The art of measuring the sides and angles of triangles, either plane or spherical, whence it is accordingly called either Plane Trigonometry, or Spherical Trigonometry.
Trigonometry is an art of the greatest use in the mathematical sciences, especially in astronomy, navigation, surveying, dialing, geography, &c. &c. By it we come to know the magnitude of the earth, the planets and stars, their distances, motions, eclipses, and almost all other useful arts and sciences. Accordingly we find this art has been cultivated from the earliest ages of mathematical knowledge.
the resolution of triangles, is founded on the mutual proportions which subsist between the sides and angles of triangles; which proportions are known by finding the relations between the radius of a circle and certain other lines drawn in and about the circle, called cords, sines, tangents, and secants. The ancients, Menelaus, Hipparchus, Ptolemy, &c. performed their trigonometry by means of the cords. As to the sines, and the common theorems relating to them, they were introduced into trigonometry by the Moors or Arabsians, from whom this art passed into Europe, with several other branches of science. The Europeans have introduced, since the 15th century, the tangents and secants, with the theorems relating to them.
The proportion of the sines, tangents, &c. to their radius, is sometimes expressed in common or natural numbers, which constitute what we call the tables of natural sines, tangents, and secants. Sometimes it is expressed in logarithms, being the logarithms of the said natural sines, tangents, &c.; and these constitute the table of artificial sines, &c. Lastly, sometimes the proportion is not expressed in numbers; but the several sines, tangents, &c. are actually laid down upon lines of scales; whence the line of sines, of tangents, &c.
In trigonometry, as angles are measured by arcs of a circle described about the angular point, so the whole circumference of the circle is divided into a great number of parts; as 360 degrees, and each degree into 60 minutes, and each minute into 60 seconds, &c.; and then any angle is said to consist of so many degrees, minutes, and seconds, as are contained in the arc that measures the angle, or that is intercepted between the legs or sides of the angle.
Now the sine, tangent, and secant, &c. of every degree and minute, &c. of a quadrant, are calculated to the radius 1, and ranged in tables for use; as also the logarithms of the same; forming the triangular canon. And these numbers, so arranged in tables, form every species of right-angled triangles; so that no such triangle can be proposed, but one similar to it may be there found, by comparison with which the proposed one may be computed by analogy or proportion.
PLANE TRIGONOMETRY.
There are usually three methods of resolving triangles, or the cases of trigonometry; viz. geometrical construction, arithmetical computation, and instrumental operation. In the first method, the triangle in question is constructed by drawing and laying down the several parts of their magnitudes given, viz. the sides from a scale of equal parts, and the angles from a scale of cords or other instrument; then the unknown parts are measured by the same scales, and so they become known.
In the second method, having stated the terms of the proportion according to rule, which terms consist partly of the numbers of the given sides, and partly of the sines, &c. of angles taken from the tables, the proportion is then resolved like all other proportions, in which a fourth term is to be found from three given terms, by multiplying the second and third together, and dividing the product by the first. Or, in working with the logarithms, adding the logarithm of the second and third terms together, and from the sum subtracting the logarithm of the first term; then the number answering to the remainder is the fourth term sought.
To work a case instrumentally, as suppose by the logarithm lines on one side of the two foot scales: Extend the compasses from the first term to the second and third, which happens to be of the same kind with it; then that extent will reach from the other term to the fourth. In this operation, for the sides of triangles, is used the line of numbers (marked Num.) and for the angles, the line of sines or tangents (marked sin. and tan.) according as the proportion respects sines or tangents. See SECTOR.
In every case of plane triangles there must be three parts, one at least of which must be a side. And then the different circumstances, as to the three parts that may be given, admit of three cases or varieties only; viz.
1st, When two of the three parts given are a side and its opposite angle. 2d, When there are given two sides and their contained angle. 3d, And, thirdly, when the three sides are given.
To each of these cases there is a particular rule or proportion adapted for resolving it by.
1st, The Rule for the 1st Case, or that in which, of the three parts that are given, an angle and its opposite side are two of them, is this, viz. that the sides are proportional to the sines of their opposite angles; that is,
As one side given : To the sine of its opposite angle : : So another side given : To the sine of its opposite angle.
Or,
As the sine of an angle given : To its opposite side : So is the sine of another angle given : To its opposite side.
