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Lesson 7-1 Geometric Mean Lesson 7-2 The Pythagorean Theorem and Its Converse Lesson 7-3 Special Right Triangles Lesson 7-4 Trigonometry Lesson 7-5 Angles of Elevation and Depression Lesson 7-6 The Law of Sines Lesson 7-7 The Law of Cosines Contents

Example 1 Geometric Mean Example 2 Altitude and Segments of the Hypotenuse Example 3 Altitude and Length of the Hypotenuse Example 4 Hypotenuse and Segment of Hypotenuse Lesson 1 Contents

Find the geometric mean between 2 and 50. Let x represent the geometric mean. Definition of geometric mean Cross products Take the positive square root of each side. Simplify. Answer: The geometric mean is 10. Example 1-1a

Find the geometric mean between 25 and 7. Let x represent the geometric mean. Definition of geometric mean Cross products Take the positive square root of each side. Simplify. Use a calculator. Answer: The geometric mean is about 13.2. Example 1-1b

a. Find the geometric mean between 3 and 12. b. Find the geometric mean between 4 and 20. Answer: 6 Answer: 8.9 Example 1-1c

Example 1-2a

Take the positive square root of each side. Cross products Take the positive square root of each side. Use a calculator. Answer: CD is about 12.7. Example 1-2b

Answer: about 8.5 Example 1-2c

KITES Ms. Alspach is constructing a kite for her son KITES Ms. Alspach is constructing a kite for her son. She has to arrange perpendicularly two support rods, the shorter of which is 27 inches long. If she has to place the short rod 7.25 inches from one end of the long rod in order to form two right triangles with the kite fabric, what is the length of the long rod? Example 1-3a

Draw a diagram of one of the right triangles formed. Let be the altitude drawn from the right angle of Example 1-3b

Cross products Divide each side by 7.25. Answer: The length of the long rod is 7.25 + 25.2, or about 32.4 inches long. Example 1-3c

AIRPLANES A jetliner has a wingspan, BD, of 211 feet AIRPLANES A jetliner has a wingspan, BD, of 211 feet. The segment drawn from the front of the plane to the tail, intersects at point E. If AE is 163 feet, what is the length of the aircraft? Answer: about 231.3 ft Example 1-3d

Find c and d in Example 1-4a

is the altitude of right triangle JKL. Use Theorem 7 is the altitude of right triangle JKL. Use Theorem 7.2 to write a proportion. Cross products Divide each side by 5. Example 1-4b

is the leg of right triangle JKL. Use the Theorem 7 is the leg of right triangle JKL. Use the Theorem 7.3 to write a proportion. Cross products Take the square root. Simplify. Use a calculator. Answer: Example 1-4c

Find e and f. f Answer: Example 1-4d

End of Lesson 1

Example 1 Find the Length of the Hypotenuse Example 2 Find the Length of a Leg Example 3 Verify a Triangle is a Right Triangle Example 4 Pythagorean Triples Lesson 2 Contents

LONGITUDE AND LATITUDE Carson City, Nevada, is located at about 120 degrees longitude and 39 degrees latitude. NASA Ames is located about 122 degrees longitude and 37 degrees latitude. Use the lines of longitude and latitude to find the degree distance to the nearest tenth degree if you were to travel directly from NASA Ames to Carson City. Example 2-1a

The change in longitude between NASA Ames and Carson City is or 2 degrees. Let this distance be a. The change in latitude is or 2 degrees latitude. Let this distance be b. Use the Pythagorean Theorem to find the distance in degrees from NASA Ames to Carson City, represented by c. Example 2-1b

Take the square root of each side. Pythagorean Theorem Simplify. Add. Take the square root of each side. Use a calculator. Answer: The degree distance between NASA Ames and Carson City is about 2.8 degrees. Example 2-1c

LONGITUDE AND LATITUDE Carson City, Nevada, is located at about 120 degrees longitude and 39 degrees latitude. NASA Dryden is located about 117 degrees longitude and 34 degrees latitude. Use the lines of longitude and latitude to find the degree distance to the nearest tenth degree if you were to travel directly from NASA Dryden to Carson City. Answer: about 5.8 degrees Example 2-1d

