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NAME -c For use with pages 527-534 Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle and use a geometric mean to solve problems Theorem 9.1 f the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. Theorem 9.2 n a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments. Theorem 9.3 n a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length-of each leg of the right triangle is a geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. Consider the right triangle shown. a. dentify the similar triangles. b. Find the heighte of MBC. SOLUT~ON 6 a. MBC- ACBD- MCD A 10 C Sketch the three similar right triangles so that the corresponding angles and sides have the same orientation. B C B ~ C D A b. Use the fact that MBC- ACED to write a proportion. CD AC CB AB h 6 10 11.6 11.6h = 6(10) h = 5.2 Corresponding Substitute. Cross product property Solve for h. A side lengths are in proportion. Practice Workbook vvith Examples Copyright McDougal Littell nc.

LESSON NAME CONTNUED For use- with pages 527-534.~~f!!.?~~f!.~.!.f!.~.~l!.'!.!!.!.~!..!.. Find the height, h, of the given right triangle. 1. 2. 3. N : w 4 Z M!'.! ~ Using a Geometric Mean Find the value of each variable. a.~ b. 10 4 SOl.UTON a. Apply Theorem 9.3. 10 + 4 x x 4 X2 = 56 x = -J56 = ~ = 2-Jl4 b. Apply Theorem 9.2. 3 l y 1 y2 = 3 y = -J3 t' '.", Copyright McDougal Littell nc. Practice Workbook with Examples

LESSON CONTNUED NAME ~------------ For use with pages 527-534 ;~~.f!!.~~~!.~.!.l!!.. ~!!.l!.'!!.!.~~.?. Find the value of each variable to the nearest tenth. 4. 5. 1 ~ x 6. Practice Workbook vvith Examples Copyright McDougal Littell nc.

NAME rectice with Examples For use with pages 535-541 GZa Find the Length,ot a Hypotenuse' Find the length of the hypotenuse of the right triangle. Tell whether the side lengths form a Pythagorean triple. SO,LUTON 6 (hypotenuse)? = (leg)2 + (leg)2 Pythagorean Theorem X2 = 36 + 64 X2 = 100 x = 10 Substitute. Multiply. Add. Find the positive square root. Because the side lengths 6, 8, and 10 are integers, they form a Pythagorean.triple..~~.~!.~ ~~~~..~C?~. ~l!.~!!.!.~.! '". Find the length of the hypotenuse of the right triangle. Tell whether the side lengths form a Pythagorean triple. 1. 8 2. 3. 7 l'. i i Copyright McDougal. Littell nc. Practice Workbook with Examples

p LESSON 9~2 CONTNUED NAME ~ ~~ ~ Proctice with Examples For use with pages 535-541 Finding the Length of a Leg Find the length of the leg of the right triangle.. SOU.JTBON (hypotenuse)? =.(leg)? + (leg)? 122 = 92 + X2 144 = 81 + X2 63 = x2.,j63 = X Pythagorean Theorem Substitute. Multiply. Subtract 81 from each side. Find the positive square root. Exercises for Example 2... -.,. Find the unknown side length. Round to the nearest tenth, if necessary. 4. ~1. L~ 24 5. 6. 5 7.1 Practice Workbook vvith Examples Copyright McDougal Littell nc. -.::~

LESSON NAME~ CONTNUED For use with pages 535-541 Finding the Area of a Triangle Find the area of the triangle to the nearest tenth. SOLUTON 4 n this case, the side of length 4 can be used as the height and the side of unknown length can be used as the base. To find the length of the unknown side, use the Pythagorean Theorem. (hypotenu~e)2 = (leg)? + (leg)? Pythagorean Theorem 15 2 = 42 + b 2 Substitute. -J209 = b Solve for b. Now find the area of the triangle. A = ~bh = ~(-J209)(4) = 28.9 square units.~~.~!.~~~~~.!.l!.~.~l!.e!.'!!!.~~.~. Find the area of the triangle to the nearest tenth. 7. 8. 9. 13 ~2---l ) i j oj -:1 l J Copyright McDougal tittell lnc. Practice WorkbOok with Examples

=='-'.. ~~~~-..-.-,;,z.:n. &l:t*m~-~~t ~~.~1*.&j G!':t::5f~-::---:=-~~i!ff5~ NAME ~----------~--------------~-- For use with pages 543--548 Use the converse of the Pythagorean Theorem to solve problems and use side lengths to classify triangles by their angle measures ' Theorem 9.5 Converse of the Pythagorean Theorem f the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle..!. Theorem 9.6.. f the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle,. is acute. i Theorem 9.7 f the square of the length of the longest side of a triangle is greater than the sum of the squares of the lengths of the other two sides, then the triangle is obtuse. Verifying Right Triangles The triangles below appear to be right triangles. Tell whether they are right triangles. a. b. ~, - Let c represent the length of the longest side of the triangle (you do not want to call this the "hypotenuse" because you do not yet know if the triangle is a right triangle). Check to see whether the side lengths satisfy the equation c 2 = a 2 + b 2.? a. 10 2 ~ 8 2 + 7 2? 100 ~ 64 + 49 100 * 113 The triangle is not a right triangle.? b. 20 2 ~ 122 + 162? 400 ~ 144 + 256 400 = 400 The triangle is a right triangle. ~ ".r. ~.< ~, ',"/ -:... Practice Workbook vvith Examples Copyright McDougal Littell nc.