So that, to find an angle, we must begin the proportion with a given side that is opposite to a given angle; and to find a side, we must begin with an angle opposite to a given side.
Example. Suppose in the triangle BDC (fig. 1) there be given the side BC = 106, DB = 65, and the angle BCD = 31° 49′; to find the angle BDC obtuse and the side CD.
1. Geometrically by Construction.
Draw the line BC equal to 106, at C make an angle of 31° 49′ by drawing CD, take 65 in your compasses, and with one foot in B lay the other upon the line CD in D; draw the line BD; 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.
2. Arithmetically by Logarithms.
As the side BD 65 log. 1.81291 Is to sine angle C 31° 49′ 9.72193 So is the side BC 106 2.02531
To fine angle D 120° 43′ 9.93438
To To find DC.
As fine ang. C $31^\circ 49'$ 9.72198 The supp. Is to the side PD 65 1.31291 So is fine ang. B $27^\circ 28'$ 9.66392
$11.47682$ $9.72198$
To the side DC $56.88$ 1.75485
Here it may be proper to observe, that if the given angle be obtuse, the angle sought will be acute; but when the given angle is acute, and opposite to a lesser given side, then the required angle is doubtful, whether acute or obtuse; it ought therefore to be determined before the operation. For it is plain the above proportion produces $50^\circ 17'$ for the required angle; but as it is obtuse, its supplement to $180^\circ$ degrees must be taken, viz. $129^\circ 43'$.
By Gunter.
"The extent from $65$ to $109$ on the line of numbers will reach from $31^\circ 49'$ to $59^\circ 17'$ on the line of sides."
"2dly," "The extent from $31^\circ 49'$ to $27^\circ 28'$ on the line of sides will reach from $65$ to $56.88$ on the line of numbers."
Case II. When there are given two sides and their contained angle, to find the rest, the rule is this:
As the sum of the two given sides: Is to the difference of the sides: So is the tangent of half the sum of the two opposite angles or cotangent of half the given angle: To tan. of half the diff. of those angles.
Then the half diff. added to the half sum, gives the greater of the two unknown angles; and subtracted leaves the less of the two angles.
Hence, the angles being now all known, the remaining 3rd side will be found by the former case.
Example. The side BC = $109$, BD = $76$ (fig. 2.), and the angle CBD $101^\circ 30'$ given, to find the angle BDC or BCD, and the side CD.
1. Geometrically by Construction.
Draw the line BC $109$, and BD, so as to make an angle with BC of $101^\circ 30'$, and make BD equal to $76$; join BC and BD with a right line, and it is done; for the angle D being measured by the cord of $60^\circ$, will be $47^\circ 32'$, angle C $30^\circ 58'$, and the side DC $144.8$, as was required.
2. Arbitrarily by Logarithms.
| Side BC | 109 | 109 | 180° | | BD | 76 | 76 | 101° 30' |
Their sum $185$ 33 their diff. $78$ sum of the ang. D and C.
$\frac{1}{2}$ Sum $39^\circ 15'$ then
To find the angles D and C.
As the sum of the sides BC and BD = $185$ Is to their difference $33$ So is tan. of $\frac{1}{2}$ the sum of the angles C and D $39^\circ 15'$ 9.91224
To the tan. of $\frac{1}{2}$ the diff. of the angles C and D $8^\circ 17'$ 9.16358
To half the sum of the angles D and C $39^\circ 15'$ Add half the difference of the angles C and D $8^\circ 17'$
Gives the greater angle D $47^\circ 32'$ Subtracted, gives the lesser angle C $30^\circ 58'$
As fine angle D $47^\circ 32'$ Is to the side BC $109$ So is fine angle B $101^\circ 30'$
$9.29119$ $12.02862$ $9.86786$
To the side DC required $144.8$
3. By Gunter.
1st, "The extent from $185$ to $33$ on the line of numbers will reach from $39^\circ 15'$ to $8^\circ 17'$ on the line of tangents. 2dly, The extent from angle D $47^\circ 32'$ to $78^\circ 30'$ (the supplement of angle B) on the line of sides, will reach from the side BC $109$ to $144.8$, the side DC required, on the line of numbers."
