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志二十一
Treatise 21
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時憲二
Shixian Calendar 2
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推步算術
Computational Methods for Calendrical Astronomy
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推步新法所用者，曰平三角形，曰弧三角形，曰橢圓形。 今撮其大旨，證立法之原，驗用數之實，都為一十六術，著於篇。
The new calendrical methods rely on plane triangles, spherical triangles, and ellipses. Here we summarize their essentials, show the principles behind their formulation, confirm the numbers used in practice, and collect sixteen methods in all for this chapter.
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平三角形者，三直線相遇而成。 其線為邊，兩線所夾空處為角。 有正角，當全圓四分之一，如甲乙丙形之甲角。 有銳角，不足四分之一，如乙、丙兩角。 有鈍角，過四分之一，如丁戊己形之戊角。 圖形尚無資料
A plane triangle is formed where three straight lines meet. The lines are its sides, and the space between two sides is the angle. There is the right angle, equal to one quarter of a full circle, as in angle A of triangle ABC. There is the acute angle, less than a quarter circle, as in angles B and C. There is the obtuse angle, greater than a quarter circle, as in angle E of triangle DEF. Diagram not yet available
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角之度無論多寡，皆有其相當之八線。 曰正弦、正矢、正割、正切，所有度與九十度相減餘度之四線也，如甲乙為本度，則丙乙為餘度。 正弦乙戊，正矢甲戊，正割庚丁，正切庚甲，餘弦乙己，餘矢丙己，餘割辛丁，餘切辛丙。 若壬癸為本度，則丑癸為餘度，正弦癸辰，正矢壬辰，餘弦癸卯，餘矢丑卯，餘割子寅，餘切丑寅。 以壬癸過九十度無正割、正切，借癸午之子未為正割，午未為正切。 若正九十度丑壬為本度，則無餘度，丑子半徑為正弦，壬子半徑為正矢，亦無正割、正切，並無餘弦、餘矢、餘割、餘切。
Every angle, whatever its measure, has a corresponding set of eight trigonometric lines. They are the sine, versed sine, secant, and tangent—the four functions for a given angle and for its complement to ninety degrees; if AB is the principal arc, then CB is the complementary arc. Sine BE, versed sine AE, secant GD, tangent GA; cosine BF, coversed sine CF, cosecant HD, cotangent HC. If RS is the principal arc, then TS is the complement: sine SC, versed sine RC, cosine SD, coversed sine TD, cosecant FE, cotangent TE. When the principal arc exceeds ninety degrees there is no direct secant or tangent; one borrows the secant and tangent of the supplementary construction instead. If the principal arc is exactly ninety degrees, there is no complement: the radius itself is the sine, and the radius is also the versed sine; secant, tangent, and all four complementary functions vanish as well.
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古定全圓周為三百六十度，四分之一稱一象限，為九十度。 每度六十分，每分六十秒，每秒六十微。 圓半徑為十萬，後改千萬。 逐度逐分求其八線，備列於表。 推算三角，在九十度內，欲用某度某線，就表取之，算得某線。 欲知某度，就表對之。 過九十度者，欲用正弦、正割、正切及四餘，以其度與半周相減餘，就表取之。 欲用正矢，取餘弦加半徑為之。 既得某線，欲知某度，就表對得其度與半周相減餘命之。
Tradition fixed the full circle at 360 degrees; one quarter, called a quadrant, is 90 degrees. Each degree divides into sixty minutes, each minute into sixty seconds, and each second into sixty micro-units. The circle's radius was first set at 100,000, later raised to 10,000,000. The eight lines are computed for every degree and minute and laid out in tables. In triangle calculation within ninety degrees, to obtain a given line for a given angle, one looks it up in the table. To find an angle, one matches the value against the table. Beyond ninety degrees, to use sine, secant, tangent, or any of the four complementary functions, one subtracts the angle from 180° and looks up the remainder in the table. For the versed sine, add the radius to the cosine. Once a line is known, to recover the angle one matches it in the table and names the complement to 180° when appropriate.