CONTNUED NAME For use with pages 543-548.~~.~!.~~~f!.~..~!.~.~l!.?!!.!.~~.!. n Exercises 1-3, determine if the triangles are right triangles. 1. 2. 3. 4 ;: ' :1 Classifying Triangles Decide whether the set of numbers can represent the side lengths of a triangle. f they can, classify the-triangle as right, acute, or obtuse. a. 58,69,80 b. 11,30,39 "';, :! ~,~'{~,\ \'~:~~s:,'" ", Copyright McDougal Littell nc. Practice Workbook with Examples

LESSON 9.3 CONTNUED NAME For use with pages 543-548 SCU..UTUON You can use the Triangle nequality.to confirm that each set of numbers can represent the side lengths of a triangle. Compare the square of the length of the longest side with the sum of the squares of the lengths of the two shorter sides. a. c 2 7 a 2 + b 2 Compare c 2 with a 2 + b 2. 80 2 7 58 2 + 69 2 64007 3364 + 4761 6400 < 8125 Substitute. Multiply. c 2 is less than a 2 + b 2. Because c 2 < a 2 + b 2, the triangle is acute. b. c 2 7 a 2 + b 2 Compare c 2 with a 2 + b 2. 392 7 112+ 3()2 1521 7 121 + 900 Substitute. Multiply. 1521 > 1021 c 2 is greater than a 2 + b 2. Because c 2 > a 2 + b 2, the triangle is obtuse. Exercises for Example 2. a Decide whether the set of numbers can represent the side lengths of a triangle. if they can, classify the triangle as right, acute, or obtuse. 4. 5,.)56, 9 5. 23,44,70 6. 12,80,87 7. 4,7, 10 Practice Workbook vvith Examples Copyright McDougal Littell nc.

NAME For use with pages 551-556 - Find the side lengths of special rjqht triangles.. ±. VOCA.BULARY Right triangles whose angle measures are 45-45 - 90 or 30-60 - 90 are called special right triangles. Theorem 9.8 The 45 45 90 Triangle Theorem n a 45-45 - 90 triangle, the hypotenuse is -fi times as long as each leg. Theorem 9.9 The 30 60 90 Triangle Theorem n a 30-60 - 90 triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is.j3 times as long as the shorter leg.. Finding Side lengths in a 45-45 -90 Triangle 'V"" ~~.= v r ;t2~"'~~'~ ~l~ Find the value of x. 7 x 1 :1 '.',,:1 By the Triangle Sum Theorem, the measure of the third angle is 45. The triangle is a 45 45-90 right triangle, so the length x of the hypotenuse is -fi times the length of a leg. Hypotenuse = -fi. leg 45-45 - 90 Triangle Theorem x = -fi. 7 x = 7-fi Substitute. Simplify. Copyright McDougal Littell nc. Practice Workbook with Examples

LESSON NAME t:- L CONTNUED Practice with' Examples For use with pages 551-556.~~.~!.~~~l!.~.!.~!.. ~l!.'!.'!!p..~f!..!.; ~ ;. Find the value of each variable. 1. 2. 3. y y Finding Side Lengths in a 30-60 -90 Triangle Find the value of x. SOLUTON Because the triangle is a 30-60 - 90 triangle, the longer leg is fi times the length x of the shorter leg. Longer leg = fi.shorter leg 22 = J3. x 22 -=x fi fi 22 -'-=x fifi 22fi' --=x 3 30-60 - 90 Triangle Theorem Substitute. Divide each side by.)3. Multiply numerator and denominator by.)3. Simplify. GeOmletrry Practice Workbook vvith Examples Copyright McDougal Littell nc.