Find d. Example 2-2a

Subtract 9 from each side. Pythagorean Theorem Simplify. Subtract 9 from each side. Take the square root of each side. Use a calculator. Answer: Example 2-2b

Find x. Answer: Example 2-2c

COORDINATE GEOMETRY Verify that is a right triangle. Example 2-3a

Use the Distance Formula to determine the lengths of the sides. Subtract. Simplify. Subtract. Simplify. Example 2-3b

Subtract. Simplify. By the converse of the Pythagorean Theorem, if the sum of the squares of the measures of two sides of a triangle equals the square of the measure of the longest side, then the triangle is a right triangle. Example 2-3c

Converse of the Pythagorean Theorem Simplify. Add. Answer: Since the sum of the squares of two sides equals the square of the longest side, is a right triangle. Example 2-3d

COORDINATE GEOMETRY Verify that is a right triangle. Answer: is a right triangle because Example 2-3e

Determine whether 9, 12, and 5 are the sides of a right triangle Determine whether 9, 12, and 5 are the sides of a right triangle. Then state whether they form a Pythagorean triple. Since the measure of the longest side is 15, 15 must be c. Let a and b be 9 and 12. Pythagorean Theorem Simplify. Add. Example 2-4a

Answer: These segments form the sides of a right triangle since they satisfy the Pythagorean Theorem. The measures are whole numbers and form a Pythagorean triple. Example 2-4b

Determine whether 21, 42, and 54 are the sides of a right triangle Determine whether 21, 42, and 54 are the sides of a right triangle. Then state whether they form a Pythagorean triple. Pythagorean Theorem Simplify. Add. Answer: Since , segments with these measures cannot form a right triangle. Therefore, they do not form a Pythagorean triple. Example 2-4c

Determine whether 4, and 8 are the sides of a right triangle Determine whether 4, and 8 are the sides of a right triangle. Then state whether they form a Pythagorean triple. Pythagorean Theorem Simplify. Add. Answer: Since 64 = 64, segments with these measures form a right triangle. However, is not a whole number. Therefore, they do not form a Pythagorean triple. Example 2-4d

Determine whether each set of measures are the sides of a right triangle. Then state whether they form a Pythagorean triple. a. 6, 8, 10 b. 5, 8, 9 c. Answer: The segments form the sides of a right triangle and the measures form a Pythagorean triple. Answer: The segments do not form the sides of a right triangle, and the measures do not form a Pythagorean triple. Answer: The segments form the sides of a right triangle, but the measures do not form a Pythagorean triple. Example 2-4e

End of Lesson 2

Example 1 Find the Measure of the Hypotenuse Example 2 Find the Measure of the Legs Example 3 30°–60°–90° Triangles Example 4 Special Triangles in a Coordinate Plan Lesson 3 Contents

WALLPAPER TILING The wallpaper in the figure can be divided into four equal square quadrants so that each square contains 8 triangles. What is the area of one of the squares if the hypotenuse of each 40°-45°-90° triangle measures millimeters? Example 3-1a

The area of one of these triangles is or 24.5 millimeters. The length of the hypotenuse of one 40°-45°-90° triangle is millimeters. The length of the hypotenuse is times as long as a leg. So, the length of each leg is 7 millimeters. The area of one of these triangles is or 24.5 millimeters. Answer: Since there are 8 of these triangles in one square quadrant, the area of one of these squares is 8(24.5) or 196 mm2. Example 3-1b

WALLPAPER TILING If each 40°-45°-90° triangle in the figure has a hypotenuse of millimeters, what is the perimeter of the entire square? Answer: 80 mm Example 3-1c

Find a. The length of the hypotenuse of a 40°-45°-90° triangle is times as long as a leg of the triangle. Example 3-2a

Rationalize the denominator. Divide each side by Rationalize the denominator. Multiply. Divide. Answer: Example 3-2b

Find b. Answer: Example 3-2c

Find QR. Example 3-3a

is the longer leg, is the shorter leg, and is the hypotenuse. Multiply each side by 2. Answer: Example 3-3b