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圖形尚無資料
Diagram not yet available
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算平三角凡五術：
Plane triangles are solved by five methods:
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一曰對邊求對角，以所知邊為一率，對角正弦為二率，所知又一邊為三率，二三相乘，一率除之，求得四率，為所不知之對角正弦。 如圖甲乙為所知邊，丁角為所知對角，乙丁為所知又一邊，甲角為所不知對角也。 此其理系兩次比例省為一次。 如圖乙丁為半徑之比，乙丙為丁角正弦之比。 法當先以半徑為一率，丁角正弦為二率，乙丁為三率，求得四率中垂線乙丙。 既得乙丙，甲乙為半徑之比，乙丙又為甲角正弦之比。 乃以甲乙為一率，乙丙為二率，半徑為三率，求得四率，自為甲角正弦。 然後合而算之，以先之一率半徑與後之一率甲乙相乘為共一率，先之二率丁角正弦與後之二率乙丙相乘為共二率，先之三率乙丁與後之三率半徑相乘為共三率，求得四率，自為先之四率乙丙與後之四率甲角正弦相乘數，仍當以乙丙除之，乃得甲角正弦。 後既當除，不如先之勿乘。 共二率內之乙丙與三率相乘者也，乘除相報，乙丙宜省。 又共三率內之半徑與二率相乘者也，共一率內之半徑又主除之，乘除相報，半徑又宜省。 故徑以甲乙為一率，丁角正弦為二率，乙丁為三率，求得四率，為甲角正弦。
First, to find an unknown angle from a known side: set the known side as the first term, the sine of the known opposite angle as the second, and the other known side as the third; multiply the second and third and divide by the first to obtain the sine of the unknown angle. In the figure, AB is the known side, angle D the known opposite angle, BD the other known side, and angle A the unknown opposite angle. The principle is to combine two successive proportions into one. In the figure, BD is to the radius as BC is to the sine of angle D. First set the radius as the first term, the sine of angle D as the second, and BD as the third; the fourth term is the perpendicular BC. Once BC is known, AB is to the radius as BC is to the sine of angle A. Then set AB as the first term, BC as the second, and the radius as the third; the fourth term is the sine of angle A. Combining the two steps yields a single proportion: AB as the first term, the sine of angle D as the second, and BD as the third; the fourth term is the sine of angle A, after canceling the intermediate perpendicular. Because a division would follow anyway, it is better to omit the intermediate multiplication. BC in the combined second term would be multiplied by the third and then divided out again, so BC can be dropped. The radius in the combined third term would be multiplied by the second while the radius in the first term divides; these cancel, so the radius drops out entirely. Hence one may proceed directly: AB as first term, sine of angle D as second, BD as third; the fourth term is the sine of angle A.
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二曰對角求對邊，以所知角正弦為一率，對邊為二率，所知又一角正弦為三率，求得四率，為所不知對邊。 此其理具對邊求對角，反觀自明。
Second, to find an unknown side from a known angle: set the sine of the known angle as the first term, the known opposite side as the second, and the sine of the other known angle as the third; the fourth term is the unknown side. This is the converse of the first method and is self-evident when read in reverse.
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三曰兩邊夾一角求不知之二角，以所知角旁兩邊相加為一率，相減餘為二率，所知角與半周相減，餘為外角，半之，取其正切為三率，求得四率，為半較角正切。 對錶得度，與半外角相加，為對所知角旁略大邊之角； 相減，餘為對所知角旁略小邊之角。 此其理一在平三角形。 三角相併，必共成半周。 如圖甲乙丙形，中垂線甲丁，分為兩正角形。 正角為長方之半，長方四角皆正九十度，正角形兩銳角斜剖長方，此角過九十度之半幾何，彼角不足九十度之半亦幾何，一線徑過，其勢然也。 故甲右邊分角必與乙角合為九十度，甲左邊分角必與丙角合為九十度。 論正角形各加丁角，皆成半周，合為銳角形。 除去丁角，三角合亦自為半周。 故既知一角之外，其餘二角雖不知各得幾何度分，必知其共得此角減半周之餘也。 一在三角同式形比例。 如圖丙庚戊形，知丙庚、丙戊兩邊及丙角。 展丙庚為丙甲，連丙戊為甲戊，兩邊相加。 截丙戊於丙丁，為戊丁，兩邊相減餘。 作庚丁虛線，丙庚、丙丁同長，庚丁向圓內二角必同度，是皆為丙角之半外角，與甲辛、辛庚之度等。 而庚向圓外之角，即本形庚角大於戊角之半，是為半外角。 以庚丁為半徑之比，則甲庚即為丁半外角正切之比。 半徑與正切恆為正角，甲庚與庚丁圓內作兩通弦，亦無不成正角故也。 又作丁己線，與甲庚平行，庚丁仍為半徑之比，丁己又為庚向圓外半較角正切之比。 而戊甲庚大形與戊丁己小形，戊甲、戊丁既在一線，甲庚、丁己又系平行，自然同式。 故甲戊兩邊相加為一率，戊丁兩邊相減餘為二率，甲庚半外角正切為三率，求得四率，自當丁己半較角正切也。
Third, given two sides and the included angle, to find the other two angles: set the sum of the adjacent sides as the first term, their difference as the second, take half the exterior angle (180° minus the known angle) and its tangent as the third; the fourth term is the tangent of the half-difference. Look up the angle in the table; added to the half exterior angle, this gives the angle opposite the slightly longer adjacent side; subtracted from it, the remainder is the angle opposite the slightly shorter adjacent side. The first part of the principle concerns plane triangles. The three angles of a triangle always sum to a semicircle (180°). In triangle ABC, the altitude AD divides it into two right triangles. A right angle is half of a rectangle