The Law of Cosines Page 1

Lecture Notes The Law of Cosines page 1

Theorem: (The Law of Cosines) In any triangle ABC,

c2 = a2 + b2 2ab cos

Notice that if is a right angle, then the law of cosines states the Pythagorean theorem. We could say that the law of cosines is a generalization of the Pythagorean theorem.

Proof of the theorem: Let us draw the height belonging to side AC. Let D denote the intersection point of the height and side AC. Using right triangle trigonometry on triangle BCD, we write

h sin = = h = a sin and a ) CD cos = = CD = a cos a ) AD = AC DC = b a cos Next, we state the Pythagorean theorem for the right triangle BDA.

2 2 2 AB = BD + DA 2 2 2 c = (a sin ) + (b a cos ) = a2 sin2 + b2 2ab cos + a2 cos2 = a2 sin2 + a2 cos2 + b2 2ab cos = a2 sin2 + cos2 + b2 2ab cos sin2 + cos2 = 1 2 2 = a + b 2ab cos

Example 1. A triangle has sides 5, 8, and 11 units long. a) Find the exact value of the cosine of the greatest angle. b) Find the exact value of the sine of the greatest angle. c) Find an approximate value of the greatest angle. Solution: a) The greatest angle in the triangle is opposite of the longest side. So we state the law of cosines for that side, and then solve for the cosine of the angle.

c2 = a2 + b2 2ab cos 112 = 52 + 82 2 5 8 cos 52 + 82 112 25 + 64 121 89 121 32 2 cos = = = = = 2 5 8 80 80 80 5

The fact that cos is negative indicates that the angle is greater than 90. b) Given the value of the cosine of the angle, we can compute the value of the sine. Because this angle is in

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a triangle, it must be between 0 and 180, therefore its sine must be positive.

2 2 4 25 4 21 p21 sin = 1 cos2 = 1 = 1 = = = s 5 r 25 r25 25 r25 5 p c) To nd the approximate value of the angle, we use the calculator.

1 2 cos 113: 578 18 5 Example 2. Suppose that in triangle ABC, we know that a = 10 unit, = 38, and c = 13 unit. Find an approximate value of side b. Solution: We can use the law of cosines to nd b.

b2 = a2 + c2 2ac cos 2 2 = 10 + 13 2 10 13 cos 38 = 100 + 169 260 cos 38 = 269 260 cos 38 64:1172 Therefore, b p64:1172 8: 0073 units long. The law of cosines is the generalization of the Pythagorean theorem. Every time we are looking for the distance between two objects and there is no suitable right angle to use, we turn to the law of cosines.

Example 3. Two cars are starting at the same point. One travels with 40 miles per hour, the other with 45 miles per hour. How far are the cars from each other two hours after they started, provided that the angle between their paths

is 30? Solution: Two hours after starting, one car is 80 miles from the starting point, the other is 90 miles from it. So the question can be reduced to: nd the third sides in a triangle where two sides are 80 miles and 90 miles long, and the angle

between these sides is 30.

2 2 2 p3 D = 80 + 90 2 80 90 cos 30 = 6400 + 8100 14400 = 14 500 720p2 D 0 2 D = 14 500 720p2 mi 116: 111 mi p

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Sample Problems

1. Solve the given triangle.

a) a = 5, c = 4, = 71 b) a = 3, b = 7, c = 6

2. Consider the triangle with sides 7, 9, and 12 units long. Let , , and denote the angles of the triangle. Compute each of the following. a) cos ( + + ) b) cos + cos + cos c) the exact value of the area of the triangle. 4 3. In triangle ABC, = 90 and cos = : Let D be the midpoint of side AC. 5

a) Find the exact value of the cosine of angle CDB. b) Compute the exact value of the cosine of angle ADB.

4. A triangle has sides of length a, b, and c; which are consecutive integers in 5 increasing order, and cos = : Find cos . 16 5. 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.

6. 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.

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.