Find BC. Answer: BC = 8 in. Example 3-3c

COORDINATE GEOMETRY is a 30°-60°-90° triangle with right angle X and as the longer leg. Graph points X(-2, 7) and Y(-7, 7), and locate point W in Quadrant III. Example 3-4a

Graph X and Y. lies on a horizontal gridline of the coordinate plane Graph X and Y. lies on a horizontal gridline of the coordinate plane. Since will be perpendicular to it lies on a vertical gridline. Find the length of Example 3-4b

is the shorter leg. is the longer leg. So, Use XY to find WX. Point W has the same x-coordinate as X. W is located units below X. Answer: The coordinates of W are or about Example 3-4b

Answer: The coordinates of S are or about COORDINATE GEOMETRY is at 30°-60°-90° triangle with right angle R and as the longer leg. Graph points T(3, 3) and R(3, 6) and locate point S in Quadrant III. Answer: The coordinates of S are or about Example 3-4d

End of Lesson 3

Example 1 Find Sine, Cosine, and Tangent Ratios Example 2 Evaluate Expressions Example 3 Use Trigonometric Ratios to Find a Length Example 4 Use Trigonometric Ratios to Find an Angle Measure Lesson 4 Contents

Find sin L, cos L, tan L, sin N, cos N, and tan N Find sin L, cos L, tan L, sin N, cos N, and tan N. Express each ratio as a fraction and as a decimal. Example 4-1a

Example 4-1b

Example 4-1c

Example 4-1d

Answer: Example 4-1e

Find sin A, cos A, tan A, sin B, cos B, and tan B Find sin A, cos A, tan A, sin B, cos B, and tan B. Express each ratio as a fraction and as a decimal. Example 4-1f

Answer: Example 4-1g

Use a calculator to find tan to the nearest ten thousandth. KEYSTROKES: 56 1.482560969 TAN ENTER Answer: Example 4-2a

Use a calculator to find tan to the nearest ten thousandth. KEYSTROKES: 90 0 COS ENTER Answer: Example 4-2b

a. Use a calculator to find sin 48° to the nearest ten thousandth. b. Use a calculator to find cos 85° to the nearest ten thousandth. Answer: Answer: Example 4-2c

EXERCISING A fitness trainer sets the incline on a treadmill to The walking surface is 5 feet long. Approximately how many inches did the trainer raise the end of the treadmill from the floor? Let y be the height of the treadmill from the floor in inches. The length of the treadmill is 5 feet, or 60 inches. Example 4-3a

Use a calculator to find y. Multiply each side by 60. Use a calculator to find y. KEYSTROKES: 60 7 7.312160604 SIN ENTER Answer: The treadmill is about 7.3 inches high. Example 4-3b

CONSTRUCTION The bottom of a handicap ramp is 15 feet from the entrance of a building. If the angle of the ramp is about how high does the ramp rise off the ground to the nearest inch? Answer: about 15 in. Example 4-3c

COORDINATE GEOMETRY Find mX in right XYZ for X(–2, 8), Y(–6, 4), and Z(–3, 1). Example 4-4a

Explore You know the coordinates of the vertices of a right triangle and that is the right angle. You need to find the measures of one of the angles. Plan Use the Distance Formula to find the measure of each side. Then use one of the trigonometric ratios to write an equation. Use the inverse to find Solve or Example 4-4b

or or Example 4-4c

Use the cosine ratio. Simplify. Solve for x. Example 4-4d

Use a calculator to find KEYSTROKES: 4 5 2ND ENTER ) Examine Use the sine ratio to check the answer. Simplify. Example 4-4e

Answer: The measure of is about 36.9. KEYSTROKES: 3 5 2ND ENTER ) Answer: The measure of is about 36.9. Example 4-4f

COORDINATE GEOMETRY Answer: about 56.3 Example 4-4g

End of Lesson 4

Example 1 Angle of Elevation Example 2 Angle of Depression Example 3 Indirect Measurement Lesson 5 Contents

CIRCUS ACTS At the circus, a person in the audience watches the high-wire routine. A 5-foot-6-inch tall acrobat is standing on a platform that is 25 feet off the ground. How far is the audience member from the base of the platform, if the angle of elevation from the audience member’s line of sight to the top of the acrobat is Make a drawing. Example 5-1a

