Square Roots and Irrational Numbers Lesson 11-1 Additional Examples Simplify each square root. a. 144 144 = 12 b.  – 81 – 81 = – 9

Square Roots and Irrational Numbers Lesson 11-1 Additional Examples You can use the formula d = 1.5h to estimate the distance d, in miles, to a horizon line when your eyes are h feet above the ground. Estimate the distance to the horizon seen by a lifeguard whose eyes are 20 feet above the ground. Use the formula. d = 1.5h Replace h with 20. d = 1.5(20) Multiply. d = 30 Find perfect squares close to 30. 25 30 36 < Find the square root of the closest perfect square. 25 = 5 The lifeguard can see about 5 miles to the horizon.

Square Roots and Irrational Numbers Lesson 11-1 Additional Examples Identify each number as rational or irrational. Explain. a.  49 rational, because 49 is a perfect square b. 0.16 rational, because it is a terminating decimal c.  3 irrational, because 3 is not a perfect square d. 0.3333 . . . rational, because it is a repeating decimal e. – 15 irrational, because 15 is not a perfect square f. 12.69 rational, because it is a terminating decimal g. 0.1234567 . . . irrational, because it neither terminates nor repeats

The Pythagorean Theorem Lesson 11-2 Additional Examples Find c, the length of the hypotenuse. c2 = a2 + b2 Use the Pythagorean Theorem. Replace a with 28, and b with 21. c2 = 282 + 212 c2 = 1,225 Simplify. c = 1,225 = 35 Find the positive square root of each side. The length of the hypotenuse is 35 cm.

The Pythagorean Theorem Lesson 11-2 Additional Examples Find the value of x in the triangle. Round to the nearest tenth. a2 + b2 = c2 49 + x2 = 196 72 + x2 = 142 Use the Pythagorean Theorem. Simplify. Replace a with 7, b with x, and c with 14. x = 147 x2 = 147 Find the positive square root of each side. Subtract 49 from each side.

The Pythagorean Theorem Lesson 11-2 Additional Examples (continued) Then use one of the two methods below to approximate . 147 Method 1: Use a calculator. is 12.124356. A calculator value for 147 Round to the nearest tenth. x 12.1  Use the table on page 800. Find the number closest to 147 in the N2 column. Then find the corresponding value in the N column. It is a little over 12. Method 2: Use a table of square roots. Estimate the nearest tenth. x 12.1  The value of x is about 12.1 in.

The Pythagorean Theorem Lesson 11-2 Additional Examples The carpentry terms span, rise, and rafter length are illustrated in the diagram. A carpenter wants to make a roof that has a span of 20 ft and a rise of 10 ft. What should the rafter length be? c2 = a2 + b2 Use the Pythagorean Theorem. Replace a with 10 (half the span), and b with 10. c2 = 102 + 102 Square 10. c2 = 100 + 100 Add. c2 = 200 Find the positive square root. c = 200 Round to the nearest tenth. c 14.1 The rafter length should be about 14.1 ft.

The Pythagorean Theorem Lesson 11-2 Additional Examples Is a triangle with sides 10 cm, 24 cm, and 26 cm a right triangle? a2 + b2 = c2 Write the equation for the Pythagorean Theorem. Replace a and b with the shorter lengths and c with the longest length. 102 + 242 262 Simplify. 100 + 576 676 676 = 676 The triangle is a right triangle.

Distance and Midpoint Formulas Lesson 11-3 Additional Examples Find the distance between T(3, –2) and V(8, 3). Use the Distance Formula. d = (x2 – x1)2 + (y2 – y1)2 Replace (x2, y2) with (8, 3) and (x1, y1) with (3, –2). d = (8 – 3)2 + (3 – (–2 ))2 Simplify. d = 52 + 52 Find the exact distance. 50 d = Round to the nearest tenth. d 7.1 The distance between T and V is about 7.1 units.

Distance and Midpoint Formulas Lesson 11-3 Additional Examples Find the perimeter of WXYZ. The points are W (–3, 2), X (–2, –1), Y (4, 0), Z (1, 5). Use the Distance Formula to find the side lengths. (–2 – (–3))2 + (–1 – 2)2 WX = 1 + 9 = 10 = Replace (x2, y2) with (–2, –1) and (x1, y1) with (–3, 2). Simplify. (4 – (–2))2 + (0 – (–1)2 XY = 36 + 1 = = Simplify. 37 Replace (x2, y2) with (4, 0) and (x1, y1) with (–2, –1).

Distance and Midpoint Formulas Lesson 11-3 Additional Examples (continued) 9 + 25 = = (1 – 4)2 + (5 – 0)2 YZ = Simplify. Replace (x2, y2) with (1, 5) and (x1, y1) with (4, 0). 34 (–3 – 1)2 + (2 – 5)2 ZW = Simplify. Replace (x2, y2) with (–3, 2) and (x1, y1) with (1, 5). = 16 + 9 = 25 = 5

Distance and Midpoint Formulas Lesson 11-3 Additional Examples (continued) perimeter = + + + 5 20.1 34 37 10 The perimeter is about 20.1 units.

