The English academy, or, A brief introduction to the seven liberal arts grammar, arithmetick, geometrie, musick, astronomie, rhetorick & logic : to which is added the necessary arts and mysteries of navigation, dyaling, surveying, mensuration, gauging & fortification, practically laid down in all their material points and particulars, highly approved to be known by the ingenious, and as such are desirous to profit, or render themselves accomplished : chiefly intended for the instruction of young scholars, who are acquainted with no other than their native language, but may also be very useful to other persons that have made some progress in the studies of the said arts / by John Newton.

CHAP. III. Of the Solution or Mensuration of plane Triangles.

IN the Solution of plane Triangles, the an∣gles only being given, the sides cannot be found, but the reason of the sides only; it is therefore necessary, that one of the sides be known.

2. In all plane Triangles, the three angles are equal to two Right: two Angles therefore being given, the third is also gi∣ven; and one of them being given, the sum of the other two is also given.

3. In a Right angled plane Triangle, one of the Acute Angles being given, the other is also given, it being the Complement of the other to a Quadrant or 90 degrees.

4. In a Right Angled plane Triangle, there are seven Cases, whose Solution shall be shewed in the Problems following.

5. The sides comprehending the Right angle we call the legs, and the side subtend∣ing the Right angle, we call the Hypothenuse.

Problem I. The Legs given, to find an Angle and the Hipothenuse.

In the right angled plane Triangle ABC, let there be given the legs.

• AB 512. To find Hypot. BC.
• AC 384. To find Angl. B and C.

Draw a line at pleasure, as AB, and upon the point A, erect the perpendicular AC, and by help of your Scale of equal parts, set off from A to B, 512, and also from A to C, 384, and draw the line BC, for the Hy∣pothenuse, which being Measured by the scale of equal Parts, will be found to be 640. and by the line of Chords, the angle at B 36.87, whose complement is the angle ACB, 53.13.

By the Tables, the Proportions are,

• 1. A•• . AC ∷ Radius . tang. B.
• 2. s•••• . AC ∷ Radius . BC.

Problem II. The Angles and one Leg given to find the Hy∣pothenuse and the other Leg.

Draw a line at pleasure, as AB, and at Right angles the point A erect the perpen∣dicular AC, and by your scale of equal parts set off from A to B 512, and upon the point B lay down the angle ABC, 36.87. and draw the line BE, till it cut the perpendi∣cular AC, then measure the lines BC and AC, by the scale of equal parts, so shall the one, to wit, BC, be the Hypothenuse, and AC, the other leg inquired.

By the Tables, the Proportions are,

3. Rad. . AB ∷ 4 B . AC.

4. Sine C . AB ∷ Rad. . BC.

Problem III. The Hypothenuse and Oblique Angles given, to find the Legs.

Draw a line at pleasure, as AB, and upon the point B protract one of the angles gi∣ven, suppose the lesser ABC, 36.87. and

draw the line BC, & by your scale of equal parts, number the given Hypothenuse from B to C 640. and from the point C to the line AB, let fall the Perpendicular AC, then is BA one, and CA, the other leg inquired.

By the Tables, the Proportion is,

5. Rad. . BC ∷ sB. AC.

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

Draw a line at pleasure, as AB, and by your scale of equal parts, number from B to A, the quantity of the given leg AB, 512. then upon the point A erect the Perpendicu∣lar AC, and opening your Compasses to the extent of your Hypothenuse BC 640, set one Foot in B, and move the other, till it touch the Perpendicular AC, and there draw BC, so shall AC be the leg inquired, and either Angle may be found by the line of Chords.

By the Tables, the Proportions are,

6. BC . Rad. ∷ AB . Sine C.

7. Rad. . BC ∷ Sine B . AC.

6. Hitherto we have spoken of Right angled plane Triangles, the Problems fol∣lowing concern such as are Oblique.

Problem V. The angles in an Oblique angled plane triangle one side given, to find the other sides.

In the Oblique angled plane Triangle BCD, let there be given the side CB 632, and the Angles DCB 11.07. D. 26.37.

