Chapter Nine Notes SN P U1C9

Transcription

1 Chapter Nine Notes SN P UC9 Name Period Section 9.: Applications Involving Right Triangles To evaluate trigonometric functions with a calculator, there are a few important things to know: On your calculator, press the MODE button to switch between radians and degrees. BE SURE TO SET IT IN THE PROPER MODE OR YOU WILL NOT GET THE RIGHT RESULTS. Be sure to use parentheses when necessary. For example, putting into your calculator sin /2" is NOT the same as putting into your calculator sin (/2). (Your calculator is very specific about using the order of operations.) To find the angle that goes with a trigonometric measure, use the inverse function of your ratio. For example, if you knew that sin = 0.5 and you wanted to find the angle corresponding with, press the 2nd button and then sin to enter sin - (0.5). This is NOT the same as (sin 0.5) -. The - indicates that the function does the inverse (or, in layman s terms, reverse function) of that which is indicated. In degree mode, sin - (0.5) = 30, meaning that a sine ratio of ½ comes from a 30 angle. To find cosecant, secant, and tangent, notice that the cosecant, secant, and tangent are reciprocals of sine, cosine, and tangent, respectively. Therefore, - o csc 30 (sin 30) sin 30 - o sec 30 (cos 30) cos 30 - o cot 30 (tan 30) tan 30 Example : Solve for. page SN P UC9

2 Example 2: Solve for x and y. There are real-life applications for the use of trigonometric functions. Example 3: You want to measure the width across a lake before you swim across it. To measure the width, you plant a stake on one side of the lake, directly across from the dock. You then walk 25 meters to the right of the dock and measure a 45 angle between the stake and the dock. What is the width w of the lake? First, draw a picture to understand the problem. There are two specific types of angles used in real-life application concerning trigonometry. One is the angle of elevation, the acute angle formed by the line of sight upward and the horizontal, meaning the angle one looks up at from the ground to see something. The angle of depression is the opposite; it is the acute angle formed by the line of sight downward and the horizontal, meaning the angle one looks down from a line parallel to the ground to see something. page 2 SN P UC9

3 Example 4: In the diagram shown below, a kite at point C is being held by a string from point A that is 60 ft long. The angle of elevation of the kite from point A is 20. An observer at point B is 0 ft from the point on the ground D directly below the kite. What is the angle of elevation of the kite from point B. Section 5.5: The Law of Sines Up to this point, you have been dealing with right triangles and finding trigonometric values for these triangles. However, trigonometric values can be expanded in the use of finding lengths of sides and measures of angles in acute and obtuse triangles. This is where the Law of Sines is introduced. (Note this will work with acute or obtuse triangles.) h sin A b h = b sin A h sin B a h = a sin B b sin A = a sin B sin A sin B a b The same process can be used to establish the following: sin B b sin C c The combination of these two equations gives us the Law of Sines: page 3 SN P UC9

4 Law of Sines: Back in geometry, you learned some congruence theorems: Angle-Side-Angle Congruence Theorem (ASA) Angle-Angle-Side Congruence Theorem (AAS) Side-Angle-Side Congruence Theorem (SAS) Side-Side-Side Congruence Theorem (SSS) For purposes of the Law of Sines, the two congruence theorems that correlate with the Law of Sines are the AAS and ASA theorems. In other words, if you know two angles and a side of a triangle, you can find its other angle and other two sides. Example: Solve the triangle. NOTE: When a problem states to solve for a triangle, it means to solve for the other angle(s) and/or side(s). #: A = 60, B = 45, b = 3.7: #2: A = 40, B = 30, b = 0: page 4 SN P UC9

5 In geometry, you learned that you can draw any triangle based on knowing two angles and a side, knowing two sides and an angle between them, or knowing three sides of a triangle. Because of this, the ASA, AAS, SAS, and SSS Congruence Theorems could be developed. However, what happens if there are two sides and an angle opposite one of the sides? These are called the ambiguous cases. When two sides and an angle opposite one of the sides is given, there are three possibilities: One triangle is formed. Two triangles can be formed. No triangle is formed. The following table illustrates how you can determine the possibilities based on triangle constructions you learned in geometry: Possible Triangles in the SSA Case In dealing with a triangle given sides a and b and angle A, if angle A is: and the relationship between a, b, and/or A is: then the number of triangles that can be formed is: Diagram of triangle construction obtuse a b obtuse a > b acute acute acute b sin A > a or h > a b sin A < a < b or h < a < b b sin A = a or h = a acute a > b page 5 SN P UC9

