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Solution of triangles (Latin: solutio triangulorum) is the main trigonometric problem of finding the characteristics of a triangle (angles and lengths of sides), when some of these are known. The triangle can be located on a plane or on a sphere. Applications requiring triangle solutions include geodesy, astronomy, construction, and navigation.
In trigonometry, Mollweide's formula is a pair of relationships between sides and angles in a triangle. [1] [2] A variant in more geometrical style was first published by Isaac Newton in 1707 and then by Friedrich Wilhelm von Oppel in 1746. Thomas Simpson published the now-standard expression in 1748.
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
The chapter on areas includes both trigonometric formulas and Heron's formula for computing the area of a triangle from its side lengths, and the chapter on inequalities includes the Erdős–Mordell inequality on sums of distances from the sides of a triangle and Weitzenböck's inequality relating the area of a triangle to that of squares on ...
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord.
Another approach is to split the triangle into two right-angled triangles. For example, take the Case 3 example where b, c, and B are given. Construct the great circle from A that is normal to the side BC at the point D. Use Napier's rules to solve the triangle ABD: use c and B to find the sides AD and BD and the angle ∠BAD.