Section 1.7 Applications
Objectives: 1. To review the slope of a line. 2. To find the inclination of a line. 3. To find the area of a triangle using trigonometry or Heron’s formula.
Definition The slope of a line is the ratio of vertical change to horizontal change when moving from one point on the line to another point on the line. Dy x1 – x2 y1 – y2 m = = Dx
Slope of a Line Dy x1 – x2 y1 – y2 m = = Dx P2(x2, y2) y P1(x1, y1)
EXAMPLE 1 Find the slope of the line that passes through the points (5, 9) and (-2, 4). x1 – x2 y1 – y2 x y m = = m = 9 – 4 5 – (-2) m = 5 7
Practice: Find the slope of the line that passes through the points (-3, -5) and (1, -3). x1 – x2 y1 – y2 x y m = = m = -5 + 3 -3 – 1 m = 1 2
Definition The inclination of a line is the angle that the line makes with a positively directed ray on the x-axis.
Angle of inclination x y tan = Dy Dx y tan = m x = tan-1 m
EXAMPLE 2 Find the angle of inclination of the line that passes through (2, 9) and (1, 2). x1 – x2 y1 – y2 x y m = = m = = 7 7 1 m = 9 - 2 2 – 1 tan α = 7 α = 81.87
Caution: If the angle is obtuse the calculator will respond with a negative angle. To bring your answer in line with the definition of the angle of inclination, add 180 to the answer the calculator gives to produce the correct obtuse angle. tan α = -2 α = -63.4 α = 116.6 (-63.4 + 180)
Practice: Find the angle of inclination of the line that passes through (2, 3) and (6, -2). m = = -1.25 5 -4 m = 3 + 2 2 – 6 x1 – x2 y1 – y2 x y m = = tan α = -1.25 α = -51.3 α = 128.7
Area of a Triangle C B A a b c h a h sin C = h = a sin C A = ½bh A = ½ab sin C
EXAMPLE 3 Find the area of ABC. 8 12 82 A = ½ ab sin C A = ½·8·12 sin 82 A = 48 sin 82 A = 47.5 sq. units
Practice: Find the area of XYZ. 10 7 58 A = ½ yz sin X A = ½·7·10 sin 58 A = 35 sin 58 A = 29.7 sq. units
EXAMPLE 4 Find the area of PQR, assuming that PQR is acute. 10 9 48 9 sin 48 10 sin Q = 10 sin 48 9 sin Q = mQ = 55.7 mP = 180 - (48 + 55.7) mP = 76.3
EXAMPLE 4 Find the area of PQR, assuming that PQR is acute. 10 9 48 A = ½ qr sin P A = ½·10·9 sin 76.3 A = 45 sin 76.3 A = 43.7 sq. units
Heron’s Formula The area of a triangle having sides of length a, b, and c is given by: A = s(s - a)(s - b)(s - c) where s = ½(a + b + c)
Homework pp. 38-39
►A. Exercises x1 – x2 y1 – y2 x y m = = m = 2 - 1 -7 - 4 m = = -11 1 Find the slope of the line that passes through the given points. 1. (2, -7), (1, 4). x1 – x2 y1 – y2 x y m = = m = 2 - 1 -7 - 4 m = = -11 1 -11
►A. Exercises m = 2 – (-5) 4 - 1 x1 – x2 y1 – y2 x y m = = m = 7 3 Find the angle of inclination of the line that passes through the given points. Round angle measure to the nearest minute. 7. (-5, 1), (2, 4). m = 2 – (-5) 4 - 1 x1 – x2 y1 – y2 x y m = = m = 7 3
►A. Exercises 7 3 tan = tan = m ÷ ø ö ç è æ 7 3 m = tan-1 Find the angle of inclination of the line that passes through the given points. Round angle measure to the nearest minute. 7. (-5, 1), (2, 4). 7 3 tan = tan = m ÷ ø ö ç è æ 7 3 m = tan-1 m = 23.2
►B. Exercises S = ½(a + b + c) S = ½(2 + 7 + 7) S = ½(16) S = 8 Find the area of triangle ABC. 15. a = 2, b = 7, c = 7 S = ½(a + b + c) S = ½(2 + 7 + 7) S = ½(16) S = 8
►B. Exercises A = s(s - a)(s - b)(s - c) A = 8(8 - 2)(8 - 7)(8 - 7) Find the area of triangle ABC. 15. a = 2, b = 7, c = 7 A = s(s - a)(s - b)(s - c) A = 8(8 - 2)(8 - 7)(8 - 7) A = 8(6)(1)(1) A = 48 = 4 3
►B. Exercises Find the slope of the line, given the angle of inclination. 19. 30°
►B. Exercises Find the slope of the line, given the angle of inclination. 21. 5 6
■ Cumulative Review 32. Under what conditions may the law of cosines be applied to a triangle?
■ Cumulative Review 33. In using the law of sines, which condition is the ambiguous case?
■ Cumulative Review 34. When can you solve a triangle given two sides without using the law of sines or the law of cosines?
■ Cumulative Review 35. Find the length of an arc of a circle if the radius is 6 ft. 3 in. and the central angle is 81.7°.
■ Cumulative Review 36. List all angles θ, such that 0° ≤ θ < 360°, that have a reference angle of 30°.