Exploring the Law of Sines in Oblique Triangles
Delve into the Law of Sines for solving oblique triangles without right angles, covering cases like AAS, ASA, SSA, and SSS. Understand how to utilize this law to find angles, sides, and areas of triangles, alongside practical applications.
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Objectives Use the Law of Sines to solve oblique triangles (AAS or ASA). Use the Law of Sines to solve oblique triangles (SSA). Find the areas of oblique triangles. Use the Law of Sines to model and solve real-life problems.
Plan for the day When to use law of sines Applying the law of sines
Introduction In this section, we will solve oblique triangles triangles that have no right angles. As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c. To solve an oblique triangle, we need to know the measure of at least one side and any two other measures of the triangle either two sides, two angles, or one angle and one side.
Introduction This breaks down into the following four cases: 1. Two angles and any side (AAS or ASA) 2. 3. Two sides and an angle opposite one of them (SSA) Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines.
Introduction The Law of Sines can also be written in the reciprocal form: .
Given Two Angles and One Side AAS For the triangle below C = 102 , B = 29 , and b = 28 feet. Find the remaining angle and sides.
Example AAS - Solution The third angle of the triangle is A = 180 B C = 180 29 102 = 49 . By the Law of Sines, you have .
Example AAS Solution cont d Using b = 28 produces and
Law of Sines For non right triangles B Law of sines a c a b c = = sin sin sin A B C A C b Try this: A = = = o o 43 , 67 , 45 B c mm
Example Single-Solution CaseSSA For the triangle below, a = 22 inches, b = 12 inches, and A = 42 . Find the remaining side and angles.
Example Solution SSA By the Law of Sines, you have B sin sin = A Reciprocal form b a (sin b A ) Multiply each side by b. B = sin a sin 42 ( ) Substitute for A, a, and b. = sin 12 B 22 o B is acute. 21 41 . B
Example Solution SSA cont d Now, you can determine that C 180 42 21.41 = 116.59 . c a = sin sin C A Then, the remaining side is 22 a 29 40 . inches = sin( 116 . 59 ) c = sin c C sin( 42 ) sin A
Objectives: Use the Law of Cosines to solve oblique triangles (SSS or SAS). Use the Law of Cosines to model and solve real-life problems.
Warm-up: B a c A C b
Introduction Four cases. 1. Two angles and any side (AAS or ASA) 2. 3. Two sides and an angle opposite one of them (SSA) Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines.
Law of Sines For non right triangles Law of sines B a b c = = sin sin sin A B C a c A C b
Law of Cosines: Introduction Two cases remain in the list of conditions needed to solve an oblique triangle SSS and SAS. If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete. In such cases, you can use the Law of Cosines.
Law of Cosines B Side, Angle, Side c a A C b = + 2 2 2 2 a b c bcCos 2 A = + 2 2 2 b a c acCos B = + 2 2 2 2 c a b abCos C
Try these: Solve ABC. Round angle measures to the nearest degree and side measures to the nearest tenth. 1. A = 52o b = 6 c = 8 2. B = 58o a = 9 c = 14
Law of Cosines B Side, Side, Side c a A C b
Law of Cosines SSS + 2 2 2 b c a = cos A 2 bc 2 + 2 2 a c b = cos B 2 ac 2 + 2 2 a b c = cos C 2 ab
Try these: Solve ABC. Round angle measures to the nearest degree. 3. a = 21 b = 16.7 c = 10.3 4. a = 19 b = 24.3 c = 21.8
Applications of the Law of Cosines The pitcher s mound on a women s softball field is 43 feet from home plate and the distance between the bases is 60 feet (The pitcher s mound is not halfway between home plate and second base.) How far is the pitcher s mound from first base?
Solution In triangle HPF, H = 45 (line HP bisects the right angle at H), f = 43, and p = 60. Using the Law of Cosines for this SAS case, you have h2 = f2 + p2 2fp cos H = 432 + 602 2(43)(60) cos 45 1800.3. So, the approximate distance from the pitcher s mound to first base is 42.43 feet.
Applications of the Law of Cosines C The leading edge of each wing of the B-2 Stealth Bomber measures 105.6 feet in length. The angle between the wing's leading edges is 109.05 . What is the wing span (the distance from A to C)? 105.6 ft A Answer: 171.99 ft. 28
Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to the triangles below, that each triangle has a height of h = b sin A. A is acute. A is obtuse.
Area of a Triangle - SAS B SAS you know two sides: b, c and the angle between: A c a h Remember area of a triangle is base height Base = b Height = c sin A Area = bc(sinA) C A b Looking at this from all three sides: Area = ab(sin C) = ac(sin B) = bc (sin A)
Example Finding the Area of a Triangular Lot Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102 . Solution: Consider a = 90 meters, b = 52 meters, and the included angle C = 102 Then, the area of the triangle is Area = ab sin C = (90)(52)(sin102 ) 2289 square meters.
Herons Area Formula The Law of Cosines can be used to establish the following formula for the area of a triangle. This formula is called Heron s Area Formula after the Greek mathematician Heron (c. 100 B.C.).
Area of a Triangle Law of Cosines Case - SSS B SSS Given all three sides c a h Heron s formula: A C b + + ( a b c s = )( )( ) A s s a s b s c = where 2
Try these Given the triangle with three sides of 6, 8, 10 find the area Given the triangle with three sides of 12, 15, 21 find the area