Draw the line CB at pleasure, and by your scale set off from C to B 632, and up∣on those points protract the given Angles DCB 11.07 CBD. 142.56, and draw the lines CD and BD, till they intersect one a∣nother, then shall the one side be CD 865, and the other DB 273.

By the Tables, the Proportion is,

1. Sine BDC . BC ∷ sDCB . DB.

Problem VI. Two sides of an Angle opposite to one of them being given, to find the other Angles and the third side, if it be known whether the Angle op∣posite to the other given side be Acute or Obtuse.

In the Oblique angled plane Triangle BCD, let there be given,

• The Sides CB 632 Ang. D. 26.37.
• The Sides CD 865 Ang. D. 26.37.

Draw the line CD at pleasure, and by your scale set off from C to D, 865, and up∣on the point D protract the Angle CDB 26.37 and draw the line DB, then open your Compasses to the length of the other side CB 632, and setting one foot in C, turn the other about till it touch the line DB, which will be in two places, in the point B or point nearest to D, if the angle opposite to the side CB be Obtuse, but in the point E, or point farthest from D if Acute; accor∣ding therefore to the Species of that Angle, you must draw either the line CB or CE, and then you may measure the other angles and the third side, as hath been shewed.

By the Tables, the Proportion is,

2. CB . Sine D ∷ CD . Sine B.

3. Sine D . CB ∷ Sine C . BD.

Problem VII. Two Sides with the Angle comprehended be∣ing given, to find the other Angles, and the third Side.

In the Oblique angled plain Triangle BCD, let there be given,

• The Sides DC 865 Angle C. 11.07.
• The Sides BC 632 Angle C. 11.07.

Draw a line at pleasure, as DC 865, and by your Scale set off from C to D, 865, then protract the Angle at C 11.07, and draw the line BC, and by your Scale set off from C to B 632, and draw the line BD, and so have you constituted the Triangle BDC, in which you measure the Angles and the third side, as hath been shewed; but to resolve this Problem by the Tables, it is somewhat more troublesome.

1. To find the Angles, the proportion is, 〈 math 〉〈 math 〉.

2. To find the third Side.

Sine D . BC ∷ Sine C . BD.

Problem VIII. The three sides given to find an Angle.

Let the given sides be DC 865. BC 632. and DB 273.

Draw a line at pleasure, as DC, and by your Scale set off from C to D 865, then o∣pen your Compasses to the extent of either of the other sides, and setting one foot of your Compasses in C, with the other draw an Occult arch, then open your Compasses to the extent of your third side, and setting one foot in D, with the other foot describe another Arch cutting the former in the point B, then will the Lines BC and DB, constitute the Triangle, whose Angles may be measured, as hath been already shewed.

To resolve this Problem by numbers, the Proportions are for the Segments of the Base.

As the base is to the sum of the other sides, so is the difference of those sides to the dif∣ference of the Segments of the Base; which being subtracted from the Base, half the re∣mainer will shew where the perpendicular must fall, suppose at F, and constitute the two Right angled Trianges BDF, and FDC, in which we have given the Hypo∣thenuses BD and DC, and the Legs BF and CF, and therefore we may find the

Angles of those Triangles, as hath been shewed in the fourth Problem.

Problem IX. To find the Superficial content of Right lined Figures.

Having shewed the Mensuration of Trian∣gular planes in respect of their sides and Angles, we will now shew how the Area or Superficial content of them, and any other plane Figures may be found: And because all many-sided Figures may be best Measu∣red by reducing them first into Right an∣gled Triangles, Quadrangles, or Trapezias, we will first shew how the Area or Superfi∣cial content of these Figures may be readily found; and first of a Right or Oblique an∣gled plane Triangle.

2. To Measure the Right angled plane Triangle BDF, in Fig. 7. Multiply BF by FD, half the Product shall be the content.

3. To Measure the Oblique angled plane Triangle BDC, let fall the Perpendicular DF, then Multiply BC by DF, half the Product shall be the content.

