Solving Triangles Lecture Notes page 1 Sample Problems 1. Solve the triangle. a = 3, b = 7, c = 6 2. Consider the triangle with sides 7, 9, and 12 units long. Let , , and Compute each of the following. a) cos ( + + ) b) cos + cos + cos denote the angles of the triangle. c) the exact value of the area of the triangle. 4 = : Let D be the midpoint of side AC 5 a) Find the exact value of the cosine of angle CDB. 3. In triangle ABC, = 90 and cos b) Compute the exact value of the cosine of angle ADB. 4. Point D is on side AB of triangle ABC; with \ACD = \BCD = 60 , AC = 5; and BC = 15. Find the length of line segment CD. 5. A triangle has sides of length a, b, and c; which are consecutive integers in increasing order, and cos Find cos . = 5 : 16 6. Suppose that ABC is a equilateral triangle with all sides of length 1 unit. We extend side AB by 1 unit beyond B to get P , we extend side BC by 1 unit beyond C to get Q and side AC beyond A to get R. Compute the length of the sides in triangle P QR. 7. All three sides of triangle ABC are 4 units long. Compute the exact value of the cosine of the angle shaded on the picture. c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Lecture Notes Solving Triangles page 2 8. Consider a square with sides 1 unit long. To the inside of each side, we draw an isosceles triangle with its greatest angle, opposite the unit long base, measures 150 . Consider all vertices of these triangles that are not on the square. If we connect these vertices, we obtain a square. Compute the exact value of the area of this square. 9. A trapezoid’s parallel sides are 18 and 24 units long. Another side is 15 units long, and the angle formed between this side and the longer parallel side is 74:5 . Compute the fourth side of the trapezoid and its angles. 10. Let ABC be a triangle with sides a, b, and c. We draw squares on all three sides. The vertices of the squares that are not on the triangle form a hexagon. Label the sides of the hexagon that are not on the square by x, y, and z. Prove that x2 + y 2 + z 2 = 3 a2 + b2 + c2 . c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Solving Triangles Lecture Notes page 3 Answers for Sample Problems 1.) 25: 208 7655 , p 2 13 3.) a) 13 96: 379 37 , p 2 13 13 b) 58: 411 8645 15 4.) = 3:75 4 2.) a) 13 5.) 20 9.) 14: 591 units, 105: 5 , 97:84440 , 82: 1556 6.) 1 p b) p c) 14 5 37 27 p 3 21 7.) 14 7 unit 8.) 2 p 3 10.) see solutions Solutions of Sample Problems 1. Solve the triangle. a = 3, b = 7, c = 6 When using the law of cosines, we will …rst …nd the largest angle in the triangle. This way, when we next apply the law of sines (the law of cosines is also an option), we know that we are solving for angles that MUST be acute. First we will use the law of cosines to …nd . b2 = a2 + c2 2 = a +c b2 a2 + c2 b2 32 + 62 72 = = = 2ac 2 3 6 2ac cos cos Then 1 = cos We can now …nd again. 2ac cos 2 1 9 1 96: 379 37 9 by using either the law of sines or the law of cosines. c2 = a2 + b2 2ab cos cos 2 We will use the law of cosines 2ab cos 2 = a +b c2 a2 + b2 c2 32 + 72 62 11 = = = 2ab 2 3 7 21 11 58: 411 8645 21 The third angle can be easily found. Thus 1 = cos = 180 And so the answer is ( + ) 25: 208 7655 , 180 (96: 379 37 + 58: 411 8645 ) = 25: 208 7655 96: 379 37 , and 58: 411 8645 . 2. Consider the triangle with sides 7, 9, and 12 units long. Let , , and Compute each of the following. a) cos ( + b) cos + ) = cos 180 = + cos denote the angles of the triangle. 