Case III. Is when the three sides are given, to find the three angles; and the method of resolving this case is, to let a perpendicular fall from the greatest angle upon the opposite side or base, dividing it into two segments, and the whole triangle into two smaller right-angled triangles: then it will be,
As the base or sum of the two segments: Is to the sum of the other two sides So is the difference of those sides To the difference of the segments of the base.
Then half this difference of the two segments added to the half sum, or half the base, gives the greater segment, and subtracted gives the less. Hence, in each of the two right-angled triangles, there are given the hypotenuse, and the base, besides the right angle, to find the other angles by the first case.
Example. The sides BC (fig. 3.) = $105$, BD = $85$, and CD = $50$, given to find the angles BDC, BCD, or CBD.
1. Geometrically by Construction.
Draw the line BC equal to $105$, take CD $50$ in your compasses, and with one foot in C describe an arch; then take BD $85$ in your compasses, and with one foot in B cut the former arch in D, join BD and DC, and it is done; for the angle B, being measured, will be found $28^\circ 4'$, angle C $53^\circ 7'$, which being added together, is $81^\circ 11'$, their sum subtracted from $180$, leaves angle D $98^\circ 49'$ as was required.
2. Arbitrarily by Logarithms.
The two shortest sides are BD ($=85$) and CD ($=50$), the sum of which is $135$, and their difference $35$. The segments of the base BC are found in this manner:
As the side BC $=105$ log. $2.02119$ Is to the sum of the sides BD & DC $=135$ So is their difference $=35$ $1.54407$ To the difference of the seg. of BC $=45$ $1.61521$
Thus the sum and difference of the segments of the base BC being known, we have only to add half this sum $=52\frac{1}{2}$ to half the difference $=22\frac{1}{2}$, and we shall obtain the greater segment, which is $=75$; which subtracted from $105$, gives $30$ = the smaller segment. Then
To find the angle BDA.
As the hypotenuse BD $=85$ log. $1.92942$ Is to radius $=10.00000$ So is the greater segment $=75$ $1.87506$ To the sum of the angle BDA $=9.94564$
The angle BDA therefore is equal to $61^\circ 56'$.
Let us now find the angle ADC, which is done thus.
As the hypotenuse DC $=50$ log. $1.69897$ Is to radius $=10.00000$ So is the smaller segment $=30$ $1.47712$ To the fine of ADC $=9.77815$
The angle ADC therefore is equal to $36^\circ 53'$, and the whole angle BDC $=98^\circ 49'$. To find the angle at B, we have only to subtract the angle BDA (61° 56') from 90°, and the rem. 28° 4' is the angle sought. The angle at C is equal to 53° 7'.
3. By Gunter.
1/8, 'The extent from 105 to 135 will reach from 35 to 45 on the line of numbers.' 2/8, 'The extent from 85 to 75, on the line of numbers, will reach from radius to 61° 56', the angle BDA on the line of fines.' 3/8, 'The extent from 50 to 30 on the line of numbers, will reach from radius to angle ADC 36° 53' on the line of fines.'
The foregoing three cases include all the varieties of plane triangles that can happen, both of right and oblique-angled triangles. But besides these, there are some other theorems that are useful upon many occasions, or suited to some particular forms of triangles, which are often more expeditious in use than the foregoing general ones; one of which, for right-angled triangles, as the case for which it serves to often occurs, may be here inferred, and is as follows.
CASE IV. When, in a right-angled triangle, there are given the angles and one leg, to find the other leg, or the hypotenuse. Then it will,
As radius : To given leg AB : So tang. adjacent the angle A : To the opposite leg BC, and : So sec. of same angle A : To hypot. AC :
Example. In the triangle ABC (fig. 4.), right-angled at B,
Given the leg AB = 162 and the angle A = 53° 7' 48" conseq. the angle C = 36° 52' 12"
1. Geometrically.—Draw the leg AB = 162; Erect the indefinite perpendicular BC; Make the angle A = 53° 7' 48", and the side AC will cut BC in C, and form the triangle ABC. Then, by measuring, there will be found AC = 270, and BC = 216.
2. Arithmetically.
As radius = 10 To AB = 162 So tang. A = 53° 7' 48" To BC = 216 So sec. A = 53° 7' 48" To AC = 270
3. By Gunter.
Extend the compasses from 45° at the end of the tangents (the radius) to the tangent of 53° 7'; then that extent will reach, on the line of numbers, from 162 to 216, for BC. Again, extend the compasses from 36° 52' to 90 on the lines; then that extent will reach, on the line of numbers, from 162 to 270 for AC.
