Properties of Triangle - Half Angle Formula
Duration: 18 min
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This video is a mathematics lecture that systematically presents the half-angle formulas for a triangle, derived from the semi-perimeter 's'. The instructor first writes out the general formulas for sin(A/2), cos(A/2), and tan(A/2) in terms of the sides a, b, c and the semi-perimeter s. The formulas are then extended to cover all three angles of the triangle. The lecture progresses to a worked example where the sides of a triangle are given as a=8, b=10, c=12, and the student is guided through the calculation of sin(A/2) and cos(A/2) using the derived formulas. The video concludes with two more problems: the first asks for the value of cot(B/2) * cot(C/2) given the condition 3a = b + c, and the second asks for the relationship between the sides a, b, and c given the values of tan(A/2) and tan(C/2). The entire lesson is delivered on a digital blackboard with the instructor visible in a small window, and the content is presented in a clear, step-by-step manner.
Chapters
0:00 – 2:00 00:00-02:00
The video begins by introducing the 'Half Angle formula' for a triangle. The instructor writes the formulas for sin(A/2), cos(A/2), and tan(A/2) on a digital blackboard. The formulas are presented as: sin(A/2) = √[(s-b)(s-c)/bc], cos(A/2) = √[s(s-a)/bc], and tan(A/2) = √[(s-b)(s-c)/s(s-a)]. The instructor explains that 's' is the semi-perimeter of the triangle, defined as s = (a+b+c)/2.
2:00 – 5:00 02:00-05:00
The instructor extends the half-angle formulas to cover all three angles of the triangle. The formulas for sin(B/2), sin(C/2), cos(B/2), cos(C/2), tan(B/2), and tan(C/2) are written out, showing the pattern of substituting the appropriate sides. For example, sin(B/2) = √[(s-a)(s-c)/ac] and tan(C/2) = √[(s-a)(s-b)/s(s-c)]. The instructor emphasizes that these formulas are derived from the standard half-angle identities and the Law of Cosines.
5:00 – 10:00 05:00-10:00
The video transitions to a worked example. The problem states: 'In ΔABC, the sides are a=8, b=10, c=12. Find the values of sin(A/2) and cos(A/2)'. The instructor first calculates the semi-perimeter s = (8+10+12)/2 = 15. Then, using the formulas, sin(A/2) = √[(15-10)(15-12)/(10*12)] = √[5*3/120] = √[15/120] = √[1/8] = 1/√8. Similarly, cos(A/2) = √[15(15-8)/(10*12)] = √[15*7/120] = √[105/120] = √[7/8].
10:00 – 15:00 10:00-15:00
The instructor presents the first problem: 'In ΔABC, if 3a = b + c, then the value of cot(B/2) * cot(C/2) is:'. The options are a) 1, b) 2, c) √3, d) √2. The instructor begins by writing the formulas for cot(B/2) and cot(C/2) as the reciprocals of tan(B/2) and tan(C/2). The product is then expressed as: cot(B/2) * cot(C/2) = √[(s-b)(s-c)/s(s-b)] * √[(s-a)(s-b)/s(s-c)] = √[(s-a)/s]. Using the given condition 3a = b + c, the instructor derives that 2s = a + b + c = a + 3a = 4a, so s = 2a. Substituting this into the expression gives √[(2a - a)/(2a)] = √[a/(2a)] = √[1/2] = 1/√2. The instructor notes that this is not among the options, indicating a possible error in the problem or options.
15:00 – 17:48 15:00-17:48
The second problem is presented: 'If tan(A/2) = 5/6 and tan(C/2) = 2/5, then find the relation between a, b, and c'. The instructor writes the formulas: tan(A/2) = √[(s-b)(s-c)/s(s-a)] = 5/6 and tan(C/2) = √[(s-a)(s-b)/s(s-c)] = 2/5. By squaring both equations and multiplying them, the instructor derives: [(s-b)(s-c)/s(s-a)] * [(s-a)(s-b)/s(s-c)] = (5/6) * (2/5) = 1/3. This simplifies to (s-b)^2 / s^2 = 1/3, so (s-b)/s = 1/√3. This leads to s - b = s/√3, and after rearranging, the relation a + c = 2b is found. The instructor confirms this by substituting s = (a+b+c)/2 into the equation.
The video provides a comprehensive lesson on the half-angle formulas for a triangle, starting with the derivation of the general formulas for sin(A/2), cos(A/2), and tan(A/2) in terms of the sides and semi-perimeter. The instructor then demonstrates the application of these formulas through a detailed worked example, calculating the values of sin(A/2) and cos(A/2) for a triangle with given side lengths. The lesson concludes with two more complex problems that require the student to apply the formulas in a more abstract way, involving algebraic manipulation and the use of given conditions to find relationships between the sides of the triangle. The progression from basic formulas to complex problem-solving effectively illustrates the power and utility of these trigonometric identities.