Math 335 Trigonometry Sec 7.1: Oblique Triangles and the Law of Sines In section 2.4, we solved right triangles. We now extend the concept to all triangles. Congruence Axioms Side-Angle-Side SAS Angle-Side-Angle ASA Side-Side-Side SSS If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent. If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent. If three sides of one triangle are equal, respectively, to three sides of a second triangle, then the triangles are congruent. Whenever SAS, ASA, or SSS is given, the triangle is unique. An oblique triangle is a triangle that is not a right triangle. Information sufficient to solve an oblique triangle: (1) one side (2) any other two measures Data Required for Solving Oblique Triangles Case 1 Case 2 Case 3 Case 4 One Side Two Angles SAA or ASA Two Sides One Angle not included between the two sides This may lead to more than one triangle SSA Two Sides One Angle included between the two sides SAS Three Sides SSS Use Law of Sines Sec 7.1 & 7.2 Use Law Cosines Sec 7.3 Derive the Law of Sines Page 1 of 19
Law of Sines In any triangle ABC, with sides a, b, and c, a sin A = b sin B = c sin C Note: The ratio above is the diameter of the circumscribed circle of the triangle. See Exercise 53. Alternative form: sin A a sin B = b = sin C c We can use the same method used to derive the Law of Sines to derive formulas for the area of a rectangle (with an unknown height). Area of a Triangle (SAS) In any triangle ABC, the area A is given by the following formulas. A = 1 2 bc sin A A = 1 2 ab sin C A = 1 ac sin B 2 Ex 1 (#4) Find the length of side a. Ex 2 (#12) Solve the triangle. B = 38 40, a = 19.7 cm, C = 91 40 Page 2 of 19
Ex 3 (#26) To determine the distance RS across a deep canyon, Rhonda lays off a distance TR = 582 yd. She then finds that T = 32 50 and R = 102 20. Find RS. Ex 4 (#30) Standing on one bank of a river flowing north, Mark notices a tree on the opposite bank at a bearing of 115.45. Lisa is on the same bank as Mark, but 428.3 m away. She notices that the bearing of the tree is 45.47. The two banks are parallel. What is the distance across the river? Ex 5 (#34) Three atoms with atomic radii of 2.0, 3.0, and 4.5 are arranged as in the figure. Find the distance between the centers of atoms A and C. Page 3 of 19
Ex 6 (#40) Find the area of the triangle using the formula A = 1 bh, and then verify that the 2 formula A = 1 ab sin C gives the same result. 2 Ex 7 (#55) Several of the exercises on right triangle applications involve a figure similar to the one shown Use the law of sines to obtain x in terms of α, β, and d. Sec 7.2: The Ambiguous Case of the Law of Sines We use the Law of Sines to solve triangles involving SAA, ASA, SSA. In the SSA case, 0, 1, or 2 such triangles may exist. This is known as the ambiguous case of the law of sines. Let θ and a be fixed and assume θ is acute. Bring string. a θ Page 4 of 19
Ex 8 (#8) Determine the number of triangles possible with the given parts. B = 54, c = 28, b = 23 Ex 9 Find the unknown angles in triangle ABC for each triangle that exists. (#18) C = 82.2, a = 10.9 km, c = 7.62 km Ex 10 Solve the triangle that exists. (#22) C = 52.3, a = 32.5 yd, c = 59.8 yd Page 5 of 19
(#30) A = 51.20, c = 7986 cm, a = 7208 cm Ex 11 (#36) The angle of elevation from the top of a building 45.0 ft high to the top of a nearby antenna tower is 15 20. From the base of the building, the angle of elevation of the tower is 29 30. Find the height of the tower. Page 6 of 19
Ex 12 (#38) A pilot flies her plane on a heading of 35 00 from point X to point Y which is 400 mi from X. Then she turns and flies on a heading of 145 00 to point Z, which is 400 mi from her starting point X. What is the heading of Z from X, and what is the distance YZ? Ex 13 (#40) Use the law of sines to prove the statement is true for any triangle ABC, with corresponding sides a, b, and c. a b sin A sin B = a + b sin A + sin B Page 7 of 19
