Section 10.2 Angles and Triangles

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1 117 Ojective #1: Section 10.2 nges n Tringes Unerstning efinitions of ifferent types of nges. In the intersection of two ines, the nges tht re cttycorner fro ech other re vertic nges. Vertic nges wi hve the se esure. n c re one pir n n re the other pir. opeentry nges re two cute nges whose su is 90. Ech is ce the copeent of the other. n re copeentry nges since + = 90. Suppeentry nges re two nges tht re ess thn 180 whose su is 180. Ech is ce the suppeent of the other. XYW n WYZ re suppeentry nges since XYW + WYZ = 180. orresponing nges re those nges tht shre the se oction in their respective intersections of two or ore ines. If two ines tht re cut y trnsvers re pre, then the corresponing nges re equ. correspons to e correspons to f c correspons to g correspons to h ternte interior nges re those nges insie the two ines shring the opposite octions, i.e., top - eft n otto - right. If the two ines re pre, then the ternte interior nges re equ. c n e re one pir of ternte interior nges n n h re the other pir. If, c = e n = h. If, = e, = f, c = g, n = h. X c c c h g h g W Y e f e f Z

2 118 ternte exterior nges re those nges outsie the two ines shring the opposite octions, i.e., top - eft n otto - right. If the two ines re pre, then the ternte exterior nges re equ. n g re one pir of ternte interior nges n n f re the other pir. If, = g n = f. c h g e f Fin the su of the nges of the tringe picture eow: Ex. 1 We egin y eing ech nge of the tringe, extening the sies of the tringe n rwing ine pre to the top sie of the tringe through the otto nge.,, & 3 for ine. This ens tht = 180. Since n t 1 is trnsvers, then 1 n re pir of ternte interior nges. Hence, 1 = n we cn repce y 1 in the foru ove to get: = 180 ut, t 2 is trnsvers, so 2 n re pir of ternte interior nges. Hence, 2 = n we cn repce y 2 to get: = 180. Thus, the su of the esures of the nges is t 1 t This is true for ny tringe, so the su of the esures of the nges of tringe is equ to 180.

3 119 In the igr eow, fin the issing nges. ssue : Ex Since 5 n 56 re vertic nges, 5 = 56. Since 1 n 43 re ternte exterior nges, 1 = 43. ut 1 n 3 re vertic nges, thus 3 = 43. 1, 2, n 56 for stright ine n 1 = 43, hence 180 = Soving for 2 yies 2 = 81. Yet, 2 n 4 re vertic nges which ens 4 = 81. Since 5 correspons to 8, 8 = 5 = 56. ut 6 n 8 re vertic nges, so 6 = n 8 re suppeentry nges, therefore 7 = = = n 9 re vertic nges n thus 9 = n 43 re vertic nges which ens 10 = 43. Since 10 n 11 re suppeentry nges, 11 = = 137. ut, 11 n 12 re vertic nges, so 12 = 137. Therefore: 1 = 43 2 = 81 3 = 43 4 = 81 5 = 56 6 = 56 7 = = 56 9 = = = = 137

4 Ojective 2: Unerstning n ssifying ifferent Types of Tringes. We cn cssify tringes y their sies: Equiter Tringe Isoscees Tringe Scene Tringe sies re equ. Two sies re equ. No sies re equ. 120 nges esure 60. The two se nges re equ. No nges re equ. n equiter tringe is so ce n equingur tringe. The thir nge in n isoscees tringe fore y the two equ sies is ce the vertex nge. We cn so cssify tringes y their nges: cute Tringe Right Tringe Otuse Tringe nges re cute. Hs on right nge. Hs one otuse nge. In right tringe, the two sies tht intersect to for the right nge re ce the egs of the right tringe whie the thir sie (the ongest sie) of right tringe is ce the hypotenuse of the right tringe. eterine wht type of tringe is picture eow: Ex. 3 Ex. 4 X R T

5 121 Since the esures of the Since TXR n 144 re suppenges of tringe tot entry nges, then 180, then TXR = = = 180 ut, TXR + T = = 180 = 90. So, R = = = 109 Since the nges re ifferent, = 71. then none of the sies re equ. So, is n Isoscees Thus, XRT is scene