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(8 – 2) Law of Sines Learning target: To solve SAA or ASA triangles To solve SSA triangles To solve applied problems We work with any triangles: Important how to label. We use capital letters for angles (vertexes) and small letters for sides corresponding. Opposite side of A is a. Opposite side of B is b. Opposite side of C is c. A B C a b c

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Cases to use the law of sines: Case 1: (AAS or ASA) two angles and one side are known. If you are given two angles and one side (ASA or AAS), the Law of Sines will nicely provide you with ONE solution for a missing side. Case 2: (SSA) two sides and an angle opposite one of them are known. Unfortunately, the Law of Sines has a problem dealing with SSA. If you are given two sides and one angle (where you must find an angle), the Law of Sines could possibly provide you with one or more solutions, or even no solution.

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Case 1: AAS case I do: Solve for all missing parts. c b 40 60 1: label all missing parts 2: find the missing angle. 3: set up the equation using law of sine, and solve for missing parts. 4

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ASA case We do: Solve for all missing parts. A B C 10 45 105 1: label all missing parts. 2: find the remaining angle. 3: set up the equation using law of sine, and solve for missing parts.

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You do: Solve for all missing parts. A B C 43 cm 52 29 1:leble all missing parts 2: find the case. 3: find the remaining angle 4:set up the equation using law of sine, and solve for missing parts.

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You do: Solve for all missing parts. A B C 94.6 m 18.7 124.1 1:leble all missing parts 2: find the case. 3: find the remaining angle 4:set up the equation using law of sine, and solve for missing parts.

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Case 2: (SSA) two sides and an angle opposite one of them are known. Unfortunately, the Law of Sines has a problem dealing with SSA. If you are given two sides and one angle (where you must find an angle), the Law of Sines could possibly provide you with one or more solutions, or even no solution. In Geometry, we found that we could prove two triangles congruent using : SAS - Side, Angle, Side ASA - Angle, Side, Angle AAS - Angle, Angle, Side SSS - Side, Side, Side HL - Hypotenuse Leg for Right Triangles. We also discovered that SSA did not work to prove triangles congruent. We politely called it the Donkey Theorem ; - )

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b A C a h no triangle is constructed a < h a = h A right triangle is constructed a (h) The arc intersects at 2distinct positive-x points. 2 triangles are constructed x y a a B BB The arc intersects at 1 positive-x point. One non-right triangle is constructed B a h h = b sinA a > b h < a < b

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Facts we need to remember: 1.

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I do: Find all missing parts. (SSA) C B 12.4 8.7 36.7 1: label all parts. 2: find angle A using law of sine. 3: find A. 4: check if there is another possible angle. We have 2 triangles. 5: write the 2 triangles. ABC & A’BC’ 6: find the remaining parts for the 2 triangles. A

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ABC: find C : C 84.9 find c using law of sine 6: A’BC’: find C’ : C’ 21.7 Find c’ using law os sine C’

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We do: Find all missing parts. (SSA) 35 86 1: label all parts. 2: find angle B using law of sine. 3: find B. B 4: check if there is another possible angle. Let We have 2 triangles. AB C 5: draw and label, and write the 2 triangles. ABC & A’BC’ 6: find the remaining parts for the 2 triangles. B’ 6

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6: find the remaining parts for the 2 triangles. ABC: find C using the sum of a triangle = 180 find c using law of sine 6: A’BC’: find B’ : Find c’ using law os sine C’ c’

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You do: Find the unknown parts. A = 29.7 , b = 41.5 ft, a = 27.2 ft.

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123 ft 184.5 ft 60 Find all missing parts.

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AB 21 P A cable car carries passengers from a point A to point P. The point A is 1.2 miles from a point B at the base of a mountain. The angles of elevation of P from A and B are 21 and 65 , respectively. 65 a)Approximate the distance between A and P. b)Approximate the height of the mountain.

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(8 – 3) Law of cosines Learning target: To solve SAS triangles To solve SSS triangles To solve applied problems. Cases to use law of cosine Case 1: (SAS) two sides and the angle between them are known. Case 2: (SSS) three sides are known.

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This may help you memorize. There is a pattern in the formulas. all letters of sides, squared letters are multiplied with 2cosine. outside letters are the same

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Law of cosine, case 1: (SAS case) I do: Solve the triangle if a = 5, c = 8, and B = 77 1: draw an triangle and label given number for each part. 2: solve for b using law of cosine. (since B = 77 is given) b 8.4 3: Find A using law of sine. 77 B 8 5 A C 4: find C using the sum of .

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Law of cosine, case 2: (SSS case) We do: Approximate the angles of the triangle if a = 90, b = 70, and c = 40 1: draw an triangle and label given number for each part. 2: find one angle using law of cosine. 3: Find another side using law of cosine again. 4: find the last side using the sum of .

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You do: In a side b = 12, side c = 20 and m A = 45º. Solve the triangle. 1: draw an triangle, and label given number for each part. 2: find the case (SAS or SSS) 3: find the missing side. 4: find other missing angles.

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a)Find the largest angle, to the nearest tenth of a degree, of a triangle whose sides are 9, 12 and 18.

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b) In a parallelogram, the adjacent sides measure 40 cm and 22 cm. If the larger angle of the parallelogram measure 116º, find the length of the larger diagonal, to the nearest integer.

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(8 – 4) Area of a Triangle Learning target: To find the area of SAS triangle To find the area of SSS triangles Heron’s Formula for the area of a triangle: If the three sides of a triangle are a, b, and c, then the area of the triangle is: where a b c

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Cases to use Heron’s formula: Case 1: (SAS) two sides and the angle between them are known. Case 2: (SSS) three sides are known. I do: Find the area of the triangle for which a = 8, b = 6, and C=30 1: draw a triangle, and label all parts. 2: find c using law of cosine. 3: find the area K using Heron’s formula. C 30 c 8 6A B

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We do: Find the area of the triangle for which a = 4, b = 5, and c = 7. 1: write the Heron’s formula. 2: Find the perimeter, and find s. 3: plug in the value of s, and find the area. = 8

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You do: find the area of the triangle. a) Given a = 154 cm, b = 179 cm, c = 183 cm.

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Solve. a) A painter needs to cover a triangular region 75 m by 68 m by 85 m. A can of paint covers 75 sq m of area. How many cans will be needed?

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b) Find the area of a triangle in a rectangular coordinate plane whose vertices are (0, 0), (3, 4), and (-8, 6) using Heron’s formula. Hint: draw a triangle, and write the coordinates, then find the each side using the distance formula.

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