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Analytical and Numerical Solving of Right Triangles with Given Angle and Altitude, Bisector Or Median of a Triangle Poland Problems of Applied Sciences, 2017, Vol. 7, pp. 031 – 046 Szczecin dr n. tech. Andrzej Antoni CZAJKOWSKI a, mgr in ż., naucz. dypl. Sławomir Wawrzyniec SNASTIN a naucz. dypl. Lech Adam REWKOWSKI b a Higher School of Technology and Economics in Szczecin, Faculty of Automotive Systems Wy ższa Szkoła Techniczno-Ekonomiczna w Szczecinie, Wydział Systemów Automotive b 18 High School in the Complex School No. 5 in Szczecin 18 Liceum Ogólnokształc ące w Zespole Szkół Nr 5 w Szczecinie ANALYTICAL AND NUMERICAL SOLVING OF RIGHT TRIANGLES WITH GIVEN ANGLE AND ALTITUDE, BISECTOR OR MEDIAN OF A TRIANGLE Abstract Introduction and aims: The paper shows the analytical models of solving right triangles with appropriate discussion. For right triangles have been discussed six cases taking into account the altitude, bisector and median of a triangle. The main aim of this paper is not only to create some analytical algorithms for solving right triangle, but also their implementation in programs MS- Excel, MathCAD and Mathematica. Material and methods: Elaboration of six analytical cases of solving right triangles has been made on the basis of the relevant trigonometric properties occurring in a right triangle. In the paper have been used some analytical and numerical methods by using MS-Excel, MathCAD and Mathematica programs. Results: As some results have been obtained numerical algorithms in the programs MS-Excel, MathCAD and Mathematica for six analytical cases of solving right triangles taking into account the altitude, bisector and median of a triangle. Conclusion: Created numerical algorithms of solving the right triangles in the programs MS- Excel, MathCAD and Mathematica allow for faster significant performance calculations than the traditional way of using logarithms and logarithmic tables. Keywords: Trigonometry, right triangle, angle, bisector, median, triangle altitude, solving, numerical algorithms, MS-Excel, MathCAD, Mathematica. (Received: 01.10.2017; Revised: 15.10.2017; Accepted: 25.10.2017) ANALITYCZNO-NUMERYCZNE ROZWI ĄZYWANIE TRÓJK ĄTÓW PROSTOK ĄTNYCH GDZIE DANE S Ą K ĄT I WYSOKO ŚĆ , DWUSIECZNA LUB ŚRODKOWA TRÓJKĄTA Streszczenie Wst ęp i cele: W pracy pokazano analityczne modele rozwi ązywania trójk ątów prostok ątnych wraz z odpowiedni ą dyskusj ą. Dla trójk ątów prostok ątnych omówiono sze ść przypadków z uwzgl ędnie- niem wysoko ści, dwusiecznej i środkowej trójk ąta. Głównym celem jest pracy jest nie tylko utwo- rzenie algorytmów analitycznych rozwi ązywania takich trójk ątów lecz równie ż ich implementacja w programach MS-Excel, MathCAD i Mathematica. Materiał i metody: Opracowanie sze ściu analitycznych przypadków rozwi ązywania trójk ątów prostok ątnych wykonano opieraj ąc si ę odpowiednich własno ściach trygonometrycznych wyst ępu- jących w trójk ącie prostok ątnym. Zastosowano metod ę analityczn ą i numeryczn ą wykorzystuj ąc programy MS-Excel, MathCAD i Mathematica. Wyniki: Otrzymano algorytmy numeryczne w programach MS-Excel, MathCAD i Mathematica dla sze ściu analitycznych przypadków rozwi ązywania trójk ątów prostok ątnych z uwzgl ędnieniem wysoko ści, dwusiecznej i środkowej trójk ąta. Wniosek: Utworzone algorytmy numeryczne rozwi ązywania trójk ątów prostok ątnych w progra- mach MS-Excel, MathCAD oraz Mathematica, pozwalaj ą na znaczne szybsze wykonanie oblicze ń ni ż drog ą tradycyjn ą z u życiem logarytmów i tablic logarytmicznych. Słowa kluczowe: Trygonometria, trójk ąt prostok ątny, kąt, dwusieczna, wysoko ść trójk ąta, algo- rytmy numeryczne, MS-Excel, MathCAD, Mathematica. (Otrzymano: 01.10.2017; Zrecenzowano: 15.10.2017; Zaakceptowano: 25.10.2017) © A.A. Czajkowski, S.W. Snastin, L.A. Rewkowski 2017 Trigonometry / Trygonometria A.A. Czajkowski, S.W. Snastin, L.A. Rewkowski 1. Introduction to trigonometry 1.1. Definitions of trigonometric functions Let be given the right triangle ABC (angle ACB is the right angle) (Fig. I). B β c a • α C A b Fig. I. The right triangle with catheti a and b and hypotenuse c Source: Elaboration of the Authors where a, b, c mean respectively the length of shorter cathetus, greater cathetus and hypote- nuse. The sides a, b and c satisfy the condition: a2 + b2 = c2 (1) and acute angles α and β satisfying the following condition [3],[4]: π α =β+ . (2) 2 Definition 1. The sine of acute angle is the ratio of the length of cathetus opposite this angle to the length of hypotenuse, which express the formulae [7], [10]: a b sin( α) = , sin( β) = . (3) c c Definition 2 . The cosine of acute angle is the ratio of the length of cathetus adjacent this angle to the length of hypotenuse, which express the formulae [7], [10]: b a cos( α) = , cos( β) = . (4) c c Definition 3 . The tangent of acute angle is the ratio of the length of cathetus