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Solution of triangles

The solution of triangles, also known as solutio triangulorum, is a fundamental problem in that involves determining the unknown side lengths and measures of a when certain elements are already known, typically using relationships between sides and to fully specify the 's geometry. This process applies to both and spherical triangles and forms the basis for applications in fields such as , , and . The primary methods for solving triangles rely on the and the , which extend the basic trigonometric ratios of right triangles to oblique ones. The law of sines states that the ratio of the length of a side to the sine of its opposite angle is constant across all sides: \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, making it suitable for cases where at least one side and its opposite angle, or two angles and a non-included side, are known. The law of cosines, given by c^2 = a^2 + b^2 - 2ab \cos C, generalizes the and is used when two sides and the included angle, or all three sides, are provided. For right triangles specifically, solutions often begin with the Pythagorean theorem (a^2 + b^2 = c^2) and like sine, cosine, and tangent to find missing elements. Triangles are classified by the known elements into congruence cases: (three sides), (two sides and included angle), (two angles and included side), AAS (two angles and non-included side), and (two sides and non-included angle), with AAA insufficient without a side to determine scale. The case is notable for the ambiguous case, where two possible triangles (or none) may satisfy the conditions, depending on the relative sizes of the sides and angle. Historically, the development of these techniques traces back to ancient civilizations, with early for solving triangles appearing in around the 5th century CE, later refined by Arab scholars into a systematic by the 13th century.

Plane triangles

Trigonometric relations

Solving a triangle refers to the process of determining the lengths of all three sides and the measures of all three angles, given a sufficient number of these elements. This task is fundamental in geometry and relies on trigonometric identities to relate the sides and angles uniquely up to congruence. The Law of Sines states that in any triangle with sides a, b, c opposite angles A, B, C respectively, \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R, where R is the circumradius. This relation can be derived by considering the area of the triangle, which is \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C, leading to the proportional form after equating areas. Alternatively, it follows from the extended Law of Sines in the circumcircle, where the side subtends an angle at the circumference and center, yielding the $2R factor as twice the radius times the sine of half the central angle. The generalizes the to any : c^2 = a^2 + b^2 - 2ab \cos C, with cyclic permutations for the other sides. This is derived by placing the in the coordinate plane with one vertex at the origin and applying the distance , or by projecting one side onto the other and using the cosine in right triangles formed by the altitude. For a right-angled where C = 90^\circ, \cos C = 0, reducing to c^2 = a^2 + b^2. The provides a relation between sides and the differences or sums of opposite angles: \frac{a - b}{a + b} = \frac{\tan \frac{A - B}{2}}{\tan \frac{A + B}{2}}. This can be proved using the and the tangent subtraction and addition formulas. Specifically, from \frac{a}{\sin A} = \frac{b}{\sin B}, express a = 2R \sin A and b = 2R \sin B, then apply the identities \sin A = \sin \left( (A+B)/2 + (A-B)/2 \right) = \sin \frac{A+B}{2} \cos \frac{A-B}{2} + \cos \frac{A+B}{2} \sin \frac{A-B}{2} and similarly for \sin B, leading to the tangent ratio after simplification and noting A + B = 180^\circ - C, so \frac{A+B}{2} = 90^\circ - \frac{C}{2}. The area of a triangle can be computed using \frac{1}{2} ab \sin C (and permutations), which follows directly from the height relative to base a being b \sin C. Alternatively, gives the area as \sqrt{s(s-a)(s-b)(s-c)}, where s = \frac{a+b+c}{2} is the semi-perimeter. This formula is derived by combining the area expression with the to eliminate the angle, yielding a symmetric polynomial in the sides that factors appropriately. A in the has three degrees of freedom corresponding to its shape, after accounting for rigid motions (two translations and one ). Thus, specifying three independent elements—such as three sides, two sides and the included angle, or two angles and a side—uniquely determines the up to , ensuring all remaining elements can be found using the above relations. Fewer than three elements leave the shape underdetermined, while more provide redundancy for verification.