8. Two sides of triangle are 8 units and 12 units long. The median belonging to the third side is 9 units long. How long is the third side of the triangle? (A median belonging to a side is the line segment connecting the midpoint of the side with the opposite vertex.)

9. 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.

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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 + y2 + z2 = 3 a2 + b2 + c2

11. (Enrichment) Consider square ABCD show on the picture. AB is a semicircle with radius 1 and center P . BD is a quarter of a circle with radius 2 and center C. Let Q be the intersection of the two arcs. g a) Find the exact value of the length of lineg segment QB. b) Suppose that the square is in a coordinate system in which AB is the x axis and the origin is point P . Find the exact value of both coordinates of point Q.

12. (Enrichment) Find the exact value of the angle formed near the carbon atom by two hydrogen atoms in the methane (CH4) molecule.

Answers

1. a) b 5:290 , 63: 354 , 45: 646 b) 25: 208 77 , 96: 379 4 , 58: 411 9 37 2p13 2p13 13 2. a) 1 b) c) 14p5 3. a) b) 4. 5. p7 unit 27 13 13 20 3p21 6. 14: 591 units, 105: 5 , 97:84440 , 82: 1556 7. 8. 2p23 9. 2 p3 10. see solutions 14 4p5 3 4 2 4 1 11. a) b) ; c) ; Hint: look at the quadrilateral CQP B. 12. cos 1 109: 471 22 5 5 5 5 5 3 c Hidegkuti, 2013 Last revised: June 6, 2019

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Solutions Sample Problems

1. Solve the given triangle.

a) a = 5, c = 4, = 71 b) a = 3, b = 7, c = 6 Solution: One of the signs that we need the law of cosines is when we cannot apply the law of sines. We can nd b rst, using the law of cosines.

b2 = a2 + c2 2ac cos b = a2 + c2 2ac cos = 52 + 42 2 5 4 cos 71 5: 290 p p We can now nd all missing quantity using the law of sines. To nd , we write

sin sin 5 sin 71 1 = = sin = 0:893 8 = = sin (0:893 8) 63: 354 a b ) 5:290 )

We are dealing with the ambigious case. 63: 354 is not the only possibility. We should also consider its supplement, 180 63: 354 = 116: 646 , as these two angles have the same sine. However, 116: 646 for will not work because with = 71, + would be over 180. So only = 63: 34 will work. Then = 180 ( + ) = 180 (63: 354 + 71 ) = 45: 646 Therefore, the solution is: b 5:290, 63: 354 , 45: 646 Note: we should know right from the start that there will be only one slution. This is because two sides and the angle between them uniquely determine a triangle. Also, once we computed b and found that b is longer than a. This means

that is greater than , therefore, cannot be greater than 90. b) We will use rst use the law of cosines to nd an agle. After that, we can use the law of sines to nd a second angle. In this case we will be dealing with the ambiguous case. We know from construction that three sides of a triangle uniquely determine the triangle. Therefore, we should only get one soluton from the ambigious case. It is good strategy to use the law of cosines to nd the greatest angle in the triangle. This way, when we apply the law of sines to nd the second angle, we know it can not be the greatest angle,

therefore, it can not exceed 90. First we will use the law of cosines to nd .

b2 = a2 + c2 2ac cos 2ac cos = a2 + c2 b2 a2 + c2 b2 32 + 62 72 1 cos = = = 2ac 2 3 6 9 1 Then = cos 1 96: 379 37 9 We can now nd by using either the law of sines or the law of cosines. If we are using the law of sines, we compute sin from cos .

1 2 1 81 1 80 p80 4p5 sin = 1 cos2 = 1 = 1 = = = = s 9 r 81 r 81 r81 9 9 p 4p5 We know that sin is positive, so sin = . 9

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sin sin We write = . Therefore, b c

4p5 6 9 c sin ! 8p5 1 8p5 sin = = = = = sin 58: 411 864 5 b 7 21 ) 21 !