1 + cos c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Solving Triangles Lecture Notes page 4 Solution: Let us label the sides a = 7, b = 9, and c = 12. We …rst use the law of cosines to compute cos a2 = b2 + c2 2 = b +c a2 2 2 b +c a2 92 + 122 72 22 = = = 2ac 2 9 12 27 2bc cos cos We compute 2bc cos 2 similarly. b2 = a2 + c2 2ac cos cos 2 2ac cos 2 = a +c b2 72 + 122 92 2 a2 + c2 b2 = = = 2ac 2 7 12 3 We will again use the law of cosines, this time to compute . c2 = a2 + b2 2ab cos = a +b c2 a2 + b2 c2 72 + 92 122 1 cos = = = 2ab 2 7 9 9 22 2 1 37 Now cos + cos + cos = + = 27 3 9 27 c) the exact value of the area of the triangle. 1 Solution: We will use the formula A = ab sin . We …rst need to compute sin . We know that sin 2 positive because is an angle in a triangle, and so 0 < < 180 . s r r p p 1 2 1 80 4 5 2 sin = 1 cos = 1 = 1 = = 9 81 81 9 2ab cos 2 2 is Now we can compute the exact value of the area of the triangle. p ! p 1 1 4 5 A = ab sin = 7 9 = 14 5 2 2 9 4 = : Let D be the midpoint of side AC 5 a) Find the exact value of the cosine of angle CDB. 3. In triangle ABC, = 90 and cos Solution: It is part of the problem to come up with a picture that depicts the data. Sometimes that is not easy and takes several attempts. However, a good picture is often an essential part of solving a geometry problem. 4 From the fact that this is right triangle and cos = , we conclude that this triangle is similar to the one 5 with sides 3, 4, and 5 units and label the sides as 3x, 4x, and 5x where x is a positive number. Since D is the midpoint of line segment AC, both AD and DC are of length 2x. c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Solving Triangles Lecture Notes page 5 We compute the length of line segment BD using the Pythagorean Theorem: q p p p BD = (2x)2 + (3x)2 = 4x2 + 9x2 = 13x2 = 13x We are now ready to …nd the cosine of angle CBD p 2x 2 2 13 cos (\CDB) = p =p = 13 13x 13 b) Compute the exact value of the cosine of angle ADB. Solution 1: Angles CDB and ADB are supplements, they add up to 180 . For all angles , cos (180 )= cos and so cos (\ADB) = cos (\ADB) = p 2 13 13 Solution 2: Consider triangle ABD. We can compute the cosine of = \ADB by stating the law of cosines for triangle ABD. (AB)2 = (AD)2 + (DB)2 2 (AD) (DB) cos p p 2 13x 2 (2x) 13x cos (5x)2 = (2x)2 + p 25x2 = 4x2 + 13x2 4 13x2 cos p 25x2 = 17 4 13 cos x2 p 25 = 17 4 13 cos p 8 = 4 13 cos 8 p = cos 4 13 p 2 2 13 p = cos = 13 13 c copyright Hidegkuti, 2013 divide by x2 Last revised: December 13, 2013 Solving Triangles Lecture Notes page 6 4. Point D is on side AB of triangle ABC; with \ACD = \BCD = 60 , AC = 5; and BC = 15. Find the length of line segment CD. Solution: It is part of the problem to come up with a picture that depicts the data. Sometimes that is not easy and takes several attempts. However, a good picture is often an essential part of solving a geometry problem. 1 Let us denote line segment CD by x. Recall the formula A = ab sin . 2 expressing the area of this triangle in two di¤erent ways. ATriangle ABC 1 (5) (15) sin 120 2 75 sin 120 75 sin 120 p 3 75 2 = ATriangle ADC + ATriangle BCD 1 1 = (5) (x) sin 60 + (15) (x) sin 60 2 2 = 5x sin 60 + 15x sin 60 We will solve this problem by multiply by 2 = 20x sin 60! p 3 = 20x 2 divide by p 3 2 75 = 20x 75 15 x = = = 3:75 20 4 5. A triangle has sides of length a, b, and c; which are consecutive integers in increasing order, and cos Find cos . = 5 : 16 Solution: First we would want to express the fact that the three sides are consecutive integers. The usual labeling, a = x, b = x + 1, and c = x + 2 would work, but we will chose a smarter labeling that will cut down on the computations, that is a = x 1, b = x, and c = x + 1. We will need to remember that x must be a positive integer. Since cos is given, we will state the law of cosines involving cos : c2 = a2 + b2 (x + 1)2 = (x 2ab cos 1)2 + x2 x2 + 2x + 1 = x2 2 (x 2x + 1 + x2 32x = 8x2 5 2 x x 8 5 x2 x 32x = 8x2 5x2 + 5x 