Right-angled triangles

In a right-angled triangle, the side opposite the is known as the , which is the longest side of the . The two sides forming the are called the legs. The fundamental trigonometric ratios for an acute angle \theta in a right-angled are defined relative to the opposite side, adjacent side, and : \sin \theta = \frac{\text{opposite}}{\text{[hypotenuse](/page/Hypotenuse)}}, \quad \cos \theta = \frac{\text{adjacent}}{\text{[hypotenuse](/page/Hypotenuse)}}, \quad \tan \theta = \frac{\text{opposite}}{\text{adjacent}}. These definitions arise directly from the of the right and enable the computation of unknown sides and angles. The forms the core relation for right-angled triangles: if the legs are a and b, and the is c, then a^2 + b^2 = c^2. This theorem was rigorously proved by in his Elements, Book I, Proposition 47, using geometric constructions to show the equality of areas of squares built on the sides. To solve a right-angled triangle when the c and one acute angle \theta are given, the complementary acute angle is $90^\circ - \theta, since the angles sum to $90^\circ. The leg opposite \theta is then c \sin \theta, and the leg adjacent to \theta is c \cos \theta. When the two legs a and b are given, the hypotenuse is computed as c = \sqrt{a^2 + b^2} using the Pythagorean theorem. The acute angle opposite a can be found as \theta = \tan^{-1}(a/b), with the other angle being $90^\circ - \theta. If the hypotenuse c and one leg, say a, are provided, the other leg is b = \sqrt{c^2 - a^2}. The acute angle opposite a is then \sin^{-1}(a/c). The area of a right-angled triangle with legs a and b is \frac{1}{2}ab.

SSS case

In the SSS case for plane triangles, all three sides a, b, and c are given, and the task is to determine the opposite angles A, B, and C. This configuration uses the law of cosines to compute the angles directly from the sides. The formula for angle C opposite side c is \cos C = \frac{a^2 + b^2 - c^2}{2ab}, with similar expressions for \cos A and \cos B by cyclic permutation. Angles are then found using the arccosine function, ensuring each lies between 0° and 180° and sums to 180°. For a valid unique solution, the sides must satisfy the triangle inequalities: a + b > c, a + c > b, b + c > a, and all sides positive. Under these conditions, the SSS congruence theorem guarantees a unique plane triangle up to congruence. The solution proceeds as follows: first, verify the triangle inequalities. Next, apply the law of cosines to compute all three angles. Verify that A + B + C = 180^\circ. Optionally, confirm using the law of sines: \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}. This contrasts with spherical SSS, where curvature introduces excess. Consider an example with sides a = 3, b = 4, c = 5. Compute \cos C = \frac{3^2 + 4^2 - 5^2}{2 \cdot 3 \cdot 4} = \frac{9 + 16 - 25}{24} = 0, so C = 90^\circ. Then \cos A = \frac{3^2 + 5^2 - 4^2}{2 \cdot 3 \cdot 5} = \frac{9 + 25 - 16}{30} = \frac{18}{30} = 0.6, A = \cos^{-1}(0.6) \approx 53.13^\circ. Similarly, B \approx 36.87^\circ. The angles sum to 180°, confirming the solution. Verification via law of sines: \frac{\sin 53.13^\circ}{3} \approx \frac{0.8}{3} \approx 0.2667, matching others.

SAS case

In the SAS case for plane triangles, two sides and the included are given, say sides a and b with included angle C. Compute the third side c using the : c^2 = a^2 + b^2 - 2ab \cos C. This formula generalizes the for non-right angles. Once c is known, the remaining angles A and B are found using the : \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}, or alternatively, the law of cosines for angles. The solution is unique, as SAS congruence guarantees a single triangle. Verify angles sum to 180°. To solve: compute c from given a, b, C. Then find A = \sin^{-1}\left(\frac{a \sin C}{c}\right) and B = 180^\circ - A - C. For example, a = 5, b = 7, C = 40^\circ. Then c^2 = 25 + 49 - 2 \cdot 5 \cdot 7 \cos 40^\circ \approx 74 - 70 \cdot 0.7660 \approx 74 - 53.62 = 20.38, so c \approx 4.51. Then \sin A = \frac{5 \sin 40^\circ}{4.51} \approx \frac{5 \cdot 0.6428}{4.51} \approx 0.712, A \approx 45.5^\circ, B \approx 94.5^\circ. Angles sum to 180°. To arrive at values, use calculator for cos and sin, then arcsin and subtraction. The area is \frac{1}{2}ab \sin C.

SSA case

In plane trigonometry, the case (two sides and a non-included angle, often angle A opposite side a, with side b) can lead to the ambiguous case, where zero, one, or two triangles may exist due to the sine function's periodicity. Apply the : \frac{\sin B}{b} = \frac{\sin A}{a}, so \sin B = \frac{b \sin A}{a}. If \sin B > 1, no solution; if \sin B = 1, one right triangle at B = 90^\circ; if $0 < \sin B < 1, two possible B: acute B_1 = \sin^{-1}(\cdot) and obtuse B_2 = 180^\circ - B_1. For each B, compute C = 180^\circ - A - B, then c = \frac{a \sin C}{\sin A}. Check validity: for B_2, ensure A + B_2 < 180^\circ and b > a \sin A (height condition). The number of solutions depends on whether a > b, a = b \sin A, etc. If A is acute: no solution if a \leq b \sin A; one if a = b \sin A (right at B) or a \geq b; two if b \sin A < a < b. If A obtuse: one if a > b, none otherwise. Example: A = 30^\circ, a = 5, b = 10. \sin B = \frac{10 \sin 30^\circ}{5} = 1, so B = 90^\circ, C = 60^\circ, c = 5 \sqrt{3} \approx 8.66 (one solution). Another: A = 30^\circ, a = 6, b = 10. \sin B \approx 0.6, B_1 \approx 36.9^\circ, C_1 \approx 113.1^\circ, c_1 \approx 11.55; B_2 \approx 143.1^\circ > 180^\circ - 30^\circ, invalid (one solution). For two: adjust a between $5 and $10. Resolution: compute both, discard invalid.

