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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.
Microsoft Math contains features that are designed to assist in solving mathematics, science, and tech-related problems, as well as to educate the user. The application features such tools as a graphing calculator and a unit converter. It also includes a triangle solver and an equation solver that provides step-by-step solutions to each problem.
In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of any triangle to the sines of its angles.According to the law, = = =, where a, b, and c are the lengths of the sides of a triangle, and α, β, and γ are the opposite angles (see figure 2), while R is the radius of the triangle's circumcircle.
Case 3: two sides and an opposite angle given (SSA). The sine rule gives C and then we have Case 7. There are either one or two solutions. Case 4: two angles and an included side given (ASA). The four-part cotangent formulae for sets (cBaC) and (BaCb) give c and b, then A follows from the sine rule. Case 5: two angles and an opposite side given ...
In geometry, calculating the area of a triangle is an elementary problem encountered often in many different situations. The best known and simplest formula is T = b h / 2 , {\displaystyle T=bh/2,} where b is the length of the base of the triangle, and h is the height or altitude of the triangle.
The only triangles with one angle being twice another and having integer sides in arithmetic progression are acute: namely, the (4,5,6) triangle and its multiples. [6] There are no acute integer-sided triangles with area = perimeter, but there are three obtuse ones, having sides [7] (6,25,29), (7,15,20), and (9,10,17).
Given any triangle ABC, and any point M on BC, construct the incircle and circumcircle of the triangle. Then construct two additional circles, each tangent to AM, BC, and to the circumcircle. Then their centers and the center of the incircle are collinear. [3] [4] Until 2003, academia thought this third problem of Thébault the most difficult ...
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.