LiiE.SSON CONTNUED!\lAME For use with pages 551-555.~~.~.~~~~~~.!.~.~.~l!.~'!!.!.~i!,.?. Find the value of each variable. 4.,\00 Y, 30 14~X 5., 6. x -~f1;~~ -\~'" Copyright McDougal Littell nc. Practice Workbook with Examples

NAME For use with pages 558-565 Find the sine, the cosine, and the tangent of an acute angle and use triqonornetrlc ratios to solve real-life problems VOCABULARY A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. The three basic trigonometric ratios are sine, cosine, and tangent, which are abbreviated as sin, cos, and tan, respectively.. The angle that your line of sight makes with a line drawn horizontally called the angle of elevation. TrigOnometric Ratios Let MBC be a right triangle. The sine, the cosine, and the tangent of the i acute angle LA are defined as follows. is sin A = side opposite LA = ~. hypotenuse' C B cas A = side adjacent LA = " hypotenuse C side opposite LA a tan A = =- side adjacent LA b A b side a opposite LA ~ ~ sideadjacenttol A ~ ~ Finding Trigonometric Ratios c Find the sine, the cosine, and the tangent of the indicated angle. a. LA b. LB ~82 B A 23.1 C SOLUTiON a. The length of the hypotenuse is 24.5. For LA, the length of the opposite side is 8.2, and the length of the adjacent side is 23.l. '. opp. 8.2 sin A = hypo = 24.5 = 0.3347 A = adj.. = 23:1 = 09429 cas hypo 24.5. tan A = opp. = 8.2 = 0.3550 ad]. 23.1.Geometiy Practice Workbook vvith Examples Copyright McDougal Littell nc.

NAME~ CONTiNUED for use with pages 558-565, J i " b. The length of the hypotenuse is 24.5. ForLB, the length of the opposite side is 23.1 and the length of the adjacent side is 8.2.. B = opp. = 23.1 = 094' 29 sin h. 245. <yp.. adj. 8.2. cos B = hypo = 24.5 = 0.3347 opp. 23.1 tanb = -. = - = 2.8171 ad]. 8.2.~~.~.~~~~l!.~..~~.~.~l!.!.!!.!.~i!..!. find the sine, cosine, and tangent of LA. 1. A 2. B 4.6 c 3. B il 'l 3 5 4 c 6 3.8 A A c ''>1 Copyright McDougal Littell nc. Practice Workbook vvith Examples

LESSQN 9.5 CONTNUED NAME ~~---------------------------------- Practice with Example~ for use with pages 558-565 ~ Estimating a Distance ' t is known that a hill frequently used forsled riding has an angle of elevation of 30 at its bottom. f the length of a sledder's ride is 52',6 feet, estimate the height of the hill. h SOLUTON Use the sine ratio for the 30 0 angle, because you have the value of the hypotenuse and you are looking for the value of the side opposite the 30 angle., ". 00,h sin 3 = 52.6 h = (52.6). sin 30 0 = (52.6). (0.5) = 26.3 feet.~~~!.~ ~~f!.~.!.c!.~.~1!:'!.r!!l!.{~.?. 4. n the sled-riding example, find the height of the hill if the angle of elevation of the hill is 42 0 5. f the angle of elevation from your position on the ground to the top of a building is 67 0 and you are standing 30 meters from the foot of the building, approximate the height of the building. Practice Vvorkbook vvith Examples Copyright McDougal Littelllnc, All rights reserved,

NAME For use with pages 567-572 Solve a right triangle VOCABULARV. To solve a right triangle means to determine the measures of all six parts (the right angle, the two acute angles, the hypotenuse, and the two legs).. Solve the right triangle. A SOUJTON Begin by using the Pythagorean Theorem to find the length of the missing side. (hypotenuse)? = (leg)" + (leg)? Pythagorean Substitute: 5~ CaB Theorem. 169 = a 2 + 25 Multiply. 144 = a 2 Subtract 25 from each side. 12 = a Then find the measure of LB. Find the positive square root. ' tanb = opp. adj. 5 tan B = 12 mlb = 22.6 Substitute. Use a calculator. 5 Finally, because LA and LB are complements, mla =:= 90 - mlb = 90-22.6 = 67.4. 12 you can write The side lengths of MBC are 5, 12, and 13. MBC has one right angle and two acute angles whose measures are about 22.6 and 67.4. Copyright McDougal Littell nc. Practice Workbook vvith Examples

. =-':'7.oo.:..:~~'f-~":"-':' ~~~~~"~7~~:4 2~ ~;...;.~.. ±:.:;;"S# :#-~ '{ :;-.o.~y~ '90. LESSON NAME' ".,, CONTNUED For use with pages 567-572 ~':.f!!.l?,~~f!.~.!.~!..~'!:!.'!!.!.~f!..!. Solve the right triangle. 1. y 2. N 3. Q x fill L Pl...----J...JR 9.4 Solve the right triangle. X SOLUTiON v~ ~ Use trigonometric ratios to find the values of x and y. Z x y sinx = opp. hypo adj. cos X =-h yp.. x sin 71 =-- 32 cos 71 = L 32 32 sin 71 = x 32 cos 71 = y 32(0.9455)= x 30.3 = x Because LX and LY are complements, you can write 1'LLY = 90 - mlx = 90-71 = 19. 32(0.3256) = y 10.4 = y The side lengths of the triangle are about 10.4, 30.3, and 32, The triangle has one right angle and two acute angles whose measures are 71 and 19. Geoli111letry Practice Workbook with Examples Copyright McDougal Littell nc.