whose corners are all 90°. A right triangle's acute angles bisect such a rectangle diagonally: whatever one exceeds 45°, the other falls short by the same amount—this follows necessarily from a straight cut. Hence the angle at A on the right must combine with angle B to make 90°, and that on the left with angle C to make 90°. Each right triangle plus angle D sums to 180°; together they reconstitute the original acute triangle. Remove angle D, and the three angles of the whole triangle still sum to 180°. Thus, given one angle, even if the other two are unknown individually, their sum must equal 180° minus the known angle. The second part lies in the proportion of similar triangles. In triangle CGE, given sides CE and CG and angle C. Extend CE to CA and connect CG as AW; their sum is one term. Mark D on CG so that CD equals CE; their difference is the other term. Draw auxiliary line ED; since CE equals CD, the two interior angles at ED are equal—each is half the exterior angle at C, matching the construction at A. The exterior angle at E is half the excess of angle E over angle G in the original triangle—the half exterior angle. ED is to the radius as AE is to the tangent of the half exterior angle at D. Radius and tangent always meet at a right angle, as do the two chords drawn within the circle at AE and ED. Draw DI parallel to AE; ED is still to the radius as DI is to the tangent of the half-difference angle at E. Triangles WAE and WDI share a line WAD and parallel sides AE and DI, so they are necessarily similar. Hence AW as first term, WD as second, tangent of the half exterior angle as third; the fourth term is the tangent of the half-difference, DI.
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四曰兩角夾一邊求不知之一角，以所知兩角相併，與半周相減，餘即得。 此其理具兩邊夾一角。
Fourth, given two angles and the included side, to find the third angle: add the two known angles and subtract from 180°. This follows directly from the case of two sides and an included angle.
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五曰三邊求角，以大邊為底，中、小二邊相併相減，兩數相乘，大邊除之，得數與大邊相加折半為分底大邊，相減餘折半為分底小邊。 乃以中邊為一率，分底大邊為二率，半徑為三率，求得四率，為對小邊角餘弦。 或以小邊為一率，分底小邊為二率，半徑為三率，求得四率，為對中邊角餘弦。 此其理在勾股弦冪相求及兩方冪相較。 如圖甲丙中邊、甲乙小邊皆為弦，乙丙大邊由丁分之，丁丙、丁乙皆為勾，中垂線甲丁為股。 勾股冪相併恆為弦冪，今甲丁股既兩形所同，則甲丙大弦冪多於甲乙小弦冪，即同丙丁大勾冪多於乙丁小勾冪。 又兩方冪相較，恆如兩方根和較相乘之數。 如圖戊寅壬庚為大方冪，減去己卯辛庚小方冪，餘戊己卯辛壬寅曲矩形。 移卯癸壬辛為癸寅丑子，成一直方形，其長戊丑，自為大方根戊寅、小方根卯辛之和； 其闊戊己，自為大方根戊庚、小方根己庚之較。 故甲乙丙形，甲丙、甲乙相加為和，相減為較。 兩數相乘，即如丙丁、丁乙和較相乘之數。 丙乙除之，自得其較。 丙午相加相減各折半，自得丙丁及乙丁，既得丙丁、乙丁，各以丙甲、乙甲為半徑之比，丙丁、乙丁自為餘弦之比矣。
Fifth, given three sides, to find an angle: with the longest side as base, multiply the sum and difference of the other two sides and divide by the longest; half the sum of this quotient and the base gives the greater segment, half the difference the lesser. Set the middle side as first term, the greater base segment as second, radius as third; the fourth is the cosine of the angle opposite the shortest side. Alternatively, set the shortest side as first, the lesser base segment as second, radius as third; the fourth is the cosine of the angle opposite the middle side. The principle rests on the Pythagorean relation among squares of sides and on the comparison of two squared areas. In the figure, AC (middle) and AB (shortest) are hypotenuses of right triangles; the longest side BC is split at D into legs DC and DB, with altitude AD as the common upright. The sum of the squares of leg and upright equals the square of the hypotenuse; since AD is shared, the excess of AC² over AB² equals the excess of DC² over DB². The difference of two squares always equals the product of the sum and difference of their roots. In the figure, the large square minus the small square leaves a gnomon-shaped region between them. Rearrange the pieces into a rectangle whose length is the sum of the two roots; and whose width is the difference of the two roots. Thus in triangle ABC, AC plus AB is the sum, AC minus AB the difference. Their product equals the product of the sum and difference of DC and DB. Dividing by BC yields the difference directly. Halve the sum and difference to obtain DC and DB; then CA and BA are to the radius as DC and DB are to the cosines of the angles at B and A.