Note. Another method, by making every side radius, is often added by the authors on trigonometry, which is thus: The given right-angled triangle being ABC, make first the hypotenuse AC radius, that is, with the extent of AC as a radius, and each of the centres A and C, describe arcs CD and AE (fig. 3.); then it is evident that each leg will represent the fine of its opposite angle, viz. the leg BC the fine of the arc CD or of the angle A, and the leg AB the fine of the arc AE or of the angle C. Again, making either leg radius, the other leg will represent the tangent of its opposite angle, and the hypotenuse the secant of the same angle; thus, with radius AB and centre A describing the arc BF, BC represents the tangent of that arc, or of the Spherical angle A, and the hypotenuse AC the secant of the same; or with the radius BC and centre C describing the arc BG, the other leg AB is the tangent of that arc BG or of the angle C, and the hypotenuse CA the secant of the same.
And then the general rule for all these cases is this, viz., that the sides bear to each other the same proportions as the parts or things which they represent. And this is called making every side radius.
SPHERICAL TRIGONOMETRY.
Spherical Trigonometry is the art whereby, from three given parts of a spherical triangle, we discover the rest; and, like plane trigonometry, is either right-angled or oblique angled. But before we give the analogies for the solution of the several cases in either, it will be proper to premise the following theorems:
THEOREM I. In all right-angled spherical triangles, the sign of the hypotenuse : radius :: fine of a leg : fine of its opposite angle. And the fine of a leg : radius :: tangent of the other leg : tangent of its opposite angle.
Demonstration. Let EDAFG (ibid. fig. 6.) represent the eighth part of a sphere, where the quadrantal planes EDFG, EDBC, are both perpendicular to the quadrantal plane ADFB; and the quadrantal plane ADGC is perpendicular to the plane EDFG; and the spherical triangle ABC is right-angled at B, where CA is the hypotenuse, and BA, BC, are the legs.
To the arches GF, CB, draw the tangents HF, OB, and the fines GM, CI, on the radii DF, DB; also draw BL the fine of the arch AB, and CK the fine of AC; and then join IK and OL. Now HF, OB, GM, CI, are all perpendicular to the plane ADFB. And HD, GK, OL lie all in the same plane ADGC. Also FD, IK, BL, lie all in the same plane ADGC. Therefore the right angled triangles HFD, CIK, ODL, having the equal angles HDF, CKI, OLB, are similar. And CK : DG :: CI : GM; that is, as the fine of the hypotenuse : radius :: fine of a leg : fine of its opposite angle. For GM is the fine of the arc GF, which measures the angle CAB. Also, LB : DF :: BO : FH; that is, as the fine of a leg : radius :: tangent of the other leg : tangent of its opposite angle.
Q.E.D.
Hence it follows, that the fines of the angles of any oblique spherical triangle ACD (fig. 7.) are to one another, directly, as the fines of the opposite sides. Hence it also follows, that, in right-angled spherical triangles, having the same perpendicular, the fines of the bases will be to each other, inversely, as the tangents of the angles at the bases.
THEOREM II. In any right-angled spherical triangle ABC (fig. 8.) it will be, As radius is to the co-fine of one leg, so is the co-fine of the other leg to the co-fine of the hypotenuse.
Hence, if two right-angled spherical triangles ABC, CBD (fig. 7.) have the same perpendicular BC, the co-fines of their hypotenuses will be to each other, directly, as the co-fines of their bases.
THEOREM III. In any spherical triangle it will be, As radius is to the fine of either angle, so is the co-fine of the adjacent leg to the co-fine of the opposite angle.
Hence, in right-angled spherical triangles, having the same perpendicular, the co-fines of the angles at the base will be to each other, directly, as the fines of the vertical angles.
THEOREM IV. In any right-angled spherical triangle **TRIGONOMETRY.**
As the sum of the sines of two unequal arches is to their difference, so is the tangent of half the sum of those arches to the tangent of half their difference; and as the sum of the co-sines is to their difference, so is the co-tangent of half the sum of the arches to the tangent of half the difference of the same arches.