Multiply both sides by x. Since QR is 25 feet and RS is 5 feet 6 inches or 5.5 feet, QS is 30.5 feet. Let x represent PQ. Multiply both sides by x. Divide both sides by tan Simplify. Example 5-1b

Answer: The audience member is about 60 feet from the base of the platform. Example 5-1c

DIVING At a diving competition, a 6-foot-tall diver stands atop the 32-foot platform. The front edge of the platform projects 5 feet beyond the ends of the pool. The pool itself is 50 feet in length. A camera is set up at the opposite end of the pool even with the pool’s edge. If the camera is angled so that its line of sight extends to the top of the diver’s head, what is the camera’s angle of elevation to the nearest degree? Answer: about Example 5-1d

SHORT-RESPONSE TEST ITEM A wheelchair ramp is 3 meters long and inclines at Find the height of the ramp to the nearest tenth centimeter. Read the Test Item The angle of depression between the ramp and the horizontal is Use trigonometry to find the height of the ramp. Solve the Test Item Method 1 The ground and the horizontal level with the platform to which the ramp extends are parallel. Therefore, since they are alternate interior angles. Example 5-2a

Answer: The height of the ramp is about 0.314 meters, Y W Mulitply each side by 3. Simplify. Answer: The height of the ramp is about 0.314 meters, Example 5-2b

Method 2 The horizontal line from the top of the platform to which the wheelchair ramp extends and the segment from the ground to the platform are perpendicular. So, and are complementary angles. Therefore, Y W Example 5-2c

Answer: The height of the ramp is about 0.314 meters, Multiply each side by 3. Simplify. Answer: The height of the ramp is about 0.314 meters, Example 5-2d

Answer: The roller coaster car was about 285 feet above the ground. SHORT-RESPONSE TEST ITEM A roller coaster car is at one of its highest points. It drops at a angle for 320 feet. How high was the roller coaster car to the nearest foot before it began its fall? Answer: The roller coaster car was about 285 feet above the ground. Example 5-2e

Vernon is on the top deck of a cruise ship and observes two dolphins following each other directly away from the ship in a straight line. Vernon’s position is 154 meters above sea level, and the angles of depression to the two dolphins are Find the distance between the two dolphins to the nearest meter. Example 5-3a

are right triangles. The distance between the dolphins is JK or Use the right triangles to find these two lengths. Because are horizontal lines, they are parallel. Thus, and because they are alternate interior angles. This means that Example 5-3b

Multiply each side by JL. Divide each side by tan Use a calculator. Example 5-3c

Multiply each side by KL. Divide each side by tan Use a calculator. Answer: The distance between the dolphins is , or about 8 meters. Example 5-3d

Madison looks out her second-floor window, which is 15 feet above the ground. She observes two parked cars. One car is parked along the curb directly in front of her window, and the other car is parked directly across the street from the first car. The angles of depression of Madison’s line of sight to the cars are Find the distance between the two cars. Answer: about 24 feet Example 5-3e

End of Lesson 5

Example 1 Use the Law of Sines Example 2 Solve Triangles Example 3 Indirect Measurement Lesson 6 Contents

Find p. Round to the nearest tenth. Example 6-1a

Law of Sines Cross products Divide each side by tan Use a calculator. Answer: Example 6-1b

to the nearest degree in , Law of Sines Cross products Divide each side by 7. Example 6-1c

Solve for L. Use a calculator. Answer: Example 6-1d

a. Find c. b. Find mT to the nearest degree in RST if r = 12, t = 7, and mT = 76. Answer: Answer: Example 6-1e

. Round angle measures to the nearest degree and side measures to the nearest tenth. We know the measures of two angles of the triangle. Use the Angle Sum Theorem to find Example 6-2a

Subtract 120 from each side. Angle Sum Theorem Add. Subtract 120 from each side. Since we know and f, use proportions involving Example 6-2b

Divide each side by sin 8°. To find d: Law of Sines Substitute. Cross products Divide each side by sin 8°. Use a calculator. Example 6-2c