Distance and Midpoint Formulas Lesson 11-3 Additional Examples Find the midpoint of TV. Use the Midpoint Formula. x1 + x2 2 y1 + y2 , Replace (x1, y1) with (4, –3) and (x2, y2) with (9, 2). = , 4 + 9 2 –3 + 2 Simplify the numerators. = , 13 2 –1 Write the fractions in simplest form. = 6 , – 1 2 The coordinates of the midpoint of TV are 6 , – . 1 2

Special Right Triangles Lesson 11-5 Additional Examples Find the length of the hypotenuse in the triangle. hypotenuse = leg • 2 Use the 45°-45°-90° relationship. y = 10 • 2 The length of the leg is 10. 14.1 Use a calculator. The length of the hypotenuse is about 14.1 cm.

Special Right Triangles Lesson 11-5 Additional Examples Patrice folds square napkins diagonally to put on a table. The side length of each napkin is 20 in. How long is the diagonal? hypotenuse = leg • 2 Use the 45°-45°-90° relationship. y = 20 • 2 The length of the leg is 20. 28.3 Use a calculator. The diagonal length is about 28.3 in.

Special Right Triangles Lesson 11-5 Additional Examples Find the missing lengths in the triangle. hypotenuse = 2 • shorter leg 14 = 2 • b The length of the hypotenuse is 14. = Divide each side by 2. 7 = b Simplify. 14 2 2b longer leg = shorter leg • 3 a = 7 • 3 The length of the shorter leg is 7. a 12.1 Use a calculator. The length of the shorter leg is 7 ft. The length of the longer leg is about 12.1 ft.

Sine, Cosine, and Tangent Ratios Lesson 11-6 Additional Examples Find the sine, cosine, and tangent of A. tan A = = = opposite adjacent 3 4 12 16 sin A = = = opposite hypotenuse 3 5 12 20 cos A = = = adjacent hypotenuse 4 5 16 20

Sine, Cosine, and Tangent Ratios Lesson 11-6 Additional Examples Find the trigonometric ratios of 18° using a scientific calculator or the table on page 779. Round to four decimal places. Scientific calculator: Enter 18 and press the key labeled SIN, COS, or TAN. cos 18° 0.9511 tan 18° 0.3249 sin 18° 0.3090 Table: Find 18° in the first column. Look across to find the appropriate ratio.

Sine, Cosine, and Tangent Ratios Lesson 11-6 Additional Examples The diagram shows a doorstop in the shape of a wedge. What is the length of the hypotenuse of the doorstop? You know the angle and the side opposite the angle. You want to find w, the length of the hypotenuse. sin A = Use the sine ratio. opposite hypotenuse sin 40° = Substitute 40° for the angle, 10 for the height, and w for the hypotenuse. 10 w w(sin 40°) = 10 Multiply each side by w. w = Divide each side by sin 40°. 10 sin 40° w 15.6 Use a calculator. The hypotenuse is about 15.6 cm long.

Angles of Elevation and Depression Lesson 11-7 Additional Examples Janine is flying a kite. She lets out 30 yd of string and anchors it to the ground. She determines that the angle of elevation of the kite is 52°. What is the height h of the kite from the ground? Draw a picture. sin A = Choose an appropriate trigonometric ratio. opposite hypotenuse sin 52° = Substitute. h 30 30(sin 52°) = h Multiply each side by 30. 24 h Simplify. The kite is about 24 yd from the ground.

Angles of Elevation and Depression Lesson 11-7 Additional Examples Greg wants to find the height of a tree. From his position 30 ft from the base of the tree, he sees the top of the tree at an angle of elevation of 61°. Greg’s eyes are 6 ft from the ground. How tall is the tree, to the nearest foot? Draw a picture. Choose an appropriate trigonometric ratio. opposite adjacent tan A = Substitute 61 for the angle measure and 30 for the adjacent side. h 30 tan 61° = 30(tan 61°) = h Multiply each side by 30. 54 h Use a calculator or a table. 54 + 6 = 60 Add 6 to account for the height of Greg’s eyes from the ground. The tree is about 60 ft tall.

Angles of Elevation and Depression Lesson 11-7 Additional Examples An airplane is flying 1.5 mi above the ground. If the pilot must begin a 3° descent to an airport runway at that altitude, how far is the airplane from the beginning of the runway (in ground distance)? Draw a picture (not to scale). tan 3° = Choose an appropriate trigonometric ratio. 1.5 d d • tan 3° = 1.5 Multiply each side by d.

Angles of Elevation and Depression Lesson 11-7 Additional Examples (continued) = Divide each side by tan 3°. d • tan 3° tan 3° 1.5 d = Simplify. 1.5 tan 3° d 28.6 Use a calculator. The airplane is about 28.6 mi from the airport.