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此五術者，有四不待算，一不可算。 對邊求對角，令所知兩邊相等，則所求角與所知角必相等。 對角求對邊，令所知兩角相等，則所求邊與所知邊必相等。 兩邊夾一角，令所知兩邊相等，則所求二角必正得所知外角之半。 三邊求角，令二邊相等，即分不等者之半為底邊； 三邊相等，即平分半周三角皆六十度，皆不待算也。 若對邊求對角，所知一邊數少，對所知一角銳； 又所知一邊數多，求所對之角，不能知其為銳、為鈍，是不可算也。 諸題求邊角未盡者，互按得之。
Of these five methods, four require no computation and one cannot be resolved by computation alone. In finding an angle from a side, if the two known sides are equal, the unknown angle equals the known opposite angle. In finding a side from an angle, if the two known angles are equal, the unknown side equals the known opposite side. For two sides and an included angle, if the known sides are equal, each unknown angle is exactly half the exterior angle. For three sides, if two are equal, the base segment is half the third side; if all three are equal, each angle is 60°—none of these require calculation. In finding an angle from a side, if the known side opposite the known acute angle is the shorter, but if the other known side is longer, the angle opposite it may be acute or obtuse and cannot be determined by calculation alone. Any remaining unknown sides or angles in a problem are found by applying the other methods in turn.
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弧三角形者，三圓周相遇而成，其邊亦以度計。 九十度為足，少於九十度為小，過九十度為大。 其角銳、鈍、正與平三角等。 算術有七：
A spherical triangle is formed where three great circles meet; its sides are also measured in degrees. A side of 90° is complete; less than 90° is small; more than 90° is large. Its acute, obtuse, and right angles follow the same definitions as in plane triangles. There are seven computational methods:
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一曰對邊求對角，以所知邊正弦為一率，對角正弦為二率，所知又一邊正弦為三率，求得四率，為所求對角正弦。 此其理亦系兩次比例省為一次。 如圖甲乙丙形，知甲乙、丙乙二邊及丙角，求甲角。 作乙辛垂弧，半徑與丙角正弦之比，同於乙丙正弦與乙辛正弦之比。 法當以半徑為一率，丙角正弦為二率，乙丙正弦為三率，求得四率，為乙辛正弦。 既得乙辛正弦，甲乙正弦與乙辛正弦之比，同於半徑與甲角正弦之比。 乃以甲乙正弦為一率，乙辛正弦為二率，半徑為三率，求得四率，為甲角正弦。 然乘除相報，可省省之。
First, to find an unknown angle from a known side: set the sine of the known side as first term, the sine of the known opposite angle as second, and the sine of the other known side as third; the fourth term is the sine of the unknown angle. Here too the principle is to combine two proportions into one. In spherical triangle ABC, given sides AB and CB and angle C, find angle A. Drop the perpendicular arc BE; radius is to sin C as sin BC is to sin BE. Set radius as first term, sin C as second, sin BC as third; the fourth term is sin BE. Once sin BE is known, sin AB is to sin BE as radius is to sin A. Then set sin AB as first term, sin BE as second, radius as third; the fourth term is sin A. Multiplication and division cancel each other, so intermediate steps may be omitted.
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二曰對角求對邊，以所知角正弦為一率，對邊正弦為二率，所知又一角正弦為三率，求得四率，為所求對邊正弦。 此其理反觀自明。
Second, to find an unknown side from a known angle: set the sine of the known angle as first term, the sine of the known opposite side as second, and the sine of the other known angle as third; the fourth term is the sine of the unknown side. This is the converse of the first method and is self-evident when read in reverse.