opposite this angle to the length of cathetus adjacent to that angle, which express the formulae [7], [10]: a b tg( α) = , tg( β) = . (5) b a Definition 4 . The cotangent of acute angle is the ratio of the length of cathetus adjacent this angle to the length of cathetus opposite to that angle, which express the formulae [7], [10]: 32 Analytical and numerical solving of right triangles with given angle and altitude, bisector or median of a triangle b a ctg( α) = , ctg( β) = . (6) a b Theorem 1 . (Basic relationships in the triangle) In any length of each side of a triangle is less than the sum of the length of the other side and higher than the absolute value of the difference in length of the sides [11], [13]-[16]: | c b | b − c | < a < b + ,c (7) | c a | a − c | < b < a + ,c (8) | b a | a − b | < c < a + b . (9) 1.2. Solving right triangles with given angle and altitude, bisector or median of a triangle For the right triangle ABC shown in figure I, there are six cases of solving which are given below (Tab. 1). Tab. 1. The discussion of solutions for the right triangles Case: Data: Solution: π Ac Ac 2A c 1 A β = − α a = b = c = c cos( α) sin( α) sin(2 α) α 2 π Ac Ac 2A c 2 A α = − β a = b = c = c sin( β) cos( β) sin(2 β) β 2 π α π α − π π α Bβ cos Bβ cos − β = − α = − 4 2 4 2 3 Bβ a Bβ cos b = = 2 4 2 α c α tg( ) sin( α) β π β β Bβ cos α = − β = = β 2 4 Bβ a Bβ cos b Bβ cos tg( ) c = 2 2 2 β β cos( ) 2M ⋅tg(α ) 2M 2M π a = b b = b c = b 5 Mb β = − α + 2 α 1+ 4tg2 ()α cos( α) 1+ 4tg2 ()α α 2 1 4tg () 2M ⋅ctg ()β 2M 2M π a = b b = b c = b 6 Mb α = − β + 2 β + 2 β sin( β) 1+ 4ctg2 ()β β 2 1 4ctg () 1 4ctg () Notes : Ac – an altitude for the triangle side c, Bβ – a bisector of the angle β, Mb – a median of the triangle side b. 33 A.A. Czajkowski, S.W. Snastin, L.A. Rewkowski 2. Solving of right triangles 2.1. Case 1: Altitude of a triangle and an angle by the greater cathetus Definition 5: The altitude of a triangle is the perpendicular dropped from a vertex onto the opposite side. Problem: Solve a right triangle with given the altitude of triangle and the angle by the big- ger cathetus. 2.1.1. Theoretical analysis Let be the right triangle as shown on the figure 1. B D β c a A c • α C A b Fig. 1. Right triangle ABC with given altitude Ac and the angle α Source: Elaboration of the Authors Data in ∆ABC: Length of the altitude Ac of a triangle and a measure of the angle α. Unknown in ∆ABC: Length of the triangle sides a, b and c and a measure of the angle β. Solution: ♦ From the properties of angles in a right triangle we obtain: π =β α− . (10) 2 ♦ From ∆BCD we obtain: A A A c = sin( β), a = c , a = c . (11) a sin( β) cos( α) ♦ From ∆ACD we have: A A c = sin( α), b = c . (12) b sin( α) ♦ From ∆ABC we know, that: a 2 + b2 = 2 ,c c = a 2 + b2 . (13) ♦ From the formulae (11)-(13) implies, that: A2 A2 A2[sin 2 (α) + cos 2 (α)] A A2 c = c + c = c = c = c . (14) sin 2 (α) cos 2 (α) sin 2 (α)cos 2 (α) sin( α)cos( α) sin( 2α) π A A A2 Answer: =β − α, a = c , b = c , c = c . 2 cos( α) sin( α) sin( 2α) 34 Analytical and numerical solving of right triangles with given angle and altitude, bisector or median of a triangle 2.1.2. Numerical algorithms in MS-Excel, MathCAD and Mathematica programs For numerical analysis we take into account: the altitude A c = 12,9 and angle α = 31 ° ♦ Program MS-Excel 7.0 [2],[8] Algorithm: Commentary: A6=12,9 Given length of the altitude A c A7=31 Given measure of the angle α [º] A8=A7*PI()/180 Command for the angle α [rad], α = 0,541 A9=90-A7 Command for the angle β [º] A10=A6/COS(A8) Command for calculation of the side a A11=A6/SIN(A8) Command for calculation of the side b A12=2*A6/SIN(2*A8) Command for calculation of the side c 59 Result: measure of the angle β [º] 15,0 Result: length of the side a 25,0 Result: length of the side b 29,2 Result: length of the side c ♦ Program MathCAD 15 [9],[12] Algorithm: Commentary: Ac:= 12.9 Given length of the altitude A c α0:= 31 Given measure of the angle α [º] α0 ⋅ π α := = 0.541052 Command for the angle α [rad] 180 β :0 = 90 − α0 = 95 Command for the angle β [º] Ac :a = = .51 049571 cos( α) Command and result: length of the side a Ac :b = = .25 046692 sin( α) Command and result: length of the side b Ac2 :a = = .29 220307 sin(2 α) Command and result: length of the side c ♦ Program Mathematica 7.0 [1],[5],[6] Algorithm: Commentary: Ac:= 12.9 Given length of the altitude A c A0:= 31 Given measure of the angle α [º] A:=A0*N[ π]/180 Command for the angle α [rad] α = 0.541036 B=90-A0 Command for the angle β [º] a=Ac/Cos[A] Command: length of the side a b=Ac/Sin[A] Command: length of the side b c=2*Ac/Sin[2*A] Command: length of the side c 59 Result: measure of the angle β [º] 15.0494 Result: length of the side a 25.0474 Result: length of the side b 29.2208 Result: length of the side c Numerical results: a =15,0 b = 25,0 c = 29,2 α = 59 °.
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