The quadrilateral ABCD= AC × 1 ( p + q).
The triangle BCE=CE × √ p; whence p

2. BCE
CE

:

The similar triangles EDD' and BEB' give p q :: BE: DE, whence DE 2. BCE X DE

B-b=B (1 - cos. A).

B. cos. A;
Since 1

b

(382) Let B= the measured inclined length, 6 this length reduced to a horizontal plane, and A the angle which the measured base makes with the horizon. Then 6 and the excess of B over b, i. e., cos. A (sin. A) [Trigonometry, Art. (9)], we have B-6=2 B (sin. A)2. Substituting for sin. †. A, its † approximate equivalent, † A × sin. 1′ [Trigonometry, Art. (5)], we obtain B — b = 2 B (A × sin. 1')2 = (sin. 1′)2. A2. B, = 0.00000004231 A2 B. By logarithms, log. (B— 6) = 2.626422 + 2 log. A+ log. B. The greater precision of this calculation than that of bB.. cos. A, arises from the slowness with which the cosines of very small angles increase or decrease in length.

(386) The exterior angle LER=LCR+CLD. Also, LER=LDR + CRD. ...LCR+CLD=LDR+CRD, LDR+CRD-CLD.

and LCR

CD

From the triangle CRD we get sin. CRD = sin. CDR ×

CR

CD

CL

From the triangle CLD we get sin. CLD=sin. LCD ×

As the angles CRD and CLD are very small, these values of the sines may called the values of the arcs which measure the angles, and we shall have

be

The last two terms are expressed in parts of radius, and to have them in seconds, they must be divided by sin. 1" [Trigonometry, Art. (5), Note], which gives the formula in the text. Otherwise, the correction being in parts of radius, may be brought into seconds by multiplying it by the length of the radius in seconds; i. e., 180° X 60 X 60

3.14159, &c.

=206264′′.80625 [Trigonometry, Art. (2)].

(391) The triangles AOB, BOC, COD, &c., give the following proportions [Trigonometry, Art. (12), Theorem I.]; AO : OB :: sin. (2) : sin. (1); OB: OC :: sin. (4) : sin. (3); OC: OD:: sin. (6): sin. (5); and so on around the polygon. Multiplying together the corresponding terms of all the proportions, the sides will all be cancelled, and there will result

1:1:: sin. (2) × sin. (4) × sin. (6) × sin. (8) × sin. (10) × sin. (12) × sin. (14): sin. (1) × sin. (3) × sin. (5) × sin. (7) X siu. (9) X sin. (11) × sin, (13). Hence the equality of the last two terms of the proportion.

c. sin S' sin. U — a. sin. S (sin. T. cos. U

In the quadrilateral ABCS, we have

BCS : 360° — ASB — BSC-ABC-BAS; or V = 360° —S—S' — B — U.

Let T360° —S—S' — B, and we have V— T— U.

Substituting this value of V, in equation [3], we get [Trig., Art. (8)],

Dividing by sin. U, we get

[4]

cos. T. sin. U)=0.

Separating this expression into two parts, and cancelling, we get

Having found U, equation [4] gives V; and either [1] or [2] gives SB; and SA and SC are then given by the familiar "Sine proportion" [Trig, Art. (12)].

DEMONSTRATIONS FOR PART VII.

CP

(403) If APC be a right angle,

=cos. CAB [Trigonometry, Art. (4)].

CA

(405) AC=PC. tan. APC; and CB= PC. tan. BPC [Trigonometry, Art. (4)]. Hence AC: CB: tan. APC : tan. BPC; and AC: AC+ CB:: tan. APC: tan. APC + tan. BPC.

Consequently, since AC + CB = AB, ACAB.

tan. APC

tan. APC+tan. BPC'

(414) The equal and supplementary angles formed prove the operation.

