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AC x sin. Ax sin. C and BD=AB x sin. A =

sin. B

sin. A x sin. C whence the area = į AC'

sin. B

(123.) Cor. If the angles ABC, ACB be equal, the area will be=į AC* x sin. A. If the angles A and

sin’. A C are equal, the area will be = 1 AC

sin. B.
But since the angle B= 180° - 2 A,
sin. B=sin. 2 A=2 sin. Ax cos. A,

sin?. A
whence the area = ACP

sin. Ax cos. A
sin. A
= 1 AC*X = \ ACox tang. A.

cos. A


(124.) Given the three angles, and the altitude of a triangle ; to find its area. Since (122) BD x sin. B= AC x sin. Ax sin. C, and 2S

:. 2Sx sin. Ax sin. C=BD x sin. B, BD

sin. B or S

AC =

= {BD x sin. A x sin. C

Prop. XXV.

(125.) Given the three sides of a triangle ; to find its area.

The area of the triangle ABC=area ADC+area BDC = AEX DE+MEX DE, (BM being parallel to DC) = AMX DE.







or s

But by similar triangles ADE, AMB,

AE : ED :: AM : MB, .:. AEX AM: EDXAM :: EDX AM : EDX MB, i.e. the area of the triangle is a mean proportional between AEX AM and EDx MB.

Now (92.) EDx MB=(P- AB).(P- AC),

and AEX AM= AL X AK = Px (P-BC),
.:. the area) =(P- AB).(P– AC). P.(P-BC)

{P.(P- AB).(P-AC).(P-BC)}. (126.) If AB=BC= AC, Sri AB. 3, the area of an equilateral triangle.

If BC=CA,S=* AB./{(2BC+BA).(2 BC-BA)}, the area of an isosceles triangle.

(127.) To construct the trigonometrical canon.

It has been proved (103) that 2 cos?. A=1 +cos. 2 A; and therefore if the cosine of any arc be known, the cosine of half that arc may be determined. Now the sine of 30° has been found to be = *, (R=1), and the

13 cosine =

if then in the formula cos. A = V{t:(1+cos. 2 A)}, A=15°, cos. 15° may be determined. In the same manner from cos. 15°, the cos. To 30' may be deduced; and so on, till after 11 divisions, cos. 52" 44" 3iv 45' is found; from which the sine of this arc may be determined. But from the nature of the circle, when the arc is very small, the ratio of the arc to the sine approaches nearly to a ratio of equality, ::52" 44" 3iv -45° : 1' :: the sine of the former arc : the sine of i'; which :. may be determined.


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The sine and cosine of l' being ascertained, the sines of 2', 3, 4', &c. may be determined (109) by making n=1, 2, 3, &c. and the cosines from (58). In this manner the sines and cosines of arcs as far as 30° may be computed. When the arc exceeds 30°, the sines may be computed by Art. (107), and the cosines as before, till the arc is 45°. And since the sine of an arc is equal to the cosine of its complement, the sines and cosines of arcs as far as 90° are determined. Also since the sines and cosines of arcs are equal to the sines and cosines of their supplements; the sines and cosines of all arcs up to 180° are known.

sin. A Since tang. A =

the tangents of all arcs may be computed. When however they exceed 45°, they are more readily computed from (115) by addition. And the tangent of an arc being equal to the tangent of its supplement, the tangents of all arcs may be determined.

Hence also the cotangents (45).

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Also the sec. A =

and .:. the cosines being

cos. A'

known, the secants may be determined. And the secants being known, the cosecants are also determined.

Also the versed sine (=1 #cosine) may be determined.

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Page 17. line 3, for finite read indefinite.

19. 3, for ACB read ACE.
37. 20, for parts read points.
38. 20, for centre read centres.
45. 14, for circle read circles.
46. In the figure join DE.

18, for circumference read circumferences.
63. 6 from the bottom, for 63 read 62.
64. 13, for 63, read 62.

line 4 from the bottom, for OC read OB.
76. In the figure join DB.

11, for made by the perpendiculars read cut off

by the perpendiculars.
90. In the figure join BC.
108. 7, for AFC read AFG.
109. 6, for A read B.
112. 12, for Bl read BL.
118. In the figure join KG.
122. last line, for a side produced read a side and a side

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7, for AG read AC.
149. line 3 from the bottom, for rectangle read rectangles.
164. 16, for 2 read 62.
217. Join CG in the figure.
245. Fig. 2, join DB.
252. 7, for C read o.

line 10, for AD read CD.


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