Divide each side by sin 8°. To find e: Law of Sines Substitute. Cross products Divide each side by sin 8°. Use a calculator. Answer: Example 6-2d

Round angle measures to the nearest degree and side measures to the nearest tenth. We know the measure of two sides and an angle opposite one of the sides. Law of Sines Cross products Example 6-2e

Subtract 116 from each side. Divide each side by 16. Solve for L. Use a calculator. Angle Sum Theorem Substitute. Add. Subtract 116 from each side. Example 6-2f

Law of Sines Cross products Divide each side by sin Use a calculator. Answer: Example 6-2g

a. Solve. Round. angle measures to the nearest degree and side a. Solve Round angle measures to the nearest degree and side measures to the nearest tenth. b. Round angle measures to the nearest degree and side measures to the nearest tenth. Answer: Answer: Example 6-2h

Draw a diagram Draw Then find the A 46-foot telephone pole tilted at an angle of from the vertical casts a shadow on the ground. Find the length of the shadow to the nearest foot when the angle of elevation to the sun is Draw a diagram Draw Then find the Example 6-3a

Since you know the measures of two angles of the triangle, and the length of a side opposite one of the angles you can use the Law of Sines to find the length of the shadow. Example 6-3b

Answer: The length of the shadow is about 75.9 feet. Law of Sines Cross products Divide each side by sin Use a calculator. Answer: The length of the shadow is about 75.9 feet. Example 6-3c

A 5-foot fishing pole is anchored to the edge of a dock A 5-foot fishing pole is anchored to the edge of a dock. If the distance from the foot of the pole to the point where the fishing line meets the water is 45 feet, about how much fishing line that is cast out is above the surface of the water? Answer: About 42 feet of the fishing line that is cast out is above the surface of the water. Example 6-3d

End of Lesson 6

Example 1 Two Sides and the Included Angle Example 2 Three Sides Example 3 Select a Strategy Example 4 Use Law of Cosines to Solve Problems Lesson 7 Contents

Use the Law of Cosines since the measures of two sides and the included angle are known. Example 7-1a

Take the square root of each side. Law of Cosines Simplify. Take the square root of each side. Use a calculator. Answer: Example 7-1b

Answer: Example 7-1c

Law of Cosines Simplify. Example 7-2a

Subtract 754 from each side. Divide each side by –270. Solve for L. Use a calculator. Answer: Example 7-2b

Answer: Example 7-2c

Determine whether the Law of Sines or the Law of Cosines should be used first to solve Then solve Round angle measures to the nearest degree and side measures to the nearest tenth. Since we know the measures of two sides and the included angle, use the Law of Cosines. Example 7-3a

Take the square root of each side. Law of Cosines Take the square root of each side. Use a calculator. Next, we can find If we decide to find we can use either the Law of Sines or the Law of Cosines to find this value. In this case, we will use the Law of Sines. Example 7-3b

Take the inverse of each side. Law of Sines Cross products Divide each side by 46.9. Take the inverse of each side. Use a calculator. Example 7-3c

Use the Angle Sum Theorem to find Subtract 168 from each side. Answer: Example 7-3d

Determine whether the Law of Sines or the Law of Cosines should be used first to solve Then solve Round angle measures to the nearest degree and side measures to the nearest tenth. Answer: Example 7-3e

Since is an isosceles triangle, AIRCRAFT From the diagram of the plane shown, determine the approximate exterior perimeter of each wing. Round to the nearest tenth meter. Since is an isosceles triangle, Example 7-4a

Use the Law of Sines to find KJ. Cross products Divide each side by sin . Simplify. Example 7-4b

Use the Law of Sines to find . Cross products Divide each side by 9. Solve for H. Use a calculator. Example 7-4c

Use the Angle Sum Theorem to find Subtract 95 from each side. Example 7-4d

Use the Law of Sines to find HK. Cross products Divide each side by sin Use a calculator. Example 7-4e

The perimeter of the wing is equal to Answer: The perimeter is about or about 67.1 meters. Example 7-4f

The rear side window of a station wagon has the shape shown in the figure. Find the perimeter of the window if the length of DB is 31 inches. Answer: about 93.5 in. Example 7-4g

End of Lesson 7