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three sides; an expression which will be found much more convenient for the application of logarithms than that contained in Art. 68.
putting в in the place of a in the formula for sina (formula (7) of the last article) we have
substituting for cos в in this, its value in (1), we have
reducing the terms under the radical to a common denominator, collecting them together and placing the common factor R2 outside the parenthesis, there results
but the difference of the squares of two quantities is equal to the product of their sum and difference, hence
b2(ac)2=(b + a—c) (b— a + c)
substituting the second member of this in place of the first in the preceding equation, and separating the 4 of the denominator into two factors 2 × 2, we have
representing b+a+c the sum of the three sides of the tri
angle by s, the second members of the two last equalities become
As the angles have each the same relations to the corresponding sides of a triangle, the same formula by a proper modification will furnish the values of the angles A and c.
It may be expressed in ordinary language thus, the sine of half either angle of a triangle is equal to radius into the square root of half the sum of the three sides minus one of the adjacent sides, into half the sum minus the other adjacent side, divided by the rectangle of the adjacent sides. To apply this to an
Let there be three places at distances from each other respectively of 50, 60 and 70 miles. Required the angle under which two roads must depart from that which is 60 and 70 distant from the other two, in the direction of these last. 60 and 70 will be the sides of a triangle adjacent the required angle, and 50 the side opposite then
log. of 20
log. of 30
ar. comp. of log. of 60
= = 8.221849
ar. comp. of log. of 70 =
sum rejecting twice 10-1.154902
Divide this sum by 2 for v, quot. ——0.5+77451
*Add 10 to multiply by R sum=9.577451=log. sin of 1 the required angle.
From the tables we find the angle to be 22° 12' 28"
44° 24' 56" 74. Before treating of the only remaining case in the solution of triangles, it will be convenient to demonstrate some additional general formulæ which shall present certain important relations of the trigonometrical lines of two different arcs; which formulæ are of frequent use in the higher analysis, are employed in the subsequent parts of the work, and will be immediately of service in deriving a formula for the last case of plane trigonometry which we have to consider.
Add together equations (5) and (7) of (Art. 69) which express the values of sin (a + b) and sin (a b) and the resulting equation, cancelling the second terms of the second members which are similar with contrary signs, is
2 sin (a+b)+sin (a – b)=
sin a cos b (1) make
a+b=p and a-b=9 add these last two equations; there results
2a=p+q whence a=3(P+9) subtracting the same equations, the second from the first,
26=P-9 whence b=1(P-9) substituting in equation (1) the value of a +b, a-b, a and b in terms of p and q, that equation becomes
2 sin p + sin q= - sin }(P +q) cos }(P-9)... (2) Which may be translated into ordinary language thus: the sum of the sines of two arcs is equal to 2 divided by rad. into the sine of half the sum into the cosine of half the difference of those arcs.
* If the r be put under the radical, and made b?, the 20 = log. of a?, must be included in the column of numbers added up above ; this will cancel the twice 10 rejected for two arithmitical complements, and the whole operation will consist in adding up the column of four numbers as above, and dividing the sum by 2.
By subtracting the latter of the same equations (5) and (7) of (Art. 69) from the former, reducing similar terms, there results
making the same substitutions as above in equation (1) this last equation becomes
or, the difference of the sines of two arcs is equal to 2 divided by radius into the sine of half their difference into the cosine of half their sum.
Divide equation (2) by equation (4) striking out the com
striking out R in the numerator and denominator the result
which may be expressed in a proportion thus: the sum of the sines of any two arcs is to the difference of their sines as the tangent of half their sum is to the tangent of half their difference.
sin a + sin B : sin a sin B: tan (A + B) : tan 1 (A — B) hence,
a+ba-b: tan (A + B): tan (A — B) That is to say, the sum of two of the sides of a plane triangle is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference.
This proportion is employed when two sides and the included angle of a triangle are given to find the other parts. Since the three angles of every triangle are together equal to two right angles or 180o, subtracting the given included angle from 180°, the remainder is the sum of the two angles opposite the given sides; then substituting for a and b in the above proportion the two given sides, three terms of it are known and the fourth may be found. After which, having half the sum and half the difference of the unknown angles, these angles themselves can be found by adding half the sum to half the difference for the greater, and subtracting half the difference from half the sum for the lesser; when all the parts of the triangle will be known, except one side, which may be found by the proportion the sines of the angles are as the opposite sides. (Art. 64.)