Sec 7.3: The Law of Cosines Triangle Side Length Restriction In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. (Sum of two shortest sides must be greater than the longest side.) Ex 14 A triangle has base of length 13 and the other two sides are equal in length. If the lengths are integers, what is the shortest possible length of the unknown side? Derive the Law of Cosines (using the distance formula) Law of Cosines In any triangle ABC with sides a, b, and c, the following hold. a 2 = b 2 + c 2 2bc cos A b 2 = a 2 + c 2 2ac cos B c 2 = a 2 + b 2 2ab cos C See Suggested Procedure for Solving Oblique Triangles on page 310. Ex 15 Determine whether SAA, ASA, SSA, SAS, or SSS is given. What law would you use? #1 #2 #3 #4 a, b, and C A, C, and c a, b, and A a, B, and C #5 #6 #7 #8 A, B, and c a, c, and A a, b, and c b, c, and A Page 8 of 19
Ex 16 (#12) Find the measure of θ. (No calculator.) 1 1 θ 3 A Ex 17 Solve each triangle. (#16) (#34) 10 8 4 C B B = 168.2 a = 15.1 cm c = 19.2 cm Ex 18 (#40) Points X and Y are on opposite sides of a ravine. From a point third point Z, the angle between the lines of sight to X and Y is 37.7. If XZ is 153 m long and YZ is 103 m long, find XY. Page 9 of 19
Ex 19 (#44) An airplane flies 180 mi from point X at a bearing of 125, and then turns and flies at a bearing of 230 for 100 mi. How far is the plane from point X. Ex 20 (#51) A weight is supported by cables attached to both ends of a balance beam, as shown in the figure. What angles are formed between the beams and cables? Page 10 of 19
Ex 21 (#57) Surveyors are often confronted with obstacles, such as trees, when measuring the boundary of a lot. One technique used to obtain an accurate measurement is the so-called triangulation method. In this technique, a triangle is constructed around the obstacle and one angle and two sides of the triangle are measured. Use this technique to find the length of the property line (the straight line between the two markers) in the figure. Heron s Formula (SSS) If a triangle has side lengths a, b, and c with semiperimeter s = 1 (a + b + c), 2 then the area A of the triangle is given by the formula: A = s(s a)(s b)(s c) Ex 22 (#66) Find the area of the triangle ABC. a = 22 in., b = 45 in., c = 31 in. Page 11 of 19
Sec 7.4: Vectors, Operations, and Dot Product Defn A vector is a physical quantity that has both magnitude and direction. Typical vector quantities include velocity, acceleration, and force. We use a directed line segment to represent a vector quantity. The length of the vector represents the magnitude and the direction of the vector, indicated by the arrowhead, represents the direction of the quantity. Vector Notation and Facts OP OP O is the initial point whereas P is the terminal point OP PO The magnitude of the vector OP is OP. OP = PO Two vectors are equal iff (if and only if) the have the same direction and same magnitude. Same Vectors The sum of two vectors is also a vector. We use the parallelogram rule to find the sum of two (geometric) vectors. Add tail to tip. Add vector A with vector B. The sum A + B is called the resultant vector. Is vector addition commutative? YES NO For every vector v, there is a vector v that has the same magnitude as v but opposite direction. Draw v, v, 2v, u v, and v v. Page 12 of 19
Ex 23 Use the vectors in the figure to graph the vector. (#14) 2u 3v + w w u v Defn A vector with its initial point at the origin is called a position vector. A position vector u with its endpoint at the point (a, b) is written a. b. So, u = a, b. a =horizontal component and b = vertical component Geometrically, a vector is a directed line segment whereas algebraically, it is like an ordered pair (whose initial point is at the origin). The positive angle formed between the positive x-axis and a position vector is called the direction angle. Magnitude and Direction Angle of a Vector a, b The magnitude (length) of vector u = a, b is given by the following. The direction angle θ satisfies u = a 2 + b 2 tan θ = b a, a 0 Horizontal and Vertical Components The horizontal and vertical components, respectively, of a vector u having magnitude u and direction angle θ are the following. That is, u = a, b = u cos θ, u sin θ a = u cos θ and b = u sin θ Parallelogram Properties 1. A parallelogram is a quadrilateral whose opposite sides are parallel. 2. The opposite sides and opposite angles of a parallelogram are equal, and adjacent angles of a parallelogram are supplementary. 3. The diagonals of a parallelogram bisect each other, but they do not necessarily bisect the angles of the parallelogram. Page 13 of 19