If using the law of cosines again, c2 = a2 + b2 2ab cos 2ab cos = a2 + b2 c2 a2 + b2 c2 32 + 72 62 11 cos = = = 2ab 2 3 7 21 11 Thus = cos 1 58: 411 8645 . The third angle can be easily found. 21

= 180 ( + ) 180 (96: 379 37 + 58: 411 8645) = 25: 208 7655 And so the answer is 25: 208 7655 , 96: 379 37 , and 58: 411 8645 . 2. Consider the triangle with sides 7, 9, and 12 units long. Let , , and denote the angles of the triangle. Compute each of the following. a) cos ( + + ) = cos 180 = 1 b) cos + cos + cos 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 2bc cos 2bc cos = b2 + c2 a2 b2 + c2 a2 92 + 122 72 22 cos = = = 2ac 2 9 12 27

We compute similarly. b2 = a2 + c2 2ac cos 2ac cos = a2 + c2 b2 a2 + c2 b2 72 + 122 92 2 cos = = = 2ac 2 7 12 3

We will again apply the law of cosines, this time to compute c2 = a2 + b2 2ab cos 2ab cos = a2 + b2 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

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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 is positive 2 because is an angle in a triangle, and so 0 < < 180.

1 2 1 80 4p5 sin = 1 cos2 = 1 = 1 = = s 9 r 81 r81 9 p Now we can compute the exact value of the area of the triangle.

1 1 4p5 A = ab sin = 7 9 = 14p5 2 2 9 !

4 3. In triangle ABC, = 90 and cos = : Let D be the midpoint of side AC. 5 a) Find the exact value of the cosine of angle CDB.

Solution: It is part of the solution 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 with sides 5 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. We compute the length of line segment BD using the Pythagorean Theorem:

BD = (2x)2 + (3x)2 = 4x2 + 9x2 = p13x2 = p13x q p We are now ready to nd the cosine of angle CBD

2x 2 2p13 cos (\CDB) = = = p13x p13 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 2p13 and so cos ( ADB) = cos ( ADB) = \ \ 13

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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 2 (5x)2 = (2x)2 + p13x 2 (2x) p13x cos 25x2 = 4x2 + 13x2 4p 13x2 cos 25x2 = 17 4p13 cos x2 divide by x2 25 = 17 4p13 cos 8 = 4p13 cos 8 2 2p13 = cos = cos = = 4p13 ) p13 13 5 4. A triangle has sides of length a, b, and c; which are consecutive integers in increasing order, and cos = : 16 Find cos . 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 2ab cos 5 (x + 1)2 = (x 1)2 + x2 2 (x 1) x 16 10 x2 + 2x + 1 = x2 2x + 1 + x2 x2 x 16 5 4x = x2 x2 x multiply by 8 8 32x = 8x2 5 x2 x 32x = 8x2 5x2 + 5x 27x = 3x2 0 = 3x2 27x 0 = 3x (x 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 three sides are a = 8, b = 9, and c = 10 units. We will compute cos using the law of cosines again:

a2 = b2 + c2 2bc cos 2bc cos = b2 + c2 a2 b2 + c2 a2 92 + 102 82 13 cos = = = 2bc 2 9 10 20

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5. 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 PBQ. Side BP is 1 unit long and side BQ is 2 units long. Angle PBQ

has measure 120 since it is the supplement of the inner angle which measures 60. We denote side PQ by x and state the law of cosines on triangle PBQ.

2 2 2 1 x = 1 + 2 2 1 2 cos 120 cos 120 = 2 1 x2 = 5 4 2 x2 = 5 + 2 x2 = 7 = x = p7 ) Since x is a distance, we discard the negative solution of the equation. The triangle P QR has sides of length p7.

6. 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. We denoted the angle by A by and the angle by B by . Then we extended line AD beyond D and line BC beyond C. 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.