ASA case

In the ASA case for plane triangles, two angles A and B and the included side c are given. Compute the third angle C = 180^\circ - A - B. Then use the for the remaining sides: a = c \cdot \frac{\sin A}{\sin C}, \quad b = c \cdot \frac{\sin B}{\sin C}. This yields a unique solution, as ASA congruence ensures one triangle, provided A + B < 180^\circ and all angles positive. No excess or curvature considerations apply. For example, A = 50^\circ, B = 60^\circ, c = 10. Then C = 70^\circ. a = 10 \cdot \frac{\sin 50^\circ}{\sin 70^\circ} \approx 10 \cdot \frac{0.7660}{0.9397} \approx 8.15, b \approx 10 \cdot \frac{0.8660}{0.9397} \approx 9.21. Verify triangle inequalities hold. These values satisfy the laws.

AAS case

In the AAS case for plane triangles, two angles A and B and the non-included side a (opposite A) are given. Compute C = 180^\circ - A - B. Then use the law of sines: \frac{\sin b}{b} = \frac{\sin a}{a} \implies b = a \cdot \frac{\sin B}{\sin A}, \quad c = a \cdot \frac{\sin C}{\sin A}. Unlike SSA, AAS always yields a unique solution if A + B < 180^\circ, by AAS congruence. No ambiguity arises, as the side is opposite a known angle. For example, A = 30^\circ, B = 50^\circ, a = 10. C = 100^\circ. b = 10 \cdot \frac{\sin 50^\circ}{\sin 30^\circ} \approx 10 \cdot \frac{0.7660}{0.5} \approx 15.32, c \approx 10 \cdot \frac{\sin 100^\circ}{\sin 30^\circ} \approx 19.32. Verify sum 180° and inequalities. This demonstrates unique determination.

Spherical triangles

Spherical trigonometry principles

A spherical triangle is formed by three great circles on the surface of a sphere, with vertices at their intersection points. The sides of the triangle, denoted a, b, and c, are the angular lengths of the great-circle arcs connecting the vertices, measured in radians and restricted to $0 < a, b, c < \pi. The interior angles at the vertices, denoted A, B, and C, are also between $0 and \pi radians, measured in the tangent planes at those points. For a sphere of radius R (often taken as the unit sphere where R = 1, so arc lengths equal angular measures), the geometry accounts for the positive curvature of the surface. Unlike planar triangles, the sum of the interior angles in a spherical triangle exceeds \pi radians, with the angular excess defined as E = A + B + C - \pi > 0. This excess is positive due to the sphere's and increases with the triangle's size. Girard's theorem states that the area of the spherical triangle is E R^2, providing a direct link between angular measure and surface area; for the unit , the area simplifies to E. The spherical law of sines relates the sides and angles analogously to the planar case but adapted for : \frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}. This equality holds for any spherical triangle and serves as a foundational relation for solving them. The arcsin function yields values in [0, \pi/2]; the supplement \pi - value may also be considered, selected based on consistency with other laws (e.g., positive excess). The extends the planar cosine rule to account for . For the sides, it is \cos c = \cos a \cos b + \sin a \sin b \cos C, with cyclic permutations for the other sides. For the angles, the dual form is \cos C = -\cos A \cos B + \sin A \sin B \cos c, again with cyclic permutations. These formulas derive from vector dot products of position vectors from the sphere's center to the vertices. A supplemental cosine law, associated with the polar triangle (formed by the poles of the original sides' great circles), provides an alternative expression: \cos c = -\cos a \cos b + \sin a \sin b \cos C. This arises from the duality between a triangle and its polar, where sides and supplementary angles interchange roles, offering utility in certain derivations. Adaptations of the , known as Napier's analogies, facilitate computations involving half-angles and differences, particularly useful for solving ambiguous cases. These include relations such as \frac{\tan\left(\frac{A+B}{2}\right)}{\tan\left(\frac{C}{2}\right)} = \frac{\cos\left(\frac{a-b}{2}\right)}{\cos\left(\frac{a+b}{2}\right)} and \frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)} = \frac{\sin\left(\frac{a-b}{2}\right)}{\sin\left(\frac{a+b}{2}\right)}, with cyclic variants for the other elements. In , specifying three elements of a generally determines it up to its antipodal counterpart on the opposite side of the sphere, as great circles intersect at diametrically opposite points; however, for small triangles where the excess is negligible, the geometry approximates that of planar triangles as the sphere's approaches .

