To make an obtufe Angle equal to 102° 20.

Draw the Line B C, and upon B defcribe an Arch as before, with a Chord of 60°, to cut B C in E, on that Arch fet off E G equal to 90°, from G fet off 12° 20' towards F: by B and F draw the Line BF D, and it is done; for the Angle CBD is an Obtufe Angle containing 102° 20′ as was required.

In like Manner an Angle may be made, which fhall contain any Number of Degrees required.

The Angles and Hypothenuse of a Right-Angled Triangle given, to find either of the Legs.

Given the Hypothenufe 121 Leagues, the Angle oppofite to the Bafe 54° 30, and confequently the other Angle 35° 30′; the Bafe and Perpendicular are required.

Draw the Lines C B and at C make an Angle equal to 35° 30%, by drawing the Line CA, take 121 Leagues in your Compaffes (from any convenient Scale of equal Parts) and fet that off from C to A; from A let fall the Perpendicular A B, C

to cut the Line C B, and it is

done, for A B being measured

30

121

70:25

B

198.5

on the fame Scale, will be 70.25 Leagues, and C B 98.5 as was required.

The Angles and one Leg of a Right-Angled Triangle being given, to find the Hypothenufe and other Leg.

The Angle A CB 33° 15', the Leg B C 274 Miles, given to find the Hypothenufe and the other Leg A B.

The Hypothenuse and one Leg given, to find the Angles and other Leg.

The Leg A B 69, the Hypothenufe 150, given to find the Angles and Leg B C.

Draw the Bafe B C, upon B erect the Perpendicular B A, upon which fet off 69, take 150 in your Compaffes, and with one Foot on A, lay the other on the Base, as at C, from C to A draw a Line, and it is done; for the Angle BCA being measured by a

133

Chord of 60° will be 27° 23', which being fubtracted from 90° leaves the Angle A 62° 37', and the Leg BC 133, as was required.

The Legs given to find the Angles and Hypothenufe.

The Leg A B 980, B C 690, given to find the Angle B A C or AC B, and the Hypothenufe A C.

PROBLEM X.

Two Angles and one Side of an Oblique-angled Triangle given, to find either of the other Legs.

The Angle BDC 101° 25', and CBD 44° 42', and the Leg BC 76 given, to find the Sides CD and B D.

Draw the Line BC equal to 76, on B describe an Arch, and make the Angle CBD 44° 42'; add the Angles B and D together, that Sum fubtracted from 180° (the Sum of the Three Angles of every Triangle) leaves the Angle C 33°53′;

BX44.42

D

101.25

761

upon C describe an Arch, and make the Angle B C D equal to 33° 53' by drawing C D, and it is done; for the fide B D will be 43.2. and D C 54.5, which was required.

Two Sides and an Angle oppofite to one of them given, to find the other oppofite Angle and the third Side.

The Side B C 106, B D 65 Miles, and the Angle С 31° 49′ given, to find the Angle D and Side CD.

Draw the Line B C equal to 106, at C make an Angle of 31° 49' by drawing C D, take 65 in your Compaffes, and with one Foot in B, lay the other upon the Line C D in D; draw the

65

706

31 49

Line B D, and it is done; for the Angle D will be 120° 43', the Angle B 27° 28', and the Side DC 56.9, as was required.

Two Sides and their contained Angle, given, to find either of the other Angles, and the third Side.

D

The Side B C 109, BD 76 Leagues, and Angle CBD 101° 30′ given, to find the Angle B D C, or BCD, and the Side CD. Draw the Line B C 109, and BD, fo as to make an Angle with B C of 101° 30′, which make equal to 76; join B C and B D with a Right Line, and it is done; for the Angle D being measured by the Chord of 60°, will be 47° 32', Angle C 30° 58', and the Side DC 144.8, as was required.

B

101.30

PROBLEM

109

XIII:

The three Sides given, to find the Angles.

The Sides B C 105, BD 85, and C D 50 Miles, given, to find the Angles BDC, BCD, or C B D.

Draw the Line B C equal to 105, take CD 50 in your Compaffes, and with one Foot in C defcribe an Arch as at D, then take B D 85 in your Compafles, and with one Foot in B cut the former Arch in D, join BD, B

8.5

105

and DC, and it is done; for the Angle B, being measured, will be found 28°4', Angle C 53° 7', which being added together, is 8111', their Sum fubtracted from 180, leaves Angle D 98° 49' as was required.

PLANE

PLANE TRIGONOMETRY,

OR THЕ

DOCTRINE of PLANE TRIANGLES,

EACHES the Menfuration of Triangles, by comparing the

T Sides and Angles together by known Analogies; whereby

Three Things being given, a Fourth may be found on Condition that one of them be a Side; in which, Right Lines are applied to the Arches of a Circle, that the Proportion they bear to the Sides of a Plane Triangle may be found.

The Right Lines applied to a Circle are Chords, Sines, Tangents, and Secants.

1.

A Chord or Subtenfe of an Arch, is a Right Line that divides a Circle into two un

Τ

Tangent

D

C

Radius R

A

Versed Sing

P

tween the Right Sine

E

and the Arch; wherefore RA is the verfed Sine of the Arch S A, and RD the verfed Sine of the Arch S D.

IV.

The Tangent of an Ah is a Right Line drawn perpendicularly from one End of the Diameter, as AT.