4. To find the Area or Superficial con∣tent of any Oblique angular Trapezium, convert it into two Oblique angled Trian∣gles, by a Diagonal, as the line BD in the Trapezium ABCD, then turn the Oblique

angled Triangles into Right, by letting fall the Perpendiculars AE and CF, then Mul∣tiply BD by the sum of AE and CF, half the Product shall be the content. In like manner may any other Irregular Multangle be also measured by turning it into Trian∣gles and Trapeziums, and computing them severally, and adding all their contents to∣gether.

Vide A.

Problem I. The Diameter of a Circle being given, to find the Circumference.

Vide B.

The Circumference of a Circle whose di∣ameter is 1, is 3.14159 and therefore,

As 1 is to 3.14159, so is any other Dia∣meter, to the Circumference answering that Diameter.

Problem II. The Diameter of a Circle being given, to find the Superficial content.

Archimedes hath Demonstrated, that the Area of a Circle is equal to the content of a Right angled plane Triangle, whose Legs

comprehending the Right angle, are one of them equal to the Semidiameter, and the o∣ther to the Circumference of a Circle. And therefore the Area or Superficial content of a Circle may be found, by Multiplying half the Circumference by half the Diameter, or the whole Diameter by the fourth part of the Circumference, they taking the Diame∣ter of a Circle to be one, and the Circum∣ference 3.14159, the Superficial content of such a Circle will be found to be 07853975.

And therefore, As 1 is to 78539, so is the Square of any other Diameter to the Super∣ficial content required.

Problem III. The Diameter of a Circle being given, to find the side of a square which may be inscribed within the same Circle.

The Chord or Subtense of the fourth of a Circle, whose Diameter is 1, is 7071067; therefore, as one, to 7071067, so is the Dia∣meter of any other Circle, to the Side required.

Problem IV. The Circumference of a Circle being given, to find the Diameter.

By the Diameter to find the Circumfe∣rence, the proportion by the tenth Problem is; As 1 to 3.14159, so the Diameter to the Circumference, and therefore putting the Circumference of a Circle to be 1.

As 3.14159 . 1 ∷ 1. 318308.

And therefore as 1 to 318308, so is any other Circumference, to the Diameter sought.

Problem V. The Circumference of a Circle being given, to find the Superficial content.

As the Square of the Circumference of a Circle given, is to the Superficial content of that Circle, so is the Square of the Circum∣ference of any other Circle, to the Superfi∣cial content of that other Circle.

And in a Circle whose Diameter is 1, the Circumference is 3.14159, and the Area 7853975, and supposing an Unite to be the Circumference of a Circle, it is, as the square of 3.14159 . 7853975 ∷ 1 . 0079578, and therefore, As 1 . 0.079578, so is the

square of any other Circumference, to the Area desired.

Problem VI. The Circumference of a Circle being given, to find the side of a square which may be inscribed within the same Circle.

As the Circumference of a Circle whose Diameter is 1, viz. 3.14159, is to 707107, the side of the inscribed square of that Cir∣cle, so is the Circumference of any other Circle, to the side inquired; and putting the Circumference to be Unity, it is, as 3.14159 . 707107 ∷ 1 . 225078, there∣fore,

As 1 to 225078, so is the Circumference given, to the side inquired.

Problem VII. The superficial content of a Circle being gi∣ven to find the Diameter.

This is the Converse of the 11. Problem, the Diameter given, to find the Content, for which the Proportion is; as 1 to 7853975, so is the square of the diameter, to the content: and therefore we must say; as 7853975 is to 1 so 1 to 1.27324; and hence, as 1 to 1.27324, so is the Area, to the Square of the Diameter.

Problem VIII. The Superficial content of a Circle being gi∣ven, to find the Circumference.

This is the Converse of the 14. Problem, the Circumference given, to find the Content.

As 1 to 079578, so Circumference square, to the Content: And therefore,

As 079578 . 1 ∷ 1 . 12.5664, and by consequence,

As 1 to 12.5664, so the Area, to the Square of the Circumference.

Vide C.

Problem IX. The Axis or Diameter of a Sphere being gi∣ven, to find the Superficial Content.

As the square of the Diameter of a Circle, which suppose 1, is to 3.14159 the Area, so is the square of the Axis given, to the Area that is required.