**Theorem V.** In any spherical triangle \(ABC\) (fig. 9 and 10,) it will be, As the co-tangent of half the sum of half their difference, so is the co-tangent of half the base to the tangent of the distance \((DE)\) of the perpendicular from the middle of the base.
Since the last proportion, by permutation, becomes co-tang. \(\frac{AC + BC}{2}\) : co-tang. \(AE\) :: tang. \(\frac{AC - BC}{2}\) : tang. \(DE\), and as the tangents of any two arches are, inversely, as their co-tangents; it follows, therefore, that tang. \(AE\) : tang. \(\frac{AC + BC}{2}\) :: tang. \(\frac{AC - BC}{2}\) : tang. \(DE\); or, that the tangent of half the base is to the tangent of half the sum of the sides, as the tangent of half the difference of the sides to the tangent of the distance of the perpendicular from the middle of the base.
**Theorem VI.** In any spherical triangle \(ABC\) (fig. 9,) it will be, As the co-tangent of half the sum of the angles at the base is to the tangent of half their difference, so is the tangent of half the vertical angle to the tangent of the angle which the perpendicular \(CD\) makes with the line \(CF\) bisecting the vertical angle.
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### Solution of the Cases of right-angled spherical Triangles, (fig. 8.)
| Case | Given | Sought | Solution | |------|-------|--------|----------| | 1 | The hyp. AC and one angle A | The opposite leg BC | As radius : fine hyp. AC : fine A : fine BC (by the former part of theor. 1.) | | 2 | The hyp. AC and one angle A | The adjacent leg AB | As radius : co-fine of A :: tang. AC : tang. AB (by the latter part of theor. 1.) | | 3 | The hyp. AC and one angle A | The other angle C | As radius : co-fine of AC :: tang. A : cotang. C (by theorem 4.) | | 4 | The hyp. AC and one leg AB | The other leg BC | As fine AC : radius :: fine AB : fine C (by the former part of theorem 1.) | | 5 | The hyp. AC and one leg AB | The opposite angle C | As tang. AC : tang. AB :: radius : co-fine A (by theorem 1.) | | 6 | One leg AB and the adjacent angle A | The other leg BC | As radius : fine AB :: tangent A : tangent BC (by theorem 4.) | | 7 | One leg AB and the adjacent angle A | The opposite angle C | As radius : fine A :: co-fine of AB : co-fine of C (by theorem 3.) | | 8 | One leg AB and the adjacent angle A | The hyp. AC | As co-fine of A : radius :: tang. AB : tang. AC (by theorem 1.) | | 9 | One leg BC and the opposite angle A | The other leg AB | As tang. A : tang. BC :: radius : fine AB (by theorem 4.) | | 10 | One leg BC and the opposite angle A | The adjacent angle C | As co-fine BC : radius :: co-fine of A : fine C (by theorem 3.) | | 11 | Both legs AB and BC | The hyp. AC | As fine A : fine BC :: radius : fine AC (by theorem 1.) | | 12 | Both legs AB and BC | An angle, suppose A | As fine AB : radius :: tang. BC : tang. A (by theorem 4.) | | 13 | Both angles A and C | A leg, suppose AB | As fine A : co-fine C :: radius : co-fine AB (by theorem 3.) | | 14 | Both angles A and C | The hyp. AC | As tang. A : co-tang. C :: radius : co-fine AC (by theorem 4.) |
*Note.* The 10th, 11th, and 12th cases are ambiguous; since it cannot be determined by the data, whether \(A, B, C,\) and \(AC,\) be greater or less than 90 degrees each. ### TRIGONOMETRY
The Solution of the Cases of oblique spherical Triangles, (fig. 9 and 10.)