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The solution of a geometry problem is often very easy after a well chosen line was drawn in. Consider the picture on the right. 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:

2 2 2 BC = 15 + 6 2 15 6 cos 74:5 BC2 212: 897 092 306 = BC 14: 590994 unit )

We can use the law of sines again to compute angle EBC (denoted by )

152 = 62 + BC2 2 6 BC cos 12BC cos = 62 + BC2 152 62 + BC2 152 62 + (14: 590994)2 152 cos = 0:1364832 12BC 12 (14: 590994) 1 cos (0:1364832) 82: 1556 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.

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 = p12 ) We will now use the law of cosines in triangle ADE to compute the length of ED.

2 2 2 ED = 2 + 3 2 2 3 cos 60 1 ED2 = 4 + 9 12 = 13 6 = 7 = ED = p7 2 )

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So now we know that EC = 1, CD = p12, and ED = p7. We need to compute the cosine of the angle oposite side EC. Let us enote this angle by .

2 2 12 = p12 + p7 2p12p7 cos 1 = 19 2p84 cos 2p84 cos = 18 18 9 9p84 9p84 3p84 3 2p21 3p21 cos ====== = 2p84 p84 84 84 28 28 14

Solution: Consider the triangle as shown on the picture. Suppose that BC = 8, AC = 12,and CD = 9 units long. Let E be the midpoint of side AC. Triangles ADE and ABC are similar, because side AB is twice as long as side AD; side AC is twice as long as side AE, and the two triangles share the angle at vertex A.

Therefore, ED is parallel to CB and its length is half of CB. Thus ED = 4. Let us denote the lengths of BD and DA by x, and the angle CAB by . We have two unknowns so far, x and . We set up two equations by stating the law of cosines for triangles ADE and ADC.

42 = x2 + 62 2 6 x cos 92 = x2 + 122 2 12 x cos This is a system in two unknowns, so we can solve it. We will eliminate cos and solve for x.

16 = x2 + 36 12x cos 81 = x2 + 144 2 12x cos 12x cos = x2 + 20 2 12x cos = x2 + 63

2 x2 + 20 = x2 + 63 2 3 2x + 40 = x + 63 x2 = 23 So the length of the third side is 2p23.

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9. 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.

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 .

2 2 2 x + x 2 (x)(x) cos 150 = 1 p3 2x2 2x2 = 1 2 ! x2 2 + p3 = 1 1 x2 = 2 + p3

1 1 2 p3 2 p3 We rationalize our value for x2. x2 = = = = 2 p3 2 + p3 2 + p3 2 p3 1 This is the area of the square. c Hidegkuti, 2013 Last revised: June 6, 2019

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proof: Consider the picture. 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.

2 2 2 x = a + c 2ac cos (180 ) x2 = a2 + c2 + 2ac cos

2 2 2 and y = a + b 2ab cos (180 ) y2 = a2 + b2 + 2ab cos

2 2 2 and z = b + c 2bc cos (180 ) z2 = b2 + c2 + 2bc cos

We add these three equations and get

x2 + y2 + z2 = a2 + c2 + 2ac cos + a2 + b2 + 2ab cos + b2 + c2 + 2bc cos = 2a2 + 2b2 + 2c2 + 2ac cos + 2ab cos + 2bc cos

We will now "get rid" of the expressions 2ac cos , 2ab cos , and 2ab cos by using the law of cosines stated on the triangle abc.

a2 = b2 + c2 2bc cos b2 = a2 + c2 2ac cos c2 = a2 + b2 2ab cos 2bc cos = b2 + c2 a2 2ac cos = a2 + c2 b2 2ab cos = a2 + b2 c2 We substitute these into our equation

x2 + y2 + z2 = 2a2 + 2b2 + 2c2 + 2ac cos + 2ab cos + 2bc cos = 2a2 + 2b2 + 2c2 + a2 + c2 b2 + a2 + b2 c2 + b2 + c2 a2 2 2 2 = 3a + 3b + 3c

This completes our proof.