Right-angled spherical triangles

In a right-angled spherical triangle, the right angle is conventionally placed at C, with legs a and b adjacent to C and c opposite C; all sides are measured as angular distances on the sphere. Unlike plane right triangles, the sum of the other two angles A and B exceeds 90°, with the spherical excess given by E = A + B - π/2. Napier's mnemonics, introduced in 1614, facilitate solving these triangles by considering five "circular parts": the legs a and b, the complement of the hypotenuse (90° - c), and the complements of the opposite angles (90° - A and 90° - B), arranged sequentially around a for mnemonic recall. The two governing rules are: the sine of any middle part equals the product of the tangents of its two adjacent parts, and the sine of any middle part equals the product of the cosines of its two opposite parts. These yield key formulas, such as: \sin a = \sin c \sin A (for the side opposite angle A), \cos c = \cos a \cos b (for the hypotenuse in terms of the legs), and \tan a = \tan c \cos B (relating a to the and adjacent ). The spherical law of sines verifies these relations in the limit of small approaching plane trigonometry. To solve when the c and a a are given, first compute A using \sin A = \sin a / \sin c, then find the other b from \cos b = \cos c / \cos a, and finally B from \sin B = \sin b / \sin c. When the two a and b are given, compute the from \cos c = \cos a \cos b, then find A from \tan A = \tan a / \sin b (or equivalently \cot A = \cot a \sin b), and B from \tan B = \tan b / \sin a. The polar triangle of a right-angled spherical triangle is also right-angled, with its sides as the supplements of the original angles (π - A, π - B, π/2) and its angles as the supplements of the original sides, providing duality for complementary solutions. For example, consider a right-angled spherical triangle with hypotenuse c = 90° and leg a = 60°. Then \sin A = \sin 60^\circ / \sin 90^\circ = \sqrt{3}/2, so A = 60°; \cos b = \cos 90^\circ / \cos 60^\circ = 0, so b = 90°; and \sin B = \sin 90^\circ / \sin 90^\circ = 1, so B = 90°. The excess is E = 60° + 90° - 90° = 60°, confirming the spherical nature.

SSS case

In the SSS case for spherical triangles, all three sides a, b, and c (measured as central angles in radians) are given, and the task is to determine the opposite angles A, B, and C. This configuration leverages the to compute the angles directly from the sides. The formula for angle C opposite side c is \cos C = \frac{\cos c - \cos a \cos b}{\sin a \sin b}, with similar expressions for \cos A and \cos B by cyclic permutation. Angles are then found using the arccosine function, ensuring each lies between 0 and \pi radians. For a valid unique solution, the sides must satisfy strict conditions: each side $0 < a, b, c < \pi, and their sum a + b + c < 2\pi. These constraints prevent ambiguities such as antipodal or supplemental configurations, where a different triangle on the sphere might satisfy the same side lengths (though such ambipolar cases are rare and typically ruled out by the sum condition). Additionally, triangle inequalities must hold in the spherical sense: a + b > c, a + c > b, b + c > a, and the excesses |a - b| < c < a + b (adjusted for the sphere's curvature). Under these, the SSS congruence theorem guarantees a unique spherical triangle up to congruence. The solution proceeds in clear steps: first, verify the side conditions to ensure validity. Next, apply the spherical law of cosines to compute all three angles. For verification, compute the spherical excess E = A + B + C - \pi (in radians), which must be positive and less than \pi; this also relates to the triangle's area via Girard's theorem as \text{Area} = E R^2, where R is the sphere's radius. Optionally, confirm consistency using the spherical law of sines: \frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}. This process contrasts with the planar SSS case, where small spherical triangles approximate flat ones using the ordinary law of cosines without excess. Consider an example with sides a = 80^\circ, b = 90^\circ, c = 100^\circ (converted to radians for computation: approximately 1.396, 1.571, and 1.745 radians, respectively). First, compute \cos C = \frac{\cos 1.745 - \cos 1.396 \cos 1.571}{\sin 1.396 \sin 1.571}. Since \cos 1.571 \approx 0 and \sin 1.571 \approx 1, this simplifies to \cos C \approx \cos 1.745 / \sin 1.396 \approx -0.1736 / 0.9848 \approx -0.1763, so C \approx \arccos(-0.1763) \approx 1.749 radians or about $100.2^\circ. Similarly, \cos A = \frac{\cos 1.396 - \cos 1.571 \cos 1.745}{\sin 1.571 \sin 1.745} \approx \frac{-0.1736 - 0 \cdot (-0.1736)}{1 \cdot 0.9848} \approx -0.1763, yielding A \approx 100.2^\circ. By symmetry or full computation, B \approx 79.6^\circ. The excess E \approx (100.2^\circ + 79.6^\circ + 100.2^\circ) - 180^\circ \approx 100^\circ or 1.745 radians, confirming the solution (area \approx 1.745 R^2). Verification via law of sines holds, as ratios match approximately 0.0173.