(421) In Fig. 285, CA: EG:: AB: GB.
AB GB GB. Hence, observing that
GB (CA — EG)
AG:

EG

By

By "division," CA-EG: EG:: AB GB AG, we shall have

(423) Art. (12), Theorem III., [Trigonometry, Appendix A,] gives; a2 + b2 — c2 cos. C= ; or c2 a2 + b2 — 2 ab. cos. C. This becomes [Trig., Art. 2 ab (6)], K being the supplement of C, c2 = a2 + b2 + 2 ab. cos. K. The series [Trig, Art. (5)] for the length of a cosine, gives, taking only its first two terms, since K is very small, cos. K=1— K2. Hence,

o2 = a2 + b2 + 2 ab — ab K2 = (a + b)o— ab K2 = (a + b)2 ( 1 — ab))

(a + b)2

Developing the quantity under the radical sign by the binomial theorem, and neglecting the terms after the second, it becomes

Substituting for K minutes, K. sin. 1′ [Trig., Art. (5)], and performing the multiplication by a + b, we obtain

c = a + b

ab K2. (sin. 1')3 ̧ Now (sin. 1')2=0.0000000423079;
2 (a + b)

whence the formula in the text, ca+b.
= a + b — 0.000000042308 X

ab K2

a + b

(430) In the triangle ABC, designate the angles as A, B, C; and the sides opposite to them as a, b, c. Let CD=d. The triangle BCD gives [Trig, Art. (12),

Theorem I], a=d

sin. BDC
sin, CBD

The triangle ACD similarly gives b—d ·

sin. ADC sin. CAD'

In the triangle ABC, we have [Trig., Art. (12), Theorem II.], tan. † (A — B) : cot. 1 C :: a —b: a+b;

Let K be an auxiliary angle, such that ba.tan. K; whence tan. K=

Dividing the second member of equation [1], above and below, by a, and substitu

Since tan. 45° — 1, we

may substitute it for 1 in the preceding equation, and ban. 45° tan. K

we get tan. † (A — B) =

cot.C.

tan. 45°+tan. K

From the expression for the tangent of the difference of two arcs [Trig., Art. (8)], the preceding fraction reduces to tan. (45° — K); and thé equation becomes tan. † (A — B) = tan. (45° — K). cot. § C.

b

α

[2]

In the equation tan. K= substitute the values of b and a from the formulas at the beginning of this investigation.

This gives

tan. K = d.

sin. ADC
sin. CAD

→d.

sin. BDC sin. ADC. sin. CBD
sin. CBD sin. CAD. sin. BDC'

(A — B) is then obtained by equation [2]; (A+B) is the supplement of C; therefore the angle A is known.

Then

c=AB=

a. sin. C
sin. A

d. sin. BDC. sin. ACB
sin. CBD. sin. CAB

The use of the auxiliary angle K, avoids the calculation of the sides a and b.. (434) In the figure on page 292, produce AD to some point F. The exterior angles, EBC=A+P; ECD=A+Q; EDF = A + R. The triangle ABE

Equating these two values of the same ratio, we get

To solve this equation of the 2d degree, with reference to x, make

x = − † (a + b) ± √ [ (a - b). tan. Kab + (a+b)*]
= (a + b) ± √ [ (a - b)2-tan.2 K + (a - b)2]
· }
—
=—† (a + b) ±(a - b) √ (tan.2 K + 1).

DEMONSTRATIONS FOR PART XI.

(493) The content being given, and the length to be n times the breadth;

Breadth Xn times breadth = content; whence, Breadth = √(content).

Given the content = c, and the difference of the length and breadth=d; to find the length 7, and the breadth b. We have 1 x bc; and l-bd. From these two equations we get l=}d+}√ (d2 + 4 c).

Given the contentc, and the sum of the length and breadth =s; to find l and b. We have 1 x be; and l+b=s; whence we get 7=1s+1√ (s2 — 4 c).

(494) The first rule is a consequence of the area of a triangle being the product of its height by half its base.

To get the second rule, call the height h; then the base =mh; and the area

=}h × mh; whence h= √ (2 X area).

For the equilateral triangle, calling its side e, the formula for the area of a triangle √ [ († s ) ( z s a) († s — b) († sc)] reduces to e2√3. Hence e2,

1.5197✓ area.

area

(495) By Art. (65), Note, . AB X BC X sin. B= content of ABC; whence,

(497) The blocks, including half of the streets and avenues around them, are 900 X 260 234000 square feet. This area gives 64 lots; then an acre, or 43560

feet, would give not quite 12 lots.

(502) The parallelogram ABDC being double the triangle ABC, the proof for Art. (495), slightly modified, applies here.

(504) Produce BC and AD to meet in E. By similar triangles,

ABE: DCE:: AB2: DC2.

ABE-DCE: ABE:: AB2 — DC2 : AB2.

Now ABE-DCE=ABCD; also, by Art. (65), Note,

Fig. 346, bis.

Multiplying extremes and means, cancelling, transposing, and extracting the square