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Class Note for MATH 1330 with Professor Flagg at UH 3


Class Note for MATH 1330 with Professor Flagg at UH 3

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 20 views.

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Date Created: 02/06/15
Math 1330 Section 73 Law of Sines and Law of Cosines The goal of this section is to solve oblique triangles An oblique triangle is one which contains no right angles We will rst develop the law of sines and the law of cosines then use them to solve triangles Law of Sines Consider triangle ABC which is not necessarily a right triangle The following relationship between the sines of the angles and the lengths of the sides opposite then is called the Law of Sines sinA sinB sinC a b c In words this law says that the sine of an angle in a triangle is proportional to the length of its opposite side Where does this equation come from Area gab sinC Remember the formula for the area of the triangle ABC is Area ac sinB Area bc sinA Since the area is the same no matter which angles and sides we use 1 l l absinC acsinB bcsinA 2 2 2 Multiplying this equation by results in the law of sines a c Law of Cosines The law of cosines is in a way a generalized version of the Pythagorean theorem For triangle ABC as above the law of cosines states a2 b2 c2 2bc cosA b2 a2 c2 2ac cosB c2 a2 b2 Zab cosC To see how these formulas are derived consider triangle ABC as above Suppose you know the lengths of b c and the measure of angle A Can you nd the length of a Put the triangle on the XaXis so angle A is in standard position 0 cosA c sinA B Drop a perpendicular line down from B to the XaXis Then tr1angle ABD is a right tr1angle so we know by the tug ratios that s1nA or in c so length1D ccosA This means other words h csinA Also cosA W c that the coordinates of the point B are c cosA csinA The point C is the point b0 The length of side a is the distance between points B and C Using the distance formula a c cosA b2 c sinA c2 a J02 cos2 A 2bc cosA b2 c2 sin2 A a J02 cos2 A sin2 A b2 2bc cosA Using the identity cos2 A sin2 A 1 and squaring both sides gives a c2 b2 2bccosA The other law of cosine formulas are derived similarly by putting angles B and then C in stande position and doing the same procedure to find the length of the side opposite Solving Oblique Triangles We use these formulas to solve oblique triangles Solving a triangle means nding the measures of all 3 angles and the lengths of all 3 sides To solve a triangle uniquely you need to be given 3 pieces of information you then nd the other three We can solve a triangle with the following information given 1 All Three sides SSS 2 Two Sides and the angle between them SAS 3 Two Sides and the angle opposite one ofthese sides SSA 4 Two angles and one side SAA Note that a triangle is not uniquely determined by just knowing all three angles Formulas used for each case 1 SSS Use the Law of Cosines 2 SAS Use the Law of Cosines 3 SSA Use the Law of Sines 4 SAA Use the Law of Sines How to Solve Case 1 SSS Notice that plugging the lengths of all three sides into the law of cosines formulas gives a value for the cosine of each one of the angles Then you use the inverse cosine function to nd the measures of the angles It is possible to be given three lengths that cannot form a triangle this happens when one number is larger than the sum of the other two The answer in this case is No Triangle is Formed Example 1 a3 b3 c7 Example 2 a5 b3 c5 Example 3 a2 bll cl2 Case 2 SAS We use the law of cosines to solve this case We are given two sides and the angle between them First we substitute these numbers into the law of cosine formula to nd the length of the third side Then with all three sides known we use the other formulas to nd the cosines of the other angles Finally we use the inverse cosine function to nd the measures of the other angles Example 4 b4 c6 and A60 Example 5 a3 b7 and C33 Case 4 SAA We will do this one rst because it is easier than case 3 Ifwe know two angles we know all three angles because the sum of the measures of all three angles ofatriangle is 180 degrees Use the law of sines to nd the lengths of the other two sides Example 6 a4 B60 and C45 Case 3 SSA This is called the ambiguous case because it is possible to form 2 triangles from some combinations of two sides and an angle opposite one side So SSA may form 0 l or 2 triangles Why Because when we use the law of sines to nd an angle we use the inverse sine function to nd the measure of an angle But for every number between 0 and 1 there is also an angle in the second quadrant with the same sine value Since a triangle can have one angle of measure greater than 90 degrees it might be possible for the angle in the second quadrant to form a triangle as well Here is an example How can you tell It is fairly simple if you use the following procedure with the law of sines Suppose you know the lengths of sides a and b and the measure of angle A By the law of sines s1nA SlngB I The n sinB 2sinA a a Procedure Step 1 Calculate sinB sinA a Step 2 Decide if there is a triangle at all formed by the rule that if the value obtained in Step 1 is greater than 1 no triangle is formed Step 3 Ifthere is atriangle formed ie sinB lt 1 use a calculator or the table of special angles to nd the measure of angle B Step 4 Calculate the extra angle 3 180 7 B If AB lt180 7 both angles B and 3 will create triangles so 2 triangles are formed If AB 2180 only angle B will make atriangle so only 1 triangle is formed Example 7 a8 b6 and A120 Example 8 a3 b7 and A110 Example 9 al b3 and A45 Example 10 a6 b8 and A40 Example 11 a7 b4 and A40


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