SAS case

In the SAS case for spherical triangles, two sides and the included angle are given, typically denoted as sides a and b with included angle C. This configuration allows direct computation of the third side c using the for sides: \cos c = \cos a \cos b + \sin a \sin b \cos C This formula accounts for the curvature of the sphere, differing from the by the inclusion of the sine terms. Once c is known, the non-included angles A and B are found by rearranging the spherical law of cosines for sides to solve for the cosines of the angles: \cos A = \frac{\cos a - \cos b \cos c}{\sin b \sin c}, \quad \cos B = \frac{\cos b - \cos a \cos c}{\sin a \sin c} Alternatively, after computing one angle (e.g., A), the law of sines can be applied to find the other: \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} or the spherical law of cosines for angles may be used, though the rearranged side formulas are typically more direct for this case. The solution is unique for spherical triangles with small excess (where the sum of angles exceeds \pi radians by a small amount), as the spherical SAS congruence theorem guarantees that two such triangles with equal corresponding elements are congruent. To solve, first compute c from the given a, b, and C. Then determine A and B using the rearranged cosine formulas. The third angle follows directly without adjustment for the planar sum, as the spherical excess E = A + B + C - \pi (in radians) is implicitly accounted for in the computations; E can be verified post-solution if needed. For example, consider a = 50^\circ, b = 60^\circ, C = 70^\circ. Compute \cos c = \cos 50^\circ \cos 60^\circ + \sin 50^\circ \sin 60^\circ \cos 70^\circ \approx 0.6428 \times 0.5 + 0.7660 \times 0.8660 \times 0.3420 \approx 0.5482, so c \approx \cos^{-1}(0.5482) \approx 56.8^\circ. Then \cos A = (\cos 50^\circ - \cos 60^\circ \cos 56.8^\circ)/(\sin 60^\circ \sin 56.8^\circ) \approx (0.6428 - 0.5 \times 0.5482)/(0.8660 \times 0.8355) \approx 0.5093, so A \approx \cos^{-1}(0.5093) \approx 59.4^\circ. Similarly, \cos B = (\cos 60^\circ - \cos 50^\circ \cos 56.8^\circ)/(\sin 50^\circ \sin 56.8^\circ) \approx (0.5 - 0.6428 \times 0.5482)/(0.7660 \times 0.8355) \approx 0.2309, so B \approx \cos^{-1}(0.2309) \approx 76.7^\circ. The excess is approximately $59.4^\circ + 76.7^\circ + 70^\circ - 180^\circ = 26.1^\circ. To arrive at these values, evaluate the trigonometric functions using standard calculators or tables, apply the inverse cosine for each angle, and sum for verification. The area of the spherical triangle can be computed using the generalized formula \frac{1}{2} R^2 \sin a \sin b \sin C for small triangles approximating the planar case, where R is the sphere's radius and angles/sides are in radians for consistency, though the exact area is R^2 E with E the excess in radians.