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Solution: a) Let us decote the smaller circle by C1 and the greater

circle by C2. CQ = CB = 2, because both CQ and CB are

radii in C2. Therefore, triangle CQB is isosceles. Similarly,

PQ = PB = 1, because both PQ and PB are radii in C1,

and so triangle P QB is also isosceles. QB is an arc in both C1

and C2. This means that the perpendicular bisector of QB will contain the centers of these circles, C and P . Therefore, PC is the perpendicular bisector of BQ, and triangles CQP and CBP

are congruent. Therefore, angle CQP = 90. Let us denote the intersection of QB and PC by T .

Triangle CQP is a right triangle with shorter sides 1 and 2. By the Pythagorean theorem, the hypotenuse, PC is then p5 long. QT is the height belonging to the hypotenuse. We nd QT using the area formula for triangle CQP .

1 1 1 1 A = PQ QC = 1 2 = 1 and also A = QT PC = QT p5 2 2 2 2 1 1 = QT p5 2 2 2 4 4p5 = QT . Then QB = 2 QT = 2 = = . p5 p5 p5 5 b) Consider triangle PBQ. We know all three sides, so we can nd all angles via the law of cosines. Let us nd the cosine of \BPQ = .

4 2 = 12 + 12 2 1 1 cos p5 16 16 2 3 = 2 2 cos = cos = 5 = 5 ) 2 5

Recall that P is the origin of the coordinate system, therefore, circle C2 is the unit circle. This means that cos is the 3 2 4 x coordinate of Q. The y coordinate is then sin = 1 = . Since is an angle in a triangle, 5 5 s 3 4 0 < < 180 , the negative value for sin is not possible. Thus the coordinates of Q are ; . 5 5

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12. (Enrichment) Find the exact value of the angle formed near the carbon atom by two hydrogen atoms in the methane (CH4) molecule.

Solution: Claim 1. Let s denote the side of a regular triangle. The distance s between the center of the triangle and a vertex of the triangle is . p3 Proof. We use the law of cosines and solve for r.

2 2 2 r + r 2 (r)(r) cos 120 = s 1 2r2 2r2 = s2 2 3r2 = s2

s2 s s r2 = = r = since r = is impossible here. 3 ) p3 p3

Consider now the the following picture. Let C represent the carbon in the middle and A; B; D; and E denote the hydrogen atoms. Let the distance between the carbon and the hydrogen atoms be denoted by 1: Let the distance between two hydrogen atoms be denoted by x.

8 Claim 2. x = 3 r

Proof. Let P be the intersection of the base plane with line AC: (This is the center of the base triangle BED:) Let a denote the distance PC. x By Claim 1, PE = . We state the Pythagorean theorem for p3 triangle CPE.

x 2 a2 + = 12 p3 x2 a2 + = 1 3

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2 2 x The Pythagorean theorem on triangle AP E is (a + 1) + = x2 p3 x2 a2 + 2a + 1 + = x2 3 2x2 a2 + 2a + 1 = 3 x2 a2 + = 1 3 We now solve the system of equations in a and x. 8 > > 2x2 < a2 + 2a + 1 = 3 > x2 > From the rst equation, = 1 a2: We substitute: this into the second equation and solve for a. 3 a2 + 2a + 1 = 2 1 a2 a2 + 2a + 1 = 2 2a2 3a2 + 2a 1 = 0 1 (3a 1) (a + 1) = 0 = a1 = a2 = 1 but a > 0, so a = 1 ) 3 6

1 x2 1 2 Thus a = : Then = 1 3 3 3 x2 8 = 3 9 8 8 x2 = = x = 3 ) r3 8 Since x > 0; we now have that x = . 3 r 1 Claim 3. Angle ECD = cos 1 3 Proof: We state the law of cosines on triangle ECD. We will denote angle ECD by .

8 12 + 12 2 1 1 cos = x2 where x = r3 8 2 2 cos = 3 1 cos = 3 1 = cos 1 109: 471 22 3

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