Properties of Triangle - Projection Formula
Duration: 13 min
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This video is a mathematics lecture that derives the projection formula for a triangle. The instructor begins by introducing the concept and drawing a triangle ABC, labeling its vertices and sides. He then drops a perpendicular from vertex A to the opposite side BC, creating a point D. Using the right-angled triangles ABD and ADC, he applies the definition of cosine to find the lengths of the segments BD and DC in terms of the sides and angles of the main triangle. This leads to the derivation of the projection formula: a = b cos C + c cos B. The lecture then presents two example problems. The first problem asks for the value of (cos C + cos A)/c + a + cos B/b, which is solved by substituting the projection formulas. The second problem involves finding the value of b cos²(C/2) + c cos²(B/2) given the perimeter k, which is solved using the half-angle identity for cosine.
Chapters
0:00 – 2:00 00:00-02:00
The video starts with a blank screen, then the instructor begins writing the title "Projection Formula:". He draws a triangle ABC, labeling the vertices and the sides opposite to them as a, b, and c. He then draws a perpendicular from vertex A to the base BC, labeling the foot of the perpendicular as D. The instructor labels the segments BD and DC, and the side BC as a, establishing the relationship a = BD + DC.
2:00 – 5:00 02:00-05:00
The instructor analyzes the right-angled triangle ABD. He writes the formula for cosine of angle B: cos B = BD/c, which he rearranges to find BD = c cos B. He then moves to the right-angled triangle ADC, writing cos C = DC/b, which he rearranges to find DC = b cos C. He substitutes these expressions into the earlier equation a = BD + DC, resulting in the projection formula: a = c cos B + b cos C.
5:00 – 10:00 05:00-10:00
The instructor writes down the three projection formulas for triangle ABC: a = b cos C + c cos B, b = c cos A + a cos C, and c = a cos B + b cos A. He then presents the first example problem, Eq 1, which asks to find the value of (cos C + cos A)/c + a + cos B/b. He substitutes the projection formulas into the expression, simplifying it to (b cos C + c cos B + c cos A + a cos C + a cos B + b cos A) / (a + b + c). After further simplification, he arrives at the final answer, which is 1.
10:00 – 13:10 10:00-13:10
The instructor presents the second example problem, Eq 2. It states that if k is the perimeter of triangle ABC, then what is the value of b cos²(C/2) + c cos²(B/2). He writes the half-angle identity cos²(θ/2) = (1 + cos θ)/2. He applies this to the expression, writing b cos²(C/2) + c cos²(B/2) = b(1 + cos C)/2 + c(1 + cos B)/2. He then expands this to (b + b cos C + c + c cos B)/2. Using the projection formula a = b cos C + c cos B, he substitutes to get (b + c + a)/2. Since k = a + b + c, the expression simplifies to k/2.
The video provides a comprehensive derivation of the projection formula for a triangle, starting from basic trigonometric principles in right-angled triangles. The core of the lesson is the derivation of the formula a = b cos C + c cos B, which is then used as a powerful tool to solve two distinct problems. The first problem is a direct algebraic manipulation of the formula, while the second problem requires the application of a trigonometric identity (the half-angle formula for cosine) in conjunction with the projection formula. The progression demonstrates how a fundamental geometric identity can be applied to solve both algebraic and trigonometric problems in triangle geometry.