last two formulæ let a 60° and b = 20o then by the first
sin 60° cos 20° + sin 20° cos 60°
1 or 1010 as the case may be
by the second
cos 60° cos 200 --sin 600 sin 200
1 or 101 or winatever R may be We shall derive expressions for sin (a - b) and cos (a−b) or the sine and cosine of the difference of two arcs in terms of the arcs themselves by making, in the formulæ just derived for sin (a + b) and cos (a + b); b – b, observing that cos (-6)= cos b and sin (-6)=-sin b (Art. 27). By this
) substitution there results
sin a cos b - sin b cos a sin(a−b)=
cos a cos 6+ sin a sin b cos (a - b) =
(8) 70. These four formulæ for the sine and cosine of the sum and difference of two arcs should be committed to memory, as they are constantly recurring in trigonometry and in the higher analysis. The four may be expressed in two by the use of the double sign, thus
sin a cos b + sin b cos a sin (a + b)
cos a cos b F sin a sin b cos (a + b) 71. From formula (5) sin(a+b)=&c., we derive one
= much used in the higher analysis for expressing twice an arc in terms of the arc itself, by simply making b=a the result is
• (1) the two terms of the second member becoming the same.
We also get an analogous expression for the cosine of twice an arc by making b = a in formula (6) of the last article cos (a + b) = &c. This expression is
2 sin a cos a sin 2a =
cos 2 a =
(2) Thus knowing the sine and cosine of 200, these last two formulæ would give us the sine and cosine of 40°.
These two formulæ may be modified so as to express the sine and cosine of an arc in terms of half the arc under which last form they are much used. This is accomplished by making a = a, which is legitimate since a is supposed to have no particular value; then 2 a becomes a and we have from (1)
2 sina cosa
. 72. By means of this last, and a very simple formula depending upon the well known property of the right angled triangle, that the square of the hypothenuse is equal to the sum of the squares of the other two sides, a formula expressing the value of the sine of half an arc in terms of the arc itself may be obtained.
The formula depending upon the property of the right angled triangle, will be found by referring to the last diagram, in which the triangle CPM is right angled at P, whence (Geom. B. 4, Prop. 11.)
clearing equation (4) of the denominator and changing the order of the members it becomes
2 sin2 a= R2. R COS a
dividing by 2 and taking the square root of both members of
this last equation, we have the formula required.
sina = √ R2.
R cos a
73. We resume the solution of triangles, having now a formula, by means of which we shall be able to derive an expression for one of the angles of a triangle in terms of the