SSA case

In spherical trigonometry, the SSA case involves determining a triangle given two sides and the angle opposite one of them, often leading to ambiguity due to the periodic nature of the sine function and the geometry of the sphere. Unlike the planar case, where ambiguity arises from the possible positions of the non-included side relative to a perpendicular height, spherical constraints such as side lengths limited to less than \pi radians and the total perimeter less than $2\pi radians can reduce the likelihood of multiple valid solutions, though up to two triangles may still exist. This ambiguity is analogous to the planar SSA condition but requires additional verification for spherical validity, including positive spherical excess E = A + B + C - \pi > 0. The arcsin yields ; supplement checked for validity. The solution begins with the : \frac{\sin A}{ \sin a } = \frac{\sin B}{ \sin b }, yielding \sin B = \frac{ \sin b \cdot \sin A }{ \sin a }. If \sin B > 1, no solution exists; if \sin B = 1, a unique right-angled solution at B may apply; otherwise, two potential values for B emerge: an acute angle B_1 = \arcsin(\sin B) and its supplement B_2 = \pi - B_1, both satisfying the equation. For each possible B, the third angle C is computed using the for angles (requiring side c first or via analogies), but practically, compute C = \arccos\left( -\cos A \cos B + \sin A \sin B \cos c \right) after finding c. The third side c is computed via the : \cos c = \cos a \cos b + \sin a \sin b \cos C, but since C depends on c, iterate or use after provisional C; alternatively, use Napier's analogies for efficiency. This process must be repeated for both B_1 and B_2. Validity checks are essential to discard invalid configurations. Each candidate triangle must satisfy: all sides less than \pi; the sum of sides less than $2\pi; positive spherical excess; and no negative cosine values indicating impossible geometries. Unlike the planar case, where a simple height comparison suffices, spherical SSA requires numerical evaluation of these conditions, as the curvature precludes straightforward geometric tests. If \sin B > 1, the case is invalid; for valid pairs, spherical triangle inequalities (e.g., each side less than the sum of the other two) must hold on the sphere. Select arcsin/supplement ensuring \cos B consistent with . For example, consider A = 40^\circ, a = 50^\circ, b = 60^\circ (angles and sides in degrees, as angular measures). Applying the law of sines gives \sin B \approx 0.7267 < 1, yielding two possible B \approx 46.7^\circ and B \approx 133.3^\circ. For each, compute C and c using the cosine law or analogies, then verify conditions: the acute B yields a valid triangle with c \approx 61.5^\circ and E > 0, while the obtuse B produces an invalid configuration (e.g., E < 0 or sum > 2\pi), discarding it. Resolution involves calculating both possibilities and retaining only those satisfying spherical constraints, potentially resulting in zero, one, or two triangles.

ASA case

In spherical trigonometry, the ASA case refers to solving a spherical triangle given two angles, say A and B, and the included side c (the side opposite the third angle C). This configuration allows direct computation of the third angle using the spherical law of cosines for angles, followed by determination of the remaining sides via the spherical law of sines. Unlike the planar case, the sum of angles exceeds \pi radians (or $180^\circ), introducing the spherical excess E = A + B + C - \pi > 0, which must be verified for the triangle's existence on the sphere. The third angle C is found using the formula \cos C = -\cos A \cos B + \sin A \sin B \cos c, where angles and sides are in radians or degrees consistently, and C is obtained as C = \arccos(\cdot) with $0 < C < \pi. This equation, known as the polar or supplemental form of the spherical law of cosines, accounts for the curvature of the sphere. With C determined, the spherical excess E is calculated to ensure E > 0, confirming the triangle is valid (for small triangles approaching the planar limit, E \approx 0). The remaining sides a (opposite A) and b (opposite B) are then computed using the spherical : \sin a = \sin c \cdot \frac{\sin A}{\sin C}, \quad \sin b = \sin c \cdot \frac{\sin B}{\sin C}, yielding a = \arcsin(\sin a) and b = \arcsin(\sin b), selecting the principal value or supplement consistent with the spherical law of cosines for sides (e.g., compute \cos a = \cos b \cos c + \sin b \sin c \cos A after provisional b, ensuring \cos a > 0 for acute cases). This approach is unambiguous, as the included side and adjacent angles uniquely determine the triangle up to congruence, provided the excess condition holds. For example, consider A = 50^\circ, B = 60^\circ, and included side c = 70^\circ. First, \cos C = -\cos 50^\circ \cos 60^\circ + \sin 50^\circ \sin 60^\circ \cos 70^\circ \approx -0.0945, so C \approx 95.4^\circ. The excess is E = 50^\circ + 60^\circ + 95.4^\circ - 180^\circ = 25.4^\circ > 0, validating the solution. Then, \sin a \approx \sin 70^\circ \cdot \frac{\sin 50^\circ}{\sin 95.4^\circ} \approx 0.7233, \quad a \approx \arcsin(0.7233) \approx 46.2^\circ (verified \cos a > 0); similarly, \sin b \approx 0.8184, b \approx 54.9^\circ. These values satisfy the laws and demonstrate the method's application in astronomical or computations.

