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Math 2201 Unit 3: Acute Triangle Trigonometry Read Learning Goals, p. 127 text. Ch. 3 Notes §3.1 Exploring Side-Angle Relationships in Acute Triangles (0.5 class) Read Goal p. 130 text. Outcomes: 1. Define an acute triangle. See notes 2. Make a conjecture about the relationship between the length of the sides of an acute triangle and the sine of the opposite angles. p. 131 3. Derive the Law of Sines (Sine Law). p. 130 This chapter is about solving acute triangles. This means finding the lengths of the missing sides and/or the measures of the missing angles. We will examine two ways to solve acute triangles – the Law of Sines (Sine Law) and the Law of Cosines (Cosine Law). In Math 1201, we labeled the sides of a right triangle with respect to the angles in the triangle. With respect to acute angle A, these sides were called the adjacent side and the opposite side. The side opposite the right angle was called the hypotenuse. hypotenuse A opposite adjacent Recall from Math 1201 that you were introduced to three special ratios that were obtained from the lengths of the three sides of a right triangle and that we gave them special names. Ratio opposite hypotenuse adjacent hypotenuse opposite adjacent Special Name sine ratio (sin) cosine ratio (cos) tangent ratio (tan) 1 In the right triangle on page 1, remember the name of the Indian below. . You can remember these ratios if you Def n Solving a triangle means finding the length(s) of the missing side(s) and/or the measure(s) of the missing angle(s). Up to now, we have used the Pythagorean Theorem and the primary trigonometric ratios to solve a right triangle. B E.g.: Solve the right triangle below. 104cm 40cm x A C Since we know the lengths of two sides of the triangle and since it is a right triangle, we can use the Pythagorean Theorem to find the length of the third side. Now we must use the trig ratios to find one of the missing angles. We will use the lengths of the two given sides just in case we made a mistake finding the length of the missing side. With respect to angle A, we are given the lengths of the opposite side and the hypotenuse so we will use the sine ratio. Make sure your graphing calculator is set to degree mode and not radian mode. 2 40 104 40 A sin 1 104 A 22.62 sin A Since m A 22.62, then m B 180 90 22.62 67.38 Solving a triangle using the Pythagorean Theorem and primary trigonometric ratios only works with a RIGHT triangle. How do we solve triangles that do NOT have a right angle? Triangles that do not have a right angle are either acute triangles or obtuse triangles. In this chapter we will deal with acute triangles only. Def n : An acute triangle is a triangle in which all the angles have a measure less than 90 . Labeling Any Triangle B c A a C b We label each vertex with an UPPERCASE letter and the side opposite the vertex with the corresponding lowercase letter. The Law of Sines (Sine Law) can be used to solve SOME acute triangles. A Conjecture about the Relationship between the Length of the Sides of an Acute Triangle and the Sine of the Opposite Angles. Measure each side to the nearest tenth of a centimeter and each angle to the nearest tenth of a degree and complete the table. 3 Measure (1 decimal place) A B C Side a 45.0 Side b 60.0 Side c 75.0 Length (cm) Calculate (1 decimal place) (4 decimal places) sin A 4.2 a sin B 5.1 b sin C 5.7 c 0.168358757 0.169---0.169 --- Make a conjecture about the relationship between the length of the sides of an acute triangle and the sine of the opposite angles. Conjecture: In an acute triangle _______________________________________________________. Derivation of the Law of Sines (Sine Law) Let’s draw any a ute triangle with an altitude (h) drawn from a vertex. A c h B D a Using the sine ratio we can write sin B sin C b C h or h c sin B . Similarly, we can write c h or h b sin C . b Since we have two expressions