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三曰兩邊夾一角，或銳或鈍，求不知之一邊。 以半徑為一率，所知角餘弦為二率，任以所知一邊正切為三率，求得四率，命為正切。 對錶得度，與所知又一邊相減，餘為分邊。 乃以前得度餘弦為一率，先用邊餘弦為二率，分邊餘弦為三率，求得四率，為不知之邊餘弦。 原角鈍，分邊大，此邊小； 分邊小，此邊大。 原角銳，分邊小，此邊小； 分邊大，此邊大。 此其理系三次比例省為二次。 如圖甲丙丁形，知甲丙、甲丁二邊及甲角，中作垂弧丙乙，半徑與甲角餘弦之比，同於甲丙正切與甲乙正切之比。 先一算為易明。 既分甲丁於乙，而得丁乙分邊，甲乙餘弦與半徑之比，同於甲丙餘弦與丙乙餘弦之比。 法當先以甲乙餘弦為一率，半徑為二率，甲丙餘弦為三率，求得四率，為丙乙餘弦。 既得丙乙餘弦，半徑與乙丁餘弦之比，同於丙乙餘弦與丁丙餘弦之比。 乃以半徑為一率，乙丁餘弦為二率，丙乙餘弦為三率，求得四率，為丁丙餘弦。 然而乘除相報，故從省。 兩邊夾一角若正，則徑以所知兩邊餘弦相乘半徑除之，即得不知邊之餘弦，理自明也。 所知兩邊俱大俱小，此邊小； 所知兩邊一小一大，此邊大。
Third, given two sides and an included angle (acute or obtuse), find the unknown side. Set the radius as first term, the cosine of the known angle as second, and the tangent of either known side as third; the fourth term is taken as a tangent value. Look up the angle in the table, subtract it from the other known side, and the remainder is the segment arc. Then set the cosine of the angle just found as first term, the cosine of the first side used as second, and the cosine of the segment as third; the fourth term is the cosine of the unknown side. If the included angle is obtuse, when the segment is large, the unknown side is small; when the segment is small, the unknown side is large. If the included angle is acute, when the segment is small, the unknown side is small; when the segment is large, the unknown side is large. The principle is to reduce three proportions to two. In spherical triangle ACD, given sides AC and AD and angle A, drop the perpendicular arc BC; radius is to cos A as tan AC is to tan AB. One preliminary calculation makes the method clear. With AD split at B to give segment DB, cos AB is to the radius as cos AC is to cos BC. First set cos AB as first term, radius as second, cos AC as third; the fourth term is cos BC. Once cos BC is known, the radius is to cos BD as cos BC is to cos DC. Then set radius as first term, cos BD as second, cos BC as third; the fourth term is cos DC. Again, multiplication and division cancel, so the abbreviated form is used. When the included angle is a right angle, multiply the cosines of the two known sides and divide by the radius to obtain the cosine of the unknown side directly. If both known sides are greater or both less than 90°, the unknown side is small; if one is small and one large, the unknown side is large.
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四曰兩角夾一邊，求不知之一角。 以角為邊，以邊為角，反求之； 得度，反取之； 求、取皆與半周相減。
Fourth, given two angles and the included side, find the third angle. Treat angles as sides and the side as an angle, and solve by the converse method; look up the value, then take its complement; both the forward lookup and the complement involve subtracting from 180°.
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五曰所知兩邊對所知兩角，或銳、或鈍，求不知之邊角。 以半徑為一率，任以所知一角之餘弦為二率，對所知又一角之邊正切為三率，求得四率，命為正切，對錶得度。 復以所知又一角、一邊如法求之，復得度。 視原所知兩角銳、鈍相同，則兩得度相加； 不同，則兩得度相減； 皆加減為不知之邊。 乃按第一術對邊求對角，即得不知之角。 原又一角鈍，對先用角之邊大於後得度，此角鈍； 對先用角之邊小於後得度，此角銳。 原又一角銳，對先用角之邊小於後得度，此角鈍； 對先用角之邊大於後得度，此角銳。 此其理系垂弧在形內與在形外之不同，及角分銳鈍，邊殊大小，前後左右俯仰向背之相應。 如圖甲乙丙形，甲乙二角俱銳，兩銳相向，故垂弧丙丁，從中取正，而在形內。 己丙庚形，己庚二角俱鈍，兩鈍相向，故垂弧戊丙亦在形內。 庚丙乙形，庚乙兩角，一銳一鈍相違，垂弧丙丁，從外補正，自在形外。 在形內者判底邊為二，兩得分邊之度，如乙丁、丁甲，合而成一底邊如乙甲，故宜相加。 在形外者，引底邊之餘，兩得分邊之度，如庚丁、乙丁，重而不揜，底邊如庚乙，故宜相減。 銳鈍大小之相應，亦如右圖審之。 所知兩邊對所知兩角有一正，則一得度即為不知之邊，理亦自明。
Fifth, given two sides opposite two known angles (acute or obtuse), find the unknown side and angle. Set the radius as first term, the cosine of either known angle as second, and the tangent of the side opposite the other known angle as third; look up the resulting tangent in the table. Repeat with the other known angle and side to obtain a second arc measure. If both known angles are acute or both obtuse, add the two arc measures; if one is acute and one obtuse, subtract them; either way yielding the unknown side. Then apply the first method to find the unknown angle from the unknown side. If the other known angle is obtuse, when the side opposite the first angle used exceeds the second arc found, the unknown angle is obtuse; when it is less, the unknown angle is acute. If the other known angle is acute, when the side opposite the first angle is less than the second arc found, the unknown angle is obtuse; when it is greater, the unknown angle is acute. The principle depends on whether the perpendicular arc falls inside or outside the triangle, and on how acute and obtuse angles, side magnitudes, and spherical orientation correspond. In triangle ABC, angles A and B are both acute and face each other, so the perpendicular arc CD falls inside the triangle. In triangle FCG, angles F and G are both obtuse and face each other, so the perpendicular arc EC also falls inside. In triangle GCB, angles G and B are one acute and one obtuse; the perpendicular arc CD is drawn from outside and lies outside the triangle. When inside, the base is split into two segments whose arcs should be added to recover the full base. When outside, the two segment arcs overlap the base extension and should be subtracted to obtain the base. The correspondence of acute, obtuse, and side magnitude follows the same logic, as the diagrams show. If one of the known angles is a right angle, one of the arc measures found is itself the unknown side, which is self-evident.