| Case | Given | Sought | Solution | |------|-------|--------|----------| | 1 | Two sides AC, BC, and an angle A opposite to one of them | The angle B opposite to the other | As fine BC : fine A :: fine AC : fine B (by theorem 1.) Note, this case is ambiguous when BC is less than AC; since it cannot be determined from the data whether B be acute or obtuse. | | 2 | Two sides AC, BC, and an angle A opposite to one of them | The included angle ACB | Upon AB produced (if need be) let fall the perpendicular CD; then (by theorem 4.) rad. : co-fine AC :: tang. A : co-tang. ACD; but (by theorem 1.) as tang. BC : tang. AC :: co-fine ACD : co-fine BCD. Whence ACB = ACD = BCD is known. | | 3 | Two sides AC, BC, and an angle opposite to one of them | The other side AB | As rad. : co-fine A :: tang. AC : tang. AD (by theor. 1.) and (by theor. 2.) as co-fine AC : co-fine BC :: co-fine AD : co-fine BD. Note, this and the last case are both ambiguous when the first is so. | | 4 | Two sides AC, AB, and the included angle A | The other side BC | As rad. : co-fine A :: tang. AC : tang. AB (by theor. 1.) whence AD is also known; then (by theor. 2.) as co-fine AD : co-fine BD :: co-fine AC : co-fine BC. | | 5 | Two sides AC, AB, and the included angle A | Either of the other angles, suppose B | As rad. : co-fine A :: tang. AC : tang. AD (by theor. 1.) whence BD is known; then (by theor. 4.) as fine BD : fine AD :: tan. A : tan. B. | | 6 | Two angles A, ACB, and the side AC betwixt them | The other angle B | As rad. : co-fine AB :: tang. A : co-tang. ACD (by theorem 4.) whence BCD is also known; then (by theor. 3.) as fine ACD : fine BCD :: co-fine A : co-fine B. | | 7 | Two angles A, ACB, and the side AC betwixt them | Either of the other sides, suppose BC | As rad. : co-fine AC :: tang. A : co-tang. ACD (by theorem 4.) whence BCD is also known; then, as co-fine BCD : co-fine ACD :: tang. AC : tang. BC (by theor. 1.) | | 8 | Two angles A, B, and a side AC opposite to one of them | The side BC opposite the other | As fine B : fine AC :: fine A : fine BC (by theorem 1.) | | 9 | Two angles A, B, and a side AC opposite to one of them | The side AB betwixt them | As rad. : co-fine A :: tang. AC : tang. AD (by theor. 1.) and as tang. B : tang. A :: fine AD : fine BD (by theorem 4.) whence AB is also known. | | 10 | Two angles A, B, and a side AC opposite to one of them | The other angle ACB | As rad. : co-fine AC :: tang. A : co-tang. ACD (by theorem 4.) and as co-fine A : co-fine B :: fine ACD : fine BCD (by theor. 3.) whence ACB is also known. | | 11 | All the three sides AB, AC, and BC | An angle, suppose A | As tang. $\frac{1}{2}$AB : tang. $\frac{AC+BC}{2}$ :: tang. $\frac{AC-BC}{2}$ : tang. DE, the distance of the perpendicular from the middle of the base (by theorem 6.) whence AD is known; then, as tang. AC : tang. AD :: rad. : co-fine A (by theor. 1.) | | 12 | All the three angles A, B, and ACB | A side, suppose AC | As co-tang. $\frac{ABC+A}{2}$ : tang. $\frac{ABC-A}{2}$ :: tang. $\frac{ACB}{2}$ : tang. of the angle included by the perpendicular and a line bisecting the vertical angles; whence ACD is also known; then (by theorem 5.) tang. A : co-tang. ACD :: rad. co-fine AC. |
The following propositions and remarks, concerning spherical triangles (selected and communicated to Dr Hutton by the reverend Nevil Maskelyne, D.D. Astronomer Royal, F.R.S.), will also render the calculation of them peripatetic, and free from ambiguity.
1. A spherical triangle is equilateral, isosceles, or scalene, according as it has its three angles all equal, or two of them equal, or all three unequal; and vice versa.
2. The greatest side is always opposite the greatest angle, and the smallest side opposite the smallest angle.
3. Any two sides taken together are greater than the third.
4. If the three angles are all acute, or all right, or all obtuse; the three sides will be, accordingly, all less than $90^\circ$, or equal to $90^\circ$, or greater than $90^\circ$; and vice versa.
5. If from the three angles A, B, C, of a triangle ABC, fig. 12, as poles, there be described, upon the surface of the sphere, three arches of a great circle DE, DF, FE, forming by their intersections a new spherical triangle DEF; each side of the new triangle will be the supplement of the angle at its pole; and each angle of the same triangle will be the supplement of the side opposite to it in the triangle ABC.
6. In any triangle ABC, or A'BC, right-angled in A, fig. 12, the angles at the hypotenuse are always of the same kind.