equal to h, we can substitute to get c sin B b sin C . Since b c 0 , we can divide both sides by bc to get c sin B b sin C bc bc sin B sin C b c 4 Now let’s use the sa e triangle above but draw a different altitude. A c h B D a Using the sine ratio we can write sin B sin A b C h or h a sin B . Similarly, we can write a h or h b sin A . b Since we have two expressions equal to h, we can substitute to get a sin B b sin A . Since a b 0 , we can divide both sides by ab to get a sin B b sin A ab ab sin B sin A b a Since sin B sin C sin B sin A and then b c b a ************* sin A sin B sin C ***************** a b c This is the Sine Law. Sine Law sine angle 1 sine angle 2 sine angle 3 length of side opposite angle 1 length of side opposite angle 2 length of side opposite angle 3 5 §3.2 Applying the Sine Law (2 classes) Read Goal p. 132 text. Outcomes: 1. Use the Sine Law to find the length of a missing side in an acute triangle. p. 134 2. Use the Sine Law to find the measure of a missing angle in an acute triangle. p. 136 3. Solve problems using the Sine Law. pp. 135-137 E.g.: Solve the triangle below. C 52 b 39 11.2cm A c B Substituting into sin A sin C gives, a c sin 39 sin 52 11.2 c Cross multiplying and solving gives c sin 39 11.2sin 52 c sin 39 11.2sin 52 sin 39 sin 39 c 14.0cm Since m A 39 and m C 52 , then m B 180 39 52 89 Substituting into sin A sin B gives, a b sin 39 sin 89 11.2 b Cross multiplying and solving gives b sin 39 11.2sin 89 b sin 39 11.2sin 89 sin 39 sin 39 b 17.8cm 6 Generally, there are two cases when you can use the Sine Law. E.g.: Find the value of x in the diagram to the right. Substituting into sin A sin B gives a b sin 63 sin 49 10 x x sin 63 10sin 49 x sin 63 10sin 49 sin 63 sin 63 x 8.5cm Do #’s 2 a, 3 a, b, c, 8, pp. 138-140 text in your homework booklet. E.g.: Find the value of x in the diagram to the right. Substituting into sin A sin B gives a b 7 sin 50 sin x 11 7 7 sin 50 11sin x 7 sin 50 11sin x 11 11 sin x 0.4874828274..... x 29.2 Do #’s 2 b, 3 d, e, f, 7, 11, pp. 138-140 text in your homework booklet. Problem Solving Using the Sine Law. We need to find the values of a and b. Substituting into sin B sin C gives b c sin 47 sin 68 a 46 46sin 47 a sin 68 46sin 47 a sin 68 sin 68 sin 68 a 36.3m Substituting into sin A sin C gives a c 8 sin 65 sin 68 b 46 46sin 65 b sin 68 46sin 65 b sin 68 sin 68 sin 68 b 45.0m So 46 36.3 45.0 127.3m of chain-link fence is needed to enclose the entire park. Substituting into sin Q sin R gives q r m P 180 60 35.3 84.7 sin 60 sin R 184.5 123 123sin 60 184.5sin R h 184.5 h 184.5sin 84.7 183.7ft sin 84.7 123sin 60 184.5sin R 184.5 184.5 sin R 0.5773502692..... R 35.3 9 Find the value of x. First let’s look at the triangle to the right. 10 sin Z sin C Substituting into gives z c z sin 30 sin10 z 50 50sin 30 z sin10 50sin 30 z sin10 sin10 sin10 z 144.0ft 30 50 feet Using right triangle trigonometry, we can write x 144 x 144 cos 40 110.3ft cos 40 Do #’s 4, 10, 13-15, pp. 139-141 text in your homework booklet. Do #’s 4-7, 9 a, p. 143 text in your homework booklet. 10 §3.3 Deriving and Applying the Cosine Law (2 classes) Read Goal p. 144 text. Outcomes: 1. Illustrate and explain situations where the Sine Law cannot be used to find a missing side or a missing angle in a triangle. p. 144 2. Derive the Law of Cosines (Cosine Law). p. 144 The Law of Sines CAN be used given two sides and the non-included angle. The Law of Sines CANNOT be used in the following situations: 1. Given two sides and the included angle. C sin 52 sin A c 19.50 19.50cm 52 11.2cm c A This is impossible to solve since there are two unknowns in the equation. C 2. Given three sides. 19.50 B 15.39 sin B sin A 15.39 19.50 11.20 A This is impossible to solve since there are two unknowns in the equation. C a 52 3. Given three angles. 35 c sin 93 sin 52 a c 93 A This is impossible to solve since there are two unknowns in the equation. Therefore, we need another formula that will enable us to solve a triangle in two of those situations. This law is the Law of Cosines. 