6 Example: Respond in one of the following ways: (a) State, Cannot be solved with the Law of Sines. (b) State, No triangle is formed. (a) Solve the triangle. #3: A = 6, a = 8, b = 2: #4: A = 36, a = 5, b = 28: Example: What is the width w of the island in the figure shown below? Section 5.6: The Law of Cosines As you learned in the previous section, the Law of Sines is used to find other measures of angles or sides when two angles and a side of the triangle are known. Because of the use of two angles and a side, the Law of Sines is related to the AAS and ASA Congruence Theorems. However, what about the SAS and SSS Congruence Theorems? In this case, the Law of Cosines is used. Let a, b, and c be the lengths of the sides of ABC as shown. Draw altitude CD having length h. If you let AD = x and DB = c x, you can use the Pythagorean theorem to find two different expressions for h 2. h 2 + (c x) 2 = a 2 x 2 + h 2 = b 2 h 2 = a 2 (c x) 2 h 2 = b 2 x 2 a 2 (c x) 2 = b 2 x 2 a 2 (c 2 2cx + x 2 ) = b 2 x 2 a 2 c 2 + 2cx x 2 = b 2 x 2 a 2 = b 2 + c 2 2cx page 6 SN P UC9

7 Since you generally will not know what x is, you can set up some trigonometric identity to replace it. Since cos A = b x, x = b cos A. a 2 = b 2 + c 2 2c(b cos A) a 2 = b 2 + c 2 2bc cos A You can do the same proof to prove b 2 and c 2. Law of Cosines: For all intents and purposes, the Law of Cosines, which is much different from the Law of Sines, is similar to the Pythagorean Theorem and is often called the generalized Pythagorean Theorem. Realize, as stated before, that the Law of Cosines is used in the SSS and SAS Congruence cases as the Law of Cosines can be used so find an angle based on knowing three sides for finding a side based on two sides and the angle between them. The Law of Cosines can be used for the SSA case (in other words, two sides and an angle opposite one of the sides), but, as in Section 9.2, one, two, or no triangle may be involved. You can use the same table from the Section 9.2 notes to identify if one, two, or no triangle is formed. If you want to solve a triangle, for the SAS case, solve the third side first, use the Law of Cosines again to figure out another angle, and then figure out the remaining angle by realizing that all angles of a triangle add up to 80. for the SSS case, solve one of the angles of the triangle using the Law of Cosines, isolating cos A, cos B, or cos C and then using the inverse cosine to solve for the angle, use the Law of Cosines to figure out a second angle, and then solve the remaining angle by realizing that all angles of a triangle add up to 80. Example: Solve the triangle. #: A = 55, b = 2, c = 7: The SAS Case: page 7 SN P UC9

8 #2: a = 3.2, b = 7.6, c = 6.4: The SSS Case: Back when you had geometry, you learned the Triangle Inequality Theorem, stating that no side of a triangle can be longer than the sum of the other two sides, or a + b > c. If one side is longer than the sum of the other two sides, the triangle cannot be formed. Example: Solve the triangle. #3: a =, b = 5, c = 4: Section 9.4: Area of a Triangle Area of a Triangle: From the figure at the beginning of the notes (as shown to the side again), sin A = b h, meaning that h = b sin A. Also recall from geometry that the area of a triangle is A bh 2. Since the base is c and h = b sin A, A cb sin A, or A bc sin A. This formula 2 2 can be expanded to all other triangles. page 8 SN P UC9

9 Example: Find the area of the triangle. #: A = 47, b = 32 ft, c = 9 ft: Another formula of finding the area of a triangle is called Heron s Formula. This formula involves simply using the sides of the triangle and works for acute, right, and obtuse triangles. Heron s Formula for Area of a Triangle: Example: Decide whether a triangle can be formed with the given side lengths. If so, use Heron s formula to find the area of the triangle. #2: a = 4, b = 5, c = 8: Example: The player waiting to receive a kickoff stands at the 5 yard line (point A) as the ball is being kicked 65 yards up the field from the opponent s 30 yard line. The kicked ball travels 73 yards at an angle of 8 to the right of the receiver, as shown in the figure (point B). Find the distance the receiver runs to catch the ball. page 9 SN P UC9