Vector Operations Let a, b, c, d, and k represent real numbers. a, b + c, d = a + c, b + d k a, b = ka, kb (k is a called a scalar) If u = a, b, then u = a, b a, b c, d = a, b + ( c, d ) = a c, b d Defn A unit vector is a vector that has magnitude 1. Two useful vectors are i = i = 1,0 and j = j = 0,1 and are graphed below. If v = a, b, then v = ai + bj. Ex 24 Given vectors u and v, find (a) 2u (b) 2u + 3v (c) v 3u (#26) (#28) u = i + 2j, v = i j u = 2, 1, v = 3, 2 The dot product of two vectors is a scalar (a real number), NOT a vector. It is also known as the inner product. Dot products are used to determine the angle between two vectors, to derive geometric theorems, and to solve physical problems. (Once we have the formula for the geometric interpretation of the dot product, we will see that we can think of the dot product u v as how much the projection of u is going in the same direction of v.) Dot Product The dot product of the two vectors u = a, b and v = c, d is denoted v, read u dot v and is given by the following. u v = ac + bd Note: The dot product is a scalar, not a vector! Page 14 of 19
Properties of the Dot Product For all vectors u, v, and w and scalars k, the following hold. (a) u v = v u (b) u (v + w) = u v + u w (c) (u + v) w = u w + v w (d) (ku) v = k(u v) = u (kv) (e) 0 u = 0 (f) u u = u 2 Derive the Formula for the Geometric Interpretation of the Dot Product (for 2-dim) (using Law Cosines) Geometric Interpretation of the Dot Product If θ is the angle between the two nonzero vectors u and v, where 0 θ 180, then the following holds. u v cos θ = u v Note: If u v > 0, the angle is acute. If u v = 0, the angle is a right angle. (The vectors are thus perpendicular or orthogonal.) If u v < 0, the angle is obtuse. Ex 25 (#30) For the pair of vectors with angle θ between them, sketch the resultant. u = 8, v = 12, θ = 20 Page 15 of 19
Ex 26 (#34) Find the magnitude and direction angle for the vector 7, 24. Ex 27 (#42) Find the magnitude of the horizontal and vertical components of v, if θ is the direction angle. θ = 146.3, v = 238 Ex 28 Write each vector in the form a, b. (#44) (#48) Ex 29 (#52) Two forces of 37.8 lb and 53.7 lb act at a point in the plane. The angle between the two forces is 68.5. Find the magnitude of the resultant force. Ex 30 (#56) Find the magnitude of the resultant force using the parallelogram rule. Ex 31 (#66) Given u = 2,5 and v = 4,3, find 2u + v 6v. Page 16 of 19
Ex 32 Find the dot product for each pair of vectors. (#74) (#76) 7, 2, 4,14 2i + 4j, j Ex 33 (#78) Find the angle between 1,7 and 1,1. Ex 34 Determine whether each pair of vectors is orthogonal. (#88) (#90) (#92) 1,1, 1, 1 3,4, 6,8 4i + 3j, 8i 6j Sec 7.5: Applications of Vectors Ex 35 (#8) Find the force required to keep a 3000-lb car parked on a hill that makes an angle of 15 with the horizontal. Page 17 of 19
Ex 36 (#21) An airline route from San Francisco to Honolulu is on a bearing of 233.0. A jet flying at 450 mph on that bearing encounters a wind blowing at 39.0 mph from direction of a 114.0. Find the resulting bearing and ground speed of the plane. Ex 37 (#25) A pilot is flying at 190.0 mph. He wants his flight path to be on a bearing of 64 30. A wind is blowing from the south at 35.0 mph. Find the bearing he should fly, and find the plane s ground speed. Page 18 of 19
Ex 38 (#29) An airplane is headed on a bearing of 174 at an airspeed of 240 km per hr. A 30-kmper-hr wind is blowing from a direction of 245. Find the ground speed and resulting bearing of the plane. Page 19 of 19