Saturday 13 October 2012

Career Mania 55: GyanCentral - The hub for engineering and law students - IIT-JEE, AIEEE, BITSAT, CLAT, AILET - 2012: Solution of triangles

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GyanCentral - The hub for engineering and law students - IIT-JEE, AIEEE, BITSAT, CLAT, AILET - 2012: Solution of triangles
Oct 13th 2012, 15:54

GyanCentral - The hub for engineering and law students - IIT-JEE, AIEEE, BITSAT, CLAT, AILET - 2012
The source for all engineering and legal education news in India
Solution of triangles
Oct 13th 2012, 15:35

In a triangle ABC, the angles are denoted by capital letters, A, B and C and the lengths of the sides opposite these angles are denoted by a, b, c respectively. Semi-perimeter of the triangle is written as s = a + b + c / 2 , and its area by S or Δ. Let R be the radius of the circumcircle of the ΔABC. Basic Formulae (i) Sine rule sin A/a = sin B/b = sinC/c = 1/2R = 2Δ/abc. . (ii) Cosine rule cos A = b2 + c2 - a2 /2abc cosB = a2 + c2 - b2/2ac , cosC = a2 + b2 - c2/2ab . (iii) Trigonometric ratios of half-angles: sin A/2 = √(s-b) (s-c)/bc, cos B/2 = √s(s - b)/ca, tan C/2 = √ (s - a) (s - b)/s(s - c) . (iv) Projection rule: a = b cosC + c cosB, b = c cosA + a cosC, c = a cosB + b cosA. (v) Area of a triangle Δ = 1/2 bc sin A = 1/2 ca sinB = 1/2 ab sin C = √s(s - a) (s - b) (s - c) = abc/4R = rs. (vi) Napier's analogy tan B - C/2 = b - c/b +c cot A/2 tan C - A/2 = c -a/ c + a cot B/2 , tan A - B/2 = a - b /a + b , cot c/2. (vii) m-n theorem If D be the point on the side BC of a triangle ABC which divide the side BC in the ratio m : n, then with respect to mentioned figure, we have: (i) (m + n) cot θ = m cot α – n cot ß. (ii)(m + n) cot θ = n cot B – m cot C. (viii) Apollonius theorem In a triangle ABC, AD is median through A, then AB2 + AC2 = 2(AD2+BD2). Process of Solution of Triangles The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles. (i) If the three sides a, b, c are given, angle A is obtained from tan A/2 = √(s - b) (s - c) / s(s - a) or cos A = b2 + c2 - a2 / 2bc . B and C can be obtained in the similar way. (ii) If two sides b and c and the included angle A are given, then tan B - c/2 = b - c/ b + c cot A/2 gives B - C/2. Also B = C/2 = 90o - A/2, so that B and C can be evaluated. The third side is given by a = b sin A/sin B or a2 = b2 + c2 – 2bc cosA. (iii) If two sides b and c and the angle B (opposite to side b) are given, then sin C = c/b sinB, A = 180o – (B + C) and b = b sin A/sinB give the remaining elements. If b < c sin B, there is no triangle possible . If b = c sinB and B is an acute angle, then only one triangle is possible. If c sinB < b < c and B is an acute angle, then there are two values of angle C . If c < b and B is an acute angle, then there is only one triangle . Alternative Method: By applying cosine rule, we have cosB = a2 + c2 - b2/2ac ⇒ a2 – (2c cosB)a + (c2 – b2) = 0 ⇒ a = c cosB + √(C cosB)2 - (c2 - b2) ⇒ a = c cosB + √b2 - (c sinB)2. This equation leads to the following cases: Case –I If b < c sinB, no triangle is possible. Case –II Let B = c sinB. There are further the following cases: (a) B is an obtuse angle ⇒ cosB is negative. There exists no such triangle. (b) B is an acute angle ⇒ cosB is positive. There exists only one triangle. Case –III Let b < c sin B. There are further following cases: (a) B is an acute angle ⇒ cosB is positive. In this case two values of a wall exists if and only if c cosB > √b2 - (c sinB)2 or c > b ⇒ Two triangles are possible. If c < b, only one triangle is possible. (b) B is an obtuse angle ⇒ cos B is negative. In this case triangle will exists if and only if √b2 - (c sinB)2 > ,c cosB, ⇒ b > c. So in this case only one triangle is possible. If b < c there exists no triangle. If one side a and angles B and C are given, then A = 180o – (B + C), and b = a sinB/sinA, c = a sinC/sinA . If the three angles A, B, C are given, we can only find the ratios of the sides a, b, c by using the sine rule (since there are infinite similar triangles possible).

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