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六曰三邊求角，以所求角旁兩邊正弦相乘為一率，半徑自乘為二率，兩邊相減餘為較弧，取其正矢與對邊之正矢相減餘為三率，求得四率，為所求角正矢。 此其理在兩次比例省為一次。 如圖甲壬乙形，求甲角，其正矢為丑丁。 法當以甲乙邊正弦乙丙為一率，半徑乙己為二率，兩邊較弧正矢乙癸與對邊正矢乙卯相減餘癸卯同辛子為三率，求得四率為壬辛。 乃以甲壬邊正弦戊辛為一率，壬辛為二率，半徑己丁為三率，求得四率為丑丁。 甲角正矢亦以乘除相報，故從省焉。
Sixth, given three sides, to find an angle: multiply the sines of the adjacent sides as first term, the square of the radius as second, and the difference of their versed sines (for the difference arc minus the opposite side) as third; the fourth term is the versed sine of the angle sought. The principle is again to combine two proportions into one. In spherical triangle ARB, to find angle A, its versed sine is CD. Set sin AB as first term, radius as second, and the difference of versed sines for the difference arc and opposite side as third; the fourth term is RN. Then set sin AR as first term, RN as second, radius as third; the fourth term is CD, the versed sine of angle A. The versed sine of angle A may likewise be found by the abbreviated form after canceling terms.
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七曰三角或銳、或鈍求邊，以角為邊，反求其角； 既得角，復取為邊； 求、取皆與半周相減。 此其理在次形，如圖甲乙丙形，甲角之度為丁戊，與半周相減為戊己，其度必同於次形子辛午之子辛邊，蓋丑卯為乙之角度丑點之交，甲乙弧必為正角，丁戊為甲之角度戊點之交，甲乙弧亦必為正角。 以一甲乙而交丑辛、戊辛二弧皆成正角，則二弧必皆九十度，弧三角之勢如此也。 戊辛既九十度，子己亦九十度，去相覆之戊子，己戊自同子辛，於是庚癸必同子午，卯未必同午辛，理皆如是矣。 而此形之餘角既皆為彼形之邊，彼形餘角不得不為此形之邊，故反取之而得焉。 若三角有一正，除正角外，以一角之正弦為一率，又一角之餘弦為二率，半徑為三率，求得四率，為對又一角之邊餘弦。 此其理亦系次形，而以正角及一角為次形之角，以又一角加減象限為次形對角之邊，取象稍異。
Seventh, for an acute or obtuse spherical triangle, to find a side: treat the angle as a side and solve for it in reverse; then take the result as a side; both steps involve subtracting from 180°. The principle lies in a secondary triangle: in ABC, the measure of angle A equals arc DE; its complement to 180° equals arc EF, which must equal a side of the secondary triangle—for the constructions at B and A each make arc AB a right angle. A single arc AB meeting two other arcs at right angles forces both to be 90°—such is the geometry of spherical triangles. With those arcs at 90°, removing the overlap shows the complementary relations; the corresponding arcs in the secondary figure match accordingly. The complementary angles of one triangle become the sides of the secondary triangle, and vice versa; hence the reverse procedure yields the side sought. If the triangle has a right angle, set the sine of one acute angle as first term, the cosine of the other as second, and the radius as third; the fourth term is the cosine of the side opposite the remaining angle. This too rests on the secondary triangle, with the right angle and one acute angle as its angles and the other acute angle adjusted by a quadrant as the opposite side—a slightly different formulation.
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凡茲七術，惟邊角相求，有銳鈍、大小不能定者，然推步無其題，不備列。 此七題中求邊角有未盡者，互按得之。
Among these seven methods, only cases where acute, obtuse, and magnitude cannot be determined from sides and angles alone are omitted, as calendrical problems do not require them. Any remaining unknowns in these seven cases are found by applying the other methods in turn.
