SOLVING RIGHT TRIANGLES

Transcription

1 PYTHAGOREAN THEOREM SOLVING RIGHT TRIANGLES An triangle tat as a rigt angle is called a RIGHT c TRIANGLE. Te two sides tat form te rigt angle, a and b, a are called LEGS, and te side opposite (tat is, across te triangle from) te rigt angle, c, is called te HYPOTENUSE. b For an rigt triangle, te sum of te squares of te legs of te triangle is equal to te square of te potenuse, tat is, a 2 + b 2 = c 2. Tis relationsip is known as te PYTHAGOREAN THEOREM. In words: (leg) 2 + (leg) 2 = (potenuse) 2. Etension: If (leg) 2 + (leg) 2 < (potenuse) 2, ten te triangle is obtuse and if (leg) 2 + (leg) 2 > (potenuse) 2, ten te triangle is acute. Eamples: Draw a diagram, ten use te Ptagorean teorem to solve eac problem. a) Solve for te missing side c c = 17 2 c = 289 c 2 = 120 c = 120 c = 2 30 c c) One end of a ten foot ladder is four feet from te base of a wall. How ig on te wall does te top of te ladder touc? b) Find te distance from ( 4, 2) to (2, 3). ( 4, 2) 5 After calculating or counting te lengt of te legs, D 2 = = 61 and so D = d) Could 3, 6 and 8 represent te lengts of te sides of a rigt triangle? Eplain ? = 8 2? = D 6 (2, 3) = = = Te ladder touces te wall about 9.2 feet above te ground. Since te Ptagorean teorem relationsip is not true for tese lengts, te cannot be te side lengts of a rigt triangle. Furtermore, since 45 < 64, tis triangle is obtuse.

2 Use te Ptagorean teorem to find te value of. Round answers to te nearest tent Solve te following word problems. Remember to draw a diagram of eac situation. 11. A 12 foot ladder is si feet from a wall. How ig on te wall does te ladder touc? 12. A foot ladder is five feet from a wall. How ig on te wall does te ladder touc? 13. A 9 foot ladder is tree feet from a wall. How ig on te wall does te ladder touc? 14. Wat is te distance from ( 1, 1) to (3, 4)?. Wat is te distance from ( 1, 3) to (4, 1)? 16. Wat is te distance from (2, 5) to ( 3, 1)? 17. Could 8, 12, and 13 represent te lengts of sides of a rigt triangle? Justif our answer. 18. Could 5, 12, and 13 represent te lengts of sides of a rigt triangle? Justif our answer. 19. Could 9, 12, and represent te lengts of sides of a rigt triangle? Justif our answer. 20. Could 10,, and 20 represent te lengts of sides of a rigt triangle? Justif our answer. 21. Wat is te longest fising pole tat could fit in a 2 foot b 3 foot b 4 foot bo? 22. Wat is te longest straigt wire tat can be stretced in a 30 foot b 30 foot b 10 foot classroom? Answers ft ft ft no (acute) 18. es 19. es 20. no (obtuse) ft ft

3 RIGHT TRIANGLE TRIGONOMETRY Te tree basic trigonometric ratios for rigt triangles are te sine (pronounced "sign"), cosine, and tangent. Eac one is used in separate situations, and te easiest wa to remember wic to use wen is te mnemonic SOH-CAH-TOA. Wit reference to one of te acute angles in a rigt triangle, Sine uses te Opposite and te Hpotenuse - SOH. Te Cosine uses te Adjacent side and te Hpotenuse - CAH, and te Tangent uses te Opposite side and te Adjacent side -TOA. In eac case, te position of te angle determines wic leg (side) is opposite or adjacent. Remember tat opposite means across from and adjacent means net to. A tan A = sin A = cos A = B C opposite leg adjacent leg = BC AC opposite leg potenuse = BC AB adjacent leg potenuse = AC AB Eample 1 Use trigonometric ratios to find te lengts of eac of te missing sides of te triangle below ft Te lengt of te adjacent side wit respect to te 42 angle is 17 ft. To find te lengt, use te tangent because is te opposite side and we know te adjacent side. tan 42 = tan 42 =.307 ft Te lengt of is approimatel.31 feet. To find te lengt, use te cosine ratio (adjacent and potenuse). cos 42 = 17 cos 42 = 17 = 17 cos ft Te potenuse is approimatel 22.9 ft long. Eample 2 Use trigonometric ratios to find te size of eac angle and te missing lengt in te triangle below. 18 ft v u 21 ft To find m u, use te tangent ratio because ou know te opposite (18 ft) and te adjacent (21 ft) sides. tanu = m u = tan Te measure of angle u is approimatel B subtraction we know tat m v Use te sine ratio for m u and te opposite side and potenuse. sin 40.6 = 18 sin 40.6 = 18 sin 40.6 = ft Te potenuse is approimatel 27.7 ft long.

4 Problems Use trigonometric ratios to solve for te variable in eac figure below z 8. z w w u v 88

5 Draw a diagram and use trigonometric ratios to solve eac of te following problems. 17. Juanito is fling a kite at te park and realizes tat all 500 feet of string are out. Margie measures te angle of te string wit te ground b using er clinometer and finds it to be How ig is Juanito s kite above te ground? 18. Nell s kite as a 350 foot string. Wen it is completel out, Ian measures te angle wit te ground to be How far would Ian need to walk to be directl under te kite? 19. Mafield Hig Scool s flagpole is feet ig. Using a clinometer, Tamara measured an angle of 11.3 to te top of te pole. Tamara is 62 inces tall. How far from te flagpole is Tamara standing? 20. Tamara took anoter sigting of te top of te flagpole from a different position. Tis time te angle is If everting else is te same, ow far from te flagpole is Tamara standing? 21. AN APPLICATION: URBAN SPRAWL As American cities epanded during te Twentiet Centur, tere were often few controls on ow te land was divided. In te town of Dr Creek, one particular tract of land ad te sape and dimensions sown in te figure at rigt. Te developer planned to build five omes per acre. Determine te number of omes tat can be built on tis tract of land. Sow all dissections and subproblems. One mile is 5,280 feet and one acre contains 43,560 sq. ft Answers 1. = sin = 8 sin = 23 cos = 37 cos = tan = 43 tan z = sin z = 18 sin w = 23 cos w = cos = tan = 91 tan = tan u = tan = tan

6 v = tan ft ft 350 ft 42.5 sin 42.5 = 500 = 500 sin ft 47.5 d ft cos 47.5 = d 350 d = 350 cos ft ft in ft feet = 180 inces, 180" 62" = 118" = inces or 49.2 ft. 62 in 58.4 ft = 118", tan58.4 = tan58.4 = 118, = 11 8, 11 8 tan 58.4 ft inces or 6.05 ft. 21. (1.3 mi.)(5,280 ft./mi.) = 6,864 ft., A( ) = 0.5(6864)(6864) = 23,557,248 sq.'; A(sm ): p. = 1.3 2, so = , b = , A = 0.5( )( ) = 20,401,175 sq.'; A(lg ): base = 1 mi., = 3 mi., A = 0.5(5280)( ) = 24,143,403; Total area = 68,101,826, divide b 43,560 = acres, mult. b 5 ouses per acre, ouses, so 7817 ouses.