4x = x2 1) x 10 2 x 16 5 16 x multiply by 8 27x = 3x2 0 = 3x2 0 = 3x (x 27x 9) =) x1 = 0; x2 = 9 Since x can not be zero, x = 9 is the only solution. Because we labeled b by x, this means that the triangle’s c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Solving Triangles Lecture Notes three sides are a = 8, b = 9, and c = 10 units. We will compute cos a2 = b2 + c2 2 page 7 using the law of cosines again: 2bc cos = b +c a2 b2 + c2 a2 92 + 102 82 13 = = = 2bc 2 9 10 20 2bc cos cos 2 6. Suppose that ABC is a equilateral triangle with all sides of length 1 unit. We extend side AB by 1 unit beyond B to get P , we extend side BC by 1 unit beyond C to get Q and side AC beyond A to get R. Compute the length of the sides in triangle P QR. Solution: By symmetry, triangle P QR is also equilateral and so we just need to compute the length of any side. Consider triangle P BQ. Side BP is 1 unit long and side BQ is 2 units long. Angle P BQ has measure 120 since it is the supplement of the inner angle which measures 60 . We denote side P Q by x and state the law of cosines on triangle P BQ. x2 = 12 + 22 2 1 2 cos 120 x2 = 5 1 2 4 cos 120 = 1 2 x2 = 5 + 2 x2 = 7 x = p 7 Since x is a distance, we discard the negative solution of the equation. The triangle P QR has sides of length p 7. c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Lecture Notes Solving Triangles page 8 7. All three sides of triangle ABC are 4 units long. Compute the exact value of the cosine of the angle shaded on the picture below. Solution: We will solve this problem by stating the law of cosines for triangle CDE. We …rst need to …nd the length of its three sides. First, side EC is clearly 1 unit long. We can compute the length of CD via the Pythagorean theorem CD2 + 22 = 42 CD2 = 12 =) CD = p 12 We will now use the law of cosines in triangle ADE to compute the length of ED. ED2 = 22 + 32 2 2 3 cos 60 p 1 ED2 = 4 + 9 12 = 13 6 = 7 =) ED = 7 2 p p So now we know that EC = 1, CD = 12, and ED = 7. We need to compute the cosine of the angle oposite side EC. Let us enote this angle by . p p 2 p p 2 12 = 12 + 7 2 12 7 cos p 1 = 19 2 84 cos p 2 84 cos = 18 p p p p p 3 84 3 2 21 3 21 18 9 9 84 9 84 p =p = = = = cos = = 84 28 28 14 84 2 84 84 8. Consider a square with sides 1 unit long. To the inside of each side, we draw an isosceles triangle with its greatest angle, opposite the unit long base, measures 150 . Consider all vertices of these triangles that are not on the square. If we connect these vertices, we obtain a square. Compute the exact value of the area of this square. c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Solving Triangles Lecture Notes page 9 Solution: This problem can be solved without the law of cosines, but it is much easier using it. Let us …rst connect the four points, denoted by E, F , G, and H as shown on the picture below. Consider …rst triangle ABF . By symmetry, AF = BF and so the triangle is isosceles. Since angle AF B = 150 , the angles F AB and ABF must measure 15 . By symmetry, angle EAD must also measure 15 . Consider now angle EAF . It must measure 90 (15 + 15 ) = 60 . Consider triangle AEF . By symmetry, AE = EF and so the angles opposite those sides are also the same. Since the third angle, \EAF = 60 , the other two angles must add up to 120 : Because they are equal, they must measure 60 and so the triangle is equilateral and AE = EF = AF . The area of the square is x2 and we can …nd its value by stating the law of cosines on triangle ABF . x2 + x2 2 (x) (x) cos 150 = 12 p ! 3 2 2 2x 2x = 1 2 p x2 2 + 3 = 1 x2 = 1 p 2+ 3 We rationalize our value for x2 . 