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橢圓形者，兩端徑長、兩腰徑短之圓面。 然必其應規，乃可推算。 作之之術，任以兩點各為心，一點為界，各用一針釘之，圍以絲線，末以鉛筆代為界之。 針引而旋轉，即成橢圓形。 如圖甲己午三點，如法作之，為丑午巳未橢圓，寅丑、寅巳為大半徑，寅午、寅未為小半徑，寅甲為兩心差，己甲為倍兩心差。 甲午數如寅巳，亦同寅丑，己午如之； 二數相和，恆與丑巳同。 令午針引至申，甲申、申己長短雖殊，共數不易。 甲午同大半徑之數如弦，兩心差如勾，小半徑如股，但知兩數，即可以勾股術得不知之一數。 若求面積，以平方面率四００００００００為一率，平圓面率三一四一五九二六五為二率，大小徑相乘成長方面為三率，求得四率為橢圓面積。 若求中率半徑，大小半徑相乘，平方開之即得。 然自甲心出線，離丑右旋，如圖至戌，甲丑、甲戌之間，有所割之面積，亦有所當之角度。
An ellipse is a figure whose diameter is long at the ends and short at the sides. It must however satisfy the defining property before it can be computed. To construct it: fix two foci with pins, loop a thread around them and a pencil, and trace the curve as the pencil moves. Rotating the thread about the pins produces the ellipse. With foci A, F, and W as in the figure, the ellipse is drawn; the major semiaxes are along the long diameter, the minor along the short, with the focal distance and its double as shown. The distance from A to W equals one semiaxis, as does that from F to W; and the sum of any two such distances always equals the major axis. With the thread at any point S on the curve, AS and SF differ in length but their sum is constant. The distance from focus to curve equals the major semiaxis as hypotenuse, the focal distance as one leg, and the minor semiaxis as the other; given any two, the third follows from the Pythagorean relation. To find the area: set 400,000,000 as the first term, π (3.14159265) as the second, and the product of the major and minor diameters as the third; the fourth term is the ellipse's area. The mean proportional semiaxis is the square root of the product of the major and minor semiaxes. From focus A, a line sweeps from one vertex rightward to another point on the curve; the sector between them has both an area and a corresponding angle.
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角積相求，爰有四術：
Angle and area may be found from each other by four methods:
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一曰以角求積，以半徑為一率，所知角度正弦為二率，倍兩心差為三率，求得四率為倍兩心差之端，垂線如己酉。 又以半徑為一率，所知角度餘弦為二率，倍兩心差為三率，求得四率為界度積線，引出之線如甲酉，倍兩心差之端垂線為勾自乘。 以引出之線，與甲戌、己戌和如巳丑大徑者相加為股弦和，除之得較。 和、較相加折半為己戌弦，與大徑相減為甲戌線。 又以半徑為一率，所知角正弦為二率，甲戌線為三率，求得四率為戌亥邊。 又以小徑為一率，大徑為二率，戌亥邊為三率，求得四率為辰亥邊。 又以大半徑寅辰同寅丑為一率，半徑為二率，辰亥邊為三率，求得四率為正弦，對錶得度。 又以半周天一百八十度化秒為一率，半圓周三一四一五九二六為二率，所得度化秒為三率，求得四率為比例弧線。 又以半徑為一率，大半徑為二率，比例弧線為三率，求得四率為辰丑弧線，與大半徑相乘折半，為寅辰丑分平圓面積。 又以大半徑為一率，小半徑為二率，分平圓面積為三率，求得四率為寅戌丑分橢圓面積。 乃以寅甲兩心差與戌亥邊相乘折半，與寅戌丑相減，為甲戌、甲丑之間所割面積。 此其理具本圖及平三角、弧三角，其法至密。
First, to find area from angle: set the radius as first term, the sine of the known angle as second, and twice the focal distance as third; the fourth gives the perpendicular at the end of the double focal distance. Again set radius as first, cosine of the angle as second, and twice the focal distance as third; the fourth is the boundary line; square the perpendicular leg at the double focal distance. Add the drawn line to the sum of two radii equal to the major diameter to get the sum of leg and hypotenuse; dividing yields their difference. Half the sum of sum and difference gives the hypotenuse; subtract from the major diameter to obtain the other leg. Set radius as first, sine of the angle as second, and the leg just found as third; the fourth is the next side in the construction. Set the minor diameter as first, major diameter as second, and the previous side as third; the fourth is the next segment. Set the major semiaxis as first, radius as second, and the segment as third; the fourth is a sine value, from which the angle is read in the table. Set 180° in seconds as first term, half the circumference (π) as second, and the angle in seconds as third; the fourth is the proportional arc length. Set radius as first, major semiaxis as second, and proportional arc as third; half the product gives the area of the corresponding circular sector. Set major semiaxis as first, minor as second, and the sector area as third; the fourth is the corresponding elliptical sector area. Halve the product of the focal distance and the constructed side, subtract from the elliptical sector area, and obtain the area between the two radii. The method rests on the figure together with plane and spherical trigonometry and is extremely rigorous.