CONTNUED NAME ------------------------------------ For use with pages 567-572.~~~!.~~~~~.!l!!..~~i!.'!!l!.~'!..?. Solve the right triangle. 4. A 5. M 6. Q N c L 41.5 R Copyright McDougal Littell nc. GeOnletlrY Practice Workbook with EXamples

NAME ~ ~---------------------- For use with pages 573-579 find the magnitude and the direction of a vector and add vectors VOCABULARY ~ The magnitude of a vector AB is the distance from the initial point A to the terminal pointb andis'written AB'. The direction of a vector is determined by the angle it makes with a horizontal line. Two vectors are equal if they have the same magnitude and direction. Two vectors are parallel i Sum of Two Vectors ~he ~: of it = (a' bl) and v = (a2, b2) is! u + v - (a l + a 2, b l + b 2 )., -- finding the Magnitude of a Vector if they have the same or opposite directions. Points P and Q are the nitial and terminal points of the vector J5Q. Draw PQ in a coordinate plane. Write the component form of the vector and find its magnitude. a. P(l, 2), Q(5, 5) b. P(O, 4), Q(-2, -4) c a. Component form = (x 2 - Xl' h - Yl) fq = (5-1,5-2) = (4,3) Use the Distance Formula to find the magnitude. 1J5Q1 ~..)(5-1)2 + (5-2)2 =,$ =5 b. Component form = (x 2 ~ xl' h - Y) J5Q = (-2-0, -4-4) = (-2, -8) Use the Distance Formula to find the magnitude. 1101 =.)(-2-0)2 + (-4-4)2 = -J68 = 8.2!, ' y, YQ ~---+-- --1---- 4h 1/ l p t! ; i! x ~-----, Y!! p 4.. 1 / 1 2 x Q! 1..a: ~.~!- :...... '. Practice VVork:book: with Examples Copyright McDougal Littell nc.

LESSON NAME "j CONTNUED For use with pages 573-579 Exercises for Example 1, ~. ---'" Draw PO in a coordinate 0 plane. Write the component form of the vector and find its magnitude. 1. P(3, 2), Q(1, 9) 2. P(-2, 1); Q(O,-5)!! '!!,!! r i! i! i!!!!! i!! i i! i,, i! "-n! 3. P(3, 8), Q( -1, 10) 4. p( -4, -11), Q(O,2) J -- ---,! ; i 1---1-1!, r i t t r : -+--+-'-1-+-1--- ---+--,--+-- i!! i i!,! j, i! i i i i i i i, r!! Describing the Direction of a Vector ': The vector AB depicts the velocity of a moving vehicle. The scale on each axis is in kilometers per hour, Find the (a) speed of the vehicle and (b) direction it is traveling relative to east. -----'" a. The magnitude of the vector AB represents the vehicle's 'speed. Use the Distance Formula. \ABl = J(20-5)2 +. (0-15)2 = J15 2 + 15 2 ' = 21.2 The speed of the vehicle is about 21.2 kilometers per hour. w N y i,a! r-, -, rt 5 -- '\. B -l 5 xi s E i Copyright McDougal All rights reserved, Littell lnc. Practice Workbook with Examples

*~~~=--'",:.~## :=a-~h- -~-M 4--*~-i - -re-~f. $ -~-3!=*- 'F (t-,lesson 9.7 CONTNUED NAME ~----------------------------~ For use with pages 573-580 l b. The tangent of the angle formed by the vector and a line drawn,parailel to the x-axis at point A is - ~~. =-l. Use a c~1culator. to find the angle measure. i - 1 i,r~f\ll = - 45 The vehicle is traveling in a direction 45 south of east...~?:.~!.?~!!.l!.~.!.l!.~.~1!.'!.!!.!.~~.? ". in Exercises 5-7, find the vehicle's magnitude and direction if points A and B are as given. 5. A(O, 0), S(6, 7) 6. A( -2,4), B(3, - 1) 7. A(2, 4), B( - 3, -1) Finding the Sum of Two Vectors Let u = (- 4, 2) and v = (3, 1). Write the component form of the sum u + v. SOB..UTU:)N To find the sum vector u + v, add the horizontal components vertical components of u and V. u +.v '= (- 4 + 3, 2 + 1) = (- 1, 3) and add the.~~.~!.?~!!,i!,~.!.l!.~.~1!.'!.'!!l!.~~.~. For the given vectors u and V, find the component the sum u + e. form of 8. u = (0, 8) and v = (- 3, 5) 9. u = (- 2, - 7) and v = (2, 10) 10. u= (3,12) and v = (- 3, -12) t;.' Geome'itt'y Practice Workbook vvith Examples Copyright McDougal Littell lnc.

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