1 1 2 p = p x = 2+ 3 2+ 3 2 2 p p 3 2 3 p = =2 1 3 p 3 This is the area of the square. c copyright Hidegkuti, 2013 Last revised: December 13, 2013 Solving Triangles Lecture Notes page 10 9. A trapezoid’s parallel sides are 18 and 24 units long. Another side is 15 units long, and the angle formed between this side and the longer parallel side is 74:5 . Compute the fourth side of the trapezoid and its angles. Solution: We will …rst prove a fact that is true for all trapezoids. The two angles formed at a trapezoid’s side conntecting parallel sides are supplements, they add up to 180 . proof: Consider the picture below. We denoted the angle by A by extended line AD beyond D and line BC beyond C. and the angle by B by . Then we These new angles formed are again and because sides AB and CD are parallel. Now it is easy to see that + \ADC = 180 and + \BCD = 180 . The solution of a geometry problem is often very easy after a well chosen line was drawn in. Consider the picture below. Let us draw a line through C that is parallel to side AC. Now AECD is a parallelogram, because it has two pairs of parallel sides. It is a propoerty of parallelograms that opposite sides have the same length. Thus AE = 18 and consequently, EB = 24 18 = 6 units. Also, angle CEB = 74:5 because AD and CE are parallel. Our smart line reduced the problem considerably. We can use the law of cosines to compute side BC: BC 2 = 152 + 62 BC c copyright Hidegkuti, 2013 2 2 15 6 cos 74:5 212: 897 092 306 =) BC 14: 590994 unit Last revised: December 13, 2013 Solving Triangles Lecture Notes page 11 We can use the law of sines again to compute angle EBC (denoted by ) 152 = 62 + BC 2 12BC cos cos 2 2 6 BC cos 2 = 6 + BC 152 62 + BC 2 152 62 + (14: 590994)2 152 = 12BC 12 (14: 590994) 1 cos (0:1364832) 82: 1556 0:1364832 The other two angles can be found easily by using the fact that the angles along a side connecting parallel sides are supplemental. \ADC = 180 74:5 = 105: 5 and \BCD = 180 = 180 82: 1556 = 97: 8444 Thus the missing side and angles are: 14: 591 units, 105: 5 , 97:84440 , and 82: 1556 . 10. Let ABC be a triangle with sides a, b, and c. We draw squares on all three sides. The vertices of the squares that are not on the triangle form a hexagon. Label the sides of the hexagon that are not on the square by x, y, and z. Prove that x2 + y 2 + z 2 = 3 a2 + b2 + c2 . proof: Consider the picture below. We …rst establish that the angles, shown on the picture are complements to , , and , correspondingly. Also, recall that for all angles , cos (180 ) = cos . Let us state the law of cosines for the triangles that include x, y, and z as a side. x2 = a2 + c2 2ac cos (180 x2 = a2 + c2 + 2ac cos y 2 = a2 + b2 2ab cos (180 y 2 = a2 + b2 + 2ab cos ) ) and z 2 = b2 + c2 z c copyright Hidegkuti, 2013 2 2 2bc cos (180 ) 2 = b + c + 2bc cos Last revised: December 13, 2013 Solving Triangles Lecture Notes page 12 We add these three equations and get x2 + y 2 + z 2 = a2 + c2 + 2ac cos + a2 + b2 + 2ab cos + b2 + c2 + 2bc cos x2 + y 2 + z 2 = 2a2 + 2b2 + 2c2 + 2ac cos + 2ab cos + 2bc cos We will now "get rid" of the expressions 2ac cos , 2ab cos , and 2ab cos on the triangle abc. a2 = b2 + c2 2bc cos 2bc cos = b2 + c2 a2 b2 = a2 + c2 2ac cos = a2 + c2 2ac cos b2 by using the law of cosines stated c2 = a2 + b2 2ab cos 2ab cos = a2 + b2 c2 We substitute these into our equation x2 + y 2 + z 2 = 2a2 + 2b2 + 2c2 + 2ac cos 2 2 2 2 + 2ab cos + 2bc cos 2 = 2a + 2b + 2c + a + c b2 + a2 + b2 c2 + b2 + c2 a2 = 3a2 + 3b2 + 3c2 This completes our proof. For more documents like this, visit our page at http://www.teaching.martahidegkuti.com and click on Lecture Notes. E-mail questions or comments to [email protected] c copyright Hidegkuti, 2013 Last revised: December 13, 2013

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