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二曰以積求角，以兩心差減大半徑餘得甲丑線自乘為一率，中率半徑自乘為二率，甲戌、甲丑之間面積為三率，求得四率為中率面積，如甲氐亢。 分橢圓面積為三百六十度，取一度之面積為法除之，即得甲戌、甲丑之間所夾角度，此其理為同式形比例。 然甲亢與甲氐同長，甲戌則長於甲丑，以所差不多，借為同數。 若引戌至心，甲丑甲心所差實多，仍須用前法求甲戌線，借甲戌甲心相近為同數求之。
Second, to find angle from area: square the leg found by subtracting the focal distance from the major semiaxis as first term, square the mean proportional semiaxis as second, and the given sector area as third; the fourth is the corresponding mean-area. Dividing the ellipse's area into 360 parts and using the area per degree as divisor yields the angle between the two radii, by similar-figure proportion. Because two constructed lines are nearly equal while the third differs only slightly, they may be treated as equal for approximation. If a radius to the center is drawn, the difference is too large for the approximation; the full method must be used, treating the nearly equal lengths as one value.
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三曰借積求積，以所知面積，如圖之辛甲丑，用一度之面積為法除之，得面積之度。 設其度為角度，於倍兩心差之端如庚己丑。 以半徑為一率，己角正弦為二率，倍兩心差為三率，求得四率為甲子垂線。 又以半徑為一率，己角餘弦為二率，倍兩心差為三率，求得四率為己子分邊。 甲子為勾自乘，己子與大徑相減餘為股弦和，除之得股弦較。 和、較相加折半得甲庚線。 又以甲庚線為一率，甲子垂線為二率，半徑為三率，求得四率為庚角正弦，得度與己角相加為庚甲丑角。 乃用以角求積法，求得庚甲丑面積，與辛甲丑面積相減餘如庚甲辛，又用以積求角法，求得度，與庚甲丑角相加，即得辛甲丑角。
Third, to find one area from another known area: divide the known area by the area per degree to obtain its angular measure. Treat that measure as an angle at the end of the double focal distance. Set radius as first, sine of that angle as second, and twice the focal distance as third; the fourth is the perpendicular from the focus. Again set radius as first, cosine as second, and twice the focal distance as third; the fourth is the segment arc. Square the leg; subtract the segment from the major diameter for the sum of leg and hypotenuse; dividing yields their difference. Half the sum of sum and difference gives the next constructed line. Set the new line as first, the perpendicular as second, radius as third; the fourth is a sine, whose angle added to the first angle gives the combined angle. Apply the angle-to-area method for the combined angle, subtract the known area, apply area-to-angle to the remainder, and add the angles to obtain the total angle sought.
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四曰借角求角，以所知面積如前法取為積度，如丑甲丁。 設其度為角度，於橢圓心如丁乙辛。 以小半徑為一率，大半徑為二率，所設角度正切為三率，求得四率為丁乙癸角正切。 對錶得度，乃於倍兩心差之端丙作丙丑線，即命丑丙甲角如癸乙丁之角度，乃將丙丑線引長至寅，使丑寅與甲丑等，則丙寅同大徑。 又作甲寅線，成甲寅丙三角形，用切線分外角法求得寅角，倍之為甲丙丑形之丑角，與丙角相加為丑甲丁角。 此其理癸乙甲角度多於丑甲丁積度，為子乙癸角度。 即以此度當前之補算辛甲庚者，蓋所差無多也。
Fourth, to find one angle from another: convert the known area to an angular measure by the previous method. Treat that measure as an angle at the ellipse's center. Set the minor semiaxis as first term, major semiaxis as second, and tangent of the assumed angle as third; the fourth is the tangent of the constructed angle. Look up the angle in the table; at the end of the double focal distance draw a line so that a constructed angle matches the tangent angle; extend it so that a segment equals another radius, making the extended line equal to the major diameter. Draw another line to form a triangle; use the tangent method for the exterior angle, double the result for one angle of the sector, and add the angle at the focal end to obtain the total angle sought. The principle is that the constructed angle exceeds the area-derived angle by a small correction angle. This value may substitute for the supplementary calculation elsewhere, as the discrepancy is slight.
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此四術內凡單言半徑者，皆八線表一千萬之數。
In all four methods, the radius refers to the value 10,000,000 from the trigonometric table.