AAS case

In the AAS case for spherical triangles, two angles A and B and the side a opposite angle A are given, with all quantities expressed in angular measure (radians or degrees, typically 0 to \pi). This configuration allows determination of the remaining elements using the spherical laws of sines and cosines, potentially with ambiguity resolution. The first step is to compute the side b opposite B via the spherical law of sines: \frac{\sin b}{\sin B} = \frac{\sin a}{\sin A} \implies \sin b = \frac{\sin a \cdot \sin B}{\sin A}. If \sin b > 1, no solution exists; if \sin b = 1, b = \pi/2; otherwise, two potential values arise: b = \arcsin\left( \frac{\sin a \cdot \sin B}{\sin A} \right) (the acute or ) and its supplement b' = \pi - b, provided both satisfy spherical conditions such as a + b > c (to be verified later) and the sum of sides less than $2\pi. Select based on \cos b consistency. For each valid b, the third angle C and side c are found using the correct Napier's analogies or . Using the second analogy: \tan\left( \frac{C}{2} \right) = \tan\left( \frac{A - B}{2} \right) \cdot \frac{\sin\left( \frac{a + b}{2} \right)}{\sin\left( \frac{a - b}{2} \right)}, and for side c, from the first analogy (rearranged): \tan\left( \frac{c}{2} \right) = \tan\left( \frac{a + b}{2} \right) \cdot \frac{\cos\left( \frac{A - B}{2} \right)}{\cos\left( \frac{A + B}{2} \right)} \cdot \tan\left( \frac{C}{2} \right), but practically, compute C first then c = 2 \arctan(\cdots). Alternatively, compute \sin C = \sin (A + B + E - \pi), but use cosine law: first provisional c from law sines \sin c = \sin a \sin C / \sin A after C, but iterate. Standard: after b, use for angle C: \cos C = -\cos A \cos B + \sin A \sin B \cos c, but solve numerically or use analogies. Validity is confirmed by computing the spherical excess E = A + B + C - \pi > 0 and ensuring no side exceeds \pi or violates spherical inequalities (e.g., each pair of sides greater than the third, adjusted for the sphere). Unlike the planar AAS case, which always yields a , the spherical version may produce zero, one, or two triangles due to the ambiguity in b and global constraints, though small triangles approximate planar . The arcsin/supplement for c chosen for . As a representative example, consider A = 30^\circ, B = 50^\circ, a = 40^\circ. Then \sin b \approx 0.8572, so b \approx 59.0^\circ or b' \approx 121.0^\circ. For b \approx 59.0^\circ, correct analogies or give c \approx 72.5^\circ and C \approx 100.0^\circ (adjusted for exact), with E \approx 0.0^\circ + E > 0 (small, \approx 5.3^\circ after precise calc), forming a valid . The supplementary b' may or may not yield another valid depending on the specific values, requiring verification.

AAA case

In spherical geometry, the AAA case—where all three angles A, B, and C of a triangle are given—determines the triangle uniquely up to , in contrast to where it specifies only similarity with infinitely many possible sizes. This uniqueness arises because the spherical excess E = A + B + C - \pi (in radians) fixes the relative area of the triangle on the sphere, constraining both shape and scale for a given spherical . The area of the triangle is E R^2, where R is the sphere's ; for the unit sphere (R = 1), the area equals E exactly. To solve for the sides a, b, and c (opposite angles A, B, and C, respectively), the standard method employs the polar triangle, a dual construction on the sphere. The polar triangle A'B'C' has sides a' = \pi - A, b' = \pi - B, and c' = \pi - C, making it an SSS case with known sides. The angles of the polar triangle relate to the original sides by A' = \pi - a, B' = \pi - b, and C' = \pi - c. The polar angles are found using the spherical law of cosines rearranged for angles from sides: \cos A' = \frac{\cos a' - \cos b' \cos c'}{\sin b' \sin c'} with A' = \arccos(\cdot); analogous formulas apply for B' and C'. The original sides are then a = \pi - A', b = \pi - B', and c = \pi - C'. This process yields the absolute angular measures of the sides, assuming a unit sphere; for other radii, physical arc lengths scale with R, but angular sides remain fixed. For instance, consider angles A = 60^\circ, B = 60^\circ, C = 120^\circ (so E = 60^\circ or \pi/3 radians). The polar sides are a' = 120^\circ, b' = 120^\circ, c' = 60^\circ. Then, \cos A' = \frac{\cos 120^\circ - \cos 120^\circ \cos 60^\circ}{\sin 120^\circ \sin 60^\circ} = \frac{-1/2 - (-1/2)(1/2)}{(\sqrt{3}/2)(\sqrt{3}/2)} = \frac{-1/4}{3/4} = -1/3, so A' = \arccos(-1/3) \approx 109.47^\circ and a = 180^\circ - 109.47^\circ \approx 70.53^\circ. By symmetry, b \approx 70.53^\circ. For the third, \cos C' = \frac{\cos 60^\circ - \cos 120^\circ \cos 120^\circ}{\sin 120^\circ \sin 120^\circ} = \frac{1/2 - (-1/2)(-1/2)}{3/4} = \frac{1/4}{3/4} = 1/3, so C' = \arccos(1/3) \approx 70.53^\circ and c = 180^\circ - 70.53^\circ \approx 109.47^\circ. The solution is unique up to reflection on the sphere (homothety not applicable, as scale is fixed by the geometry).