11 Derivation of the Law of Cosines (Cosine Law) B a c h A P C b Using right triangle APB and the Pythagorean Theorem, we can write Similarly, using right triangle BPC and the Pythagorean Theorem, we can write Since the expressions on the left hand side (LHS) of equation 1 and equation 2 equal to each other. By substitution, Solving for gives But m is not a side of the triangle. Using right triangle APB again and the trig ratios, we can write Therefore, substituting for m in we get, This is the Law of Cosines. ******* Note that the Law of Cosines can be written in three ways. 1. 2. 3. 12 , they must be equal E.g.: Solve the following triangle. B This is not a right triangle and we cannot use the Law of Sines so we must use the Law of Cosines to solve the triangle. Now we have the choice of using the Law of Sines or the Law of Cosines to find the measure of angle B. We will use the Law of Cosines again for practice. The measure of angle A is Like the Sine Law, we can use the Cosine Law to find the length of a missing side or the measure of a missing angle. 13 E.g.: How long is the tunnel in the diagram to the right? Substituting into a2 b2 c2 2bc cos A gives BC 2 3002 2002 2 300 200 cos80 BC 2 90000 40000 120000 cos80 BC 2 130000 120000 cos80 BC 2 130000 20837.78132..... BC 2 109162.2187... BC 109162.2187... BC 330.4 BC 330.4ft So the length of the tunnel is about 330.4ft long. Draw a triangle which illustrates the information in the workings to the right. Do #’s 2, 4 a, 6 a, p. 151 text in your homework booklet. E.g.: Find the value of A . Substituting into a2 b2 c2 2bc cos A gives 162 92 192 2 9 19 cos A 256 81 361 342 cos A 256 442 342 cos A 256 442 442 442 342 cos A 186 342 cos A 186 342 cos A 342 342 cos A 0.5438596491... A cos 1 0.5438596491... A 57.1 Do #’s 3, 5 a, 6 c, 7 a, c, pp. 151-152 text in your homework booklet. 14 Problem Solving Using the Cosine Law. E.g.: Aircraft 1 flies at 400km/h and aircraft 2 flies at 350km/h. If the angle between their paths is 49 , how far apart are the aircraft after 2h? Substituting into a2 b2 c2 2bc cos A gives d 2 8002 7002 2 800 700 cos 49 d 2 640000 490000 1120000 cos 49 d 2 1130000 1120000 cos 49 d 2 1130000 734786.1125..... d 2 395213.8875... d 395213.8875... d 628.66 d 628.66km The aircraft are about 628.66km apart. E.g.: How far apart are the two trees in the diagram to the right. Substituting into a2 b2 c2 2bc cos A gives d 2 752 1002 2 75 100 cos 32 d 2 5625 10000 15000 cos 32 d 2 15625 15000 cos 32 d 2 15625 12720.72144..... d 2 2904.278558... d 2904.278558... d 53.89 d 53.89m The trees are about 53.89m apart. 15 E.g.: What is the angle between the 22ft and the 18ft sides? Substituting into a2 b2 c2 2bc cos B gives 122 182 222 2 18 22 cos B 144 324 484 792 cos B 144 808 792 cos b 144 808 808 808 792 cos B 664 792 cos B 664 792 cos B 792 792 cos B 0.83 B cos 1 0.83 B 33.0 The angle between the 22ft and 18ft sides is about 33.0 16 Errors Using the Cosine Law E.g.: Find the error in the solution below and correct it. Step 1: BC 2 3752 4002 2 375 400 cos 35 Step 2: BC 2 140625 160000 300000 cos 35 Step 3: BC 2 300625 300000 cos 35 Step 4: BC 2 625cos 35 Step 5: BC 2 511.9700277... Step 6: BC 511.9700277... Step 7: BC 22.63 Step 8: BC 22.63m The error occurs in Step ____ where ________________________________________________________. In this step, ____________________________________________________________________________. Do #’s 9, 13, pp. 152-153 text in your homework booklet. Do #’s 1, 5, 6, 7, 9, 13, pp. 161-163 text in your homework booklet. Do #’s 1-7, 9, 10 p. 168 text in your homework booklet. Pythagorean Theorem Primary Trigonometric Ratios Sine Law Formulae c 2 a 2 b2 o a o sin ; cos ; tan h h a sin A sin B sin C a b c 2 2 2 a b c 2bc cos A b 2 a 2 c 2 2ac cos B Cosine Law c 2 a 2 b 2 2ab cos C 17