Applications

Triangulation methods

Triangulation in land surveying relies on dividing a region into a connected network of triangles, beginning with a precisely measured baseline between two known points. Angles are then measured at these points to additional stations, allowing the determination of unknown side lengths through solutions to ASA (angle-side-angle) or AAS (angle-angle-side) triangle cases using trigonometric relations. This method gained prominence historically through the of India, launched by William Lambton in 1802 near Madras and extended under George Everest's leadership until 1871, which systematically mapped over 2,400 kilometers of the subcontinent's arc to establish accurate geodetic control. The surveying process commences with establishing a , a directly measured side using instruments like tapes or electronic distance meters for high precision. From the endpoints, horizontal and vertical angles are observed to subsequent points using theodolites, forming initial . Distances to new points are then calculated by applying the within each , propagating the known length through a chain of interconnected to cover the area. In such networks, angular measurements achieve greater precision (often to seconds of arc) than the derived side lengths, but errors propagate cumulatively as computations extend outward, with angular inaccuracies amplifying distances in remote triangles. Adjustments, such as least-squares optimization and corrections for or minor curvature effects, are applied in larger networks to minimize distortions. Today, while (GPS) technologies provide direct coordinate fixes and supplement by establishing initial control points with sub-meter accuracy over long , the core principles of underpin cadastral mapping for property boundaries and , ensuring robust horizontal networks. A practical example involves estimating a mountain's : surveyors measure a on level plain below the peak and record elevation angles from each end to the summit, approximating right-angled triangles to compute the vertical rise via relations, as done in early surveys over distances up to 160 kilometers. The plane approximation in triangulation holds for small areas where Earth's curvature is negligible, typically for areas up to about 250 square kilometers (linear extents around 15-20 km), beyond which spherical trigonometry becomes necessary to avoid systematic distortions in positions and distances.

Geodesic distance calculations

Points on the Earth's surface are commonly specified using spherical coordinates, namely latitude \phi (ranging from -90^\circ to $90^\circ) and longitude \lambda (ranging from -180^\circ to $180^\circ). The great-circle distance, which represents the shortest path between two such points on a sphere, can be computed using the haversine formula: d = 2R \arcsin\left(\sqrt{\sin^2\left(\frac{\Delta\phi}{2}\right) + \cos\phi_1 \cos\phi_2 \sin^2\left(\frac{\Delta\lambda}{2}\right)}\right), where R is the Earth's mean radius (approximately 6371 km), \Delta\phi = \phi_2 - \phi_1, and \Delta\lambda = \lambda_2 - \lambda_1. This formula derives from the spherical law of cosines and avoids issues with numerical instability near small distances. To frame this calculation within the solution of spherical triangles, consider two points P_1(\phi_1, \lambda_1) and P_2(\phi_2, \lambda_2), along with the N. These form a spherical triangle N-P_1-P_2 with sides a = 90^\circ - \phi_2 ( of P_2), b = 90^\circ - \phi_1 ( of P_1), and included angle C = |\Delta\lambda| at N. The third side c corresponds to the \delta between P_1 and P_2, solved using the spherical SAS case via the cosine law: \cos c = \cos a \cos b + \sin a \sin b \cos C, which simplifies to \cos \delta = \sin \phi_1 \sin \phi_2 + \cos \phi_1 \cos \phi_2 \cos \Delta\lambda. The linear distance is then d = R \delta (with \delta in radians). The initial azimuth (bearing) from P_1 to P_2 is the angle at P_1 in the triangle, computed using the spherical : \sin A = \frac{\sin \Delta\lambda \cos \phi_2}{\sin \delta}, with quadrant adjustments based on the signs of \sin A and \cos A = \frac{\sin \phi_2 - \sin \phi_1 \cos \delta}{\cos \phi_1 \sin \delta}. This yields the direction for along the . For greater accuracy on the oblate , Vincenty's formulae provide an iterative solution for distances on an , reducing errors to within 0.5 mm compared to spherical approximations. These account for the 's and are implemented in the to find distances and azimuths between latitude-longitude pairs. As an example, consider the distance from (\phi_1 = 40.7^\circ N, \lambda_1 = -74.0^\circ) to (\phi_2 = 51.5^\circ N, \lambda_2 = -0.1^\circ), with \Delta\lambda \approx 73.9^\circ. Using the cosine law, \cos \delta = \sin 40.7^\circ \sin 51.5^\circ + \cos 40.7^\circ \cos 51.5^\circ \cos 73.9^\circ \approx 0.607, so \delta \approx 52.6^\circ or d \approx 5570 km (with R = 6371 km). The initial azimuth from New York is approximately $52^\circ east of north. These methods underpin applications in route planning, where great-circle paths minimize fuel use, and GPS systems, which compute positions via the (solving for locations given distances and bearings). Unlike planar approximations, which treat the as flat and introduce errors exceeding 0.25% (over 250 m) for distances beyond 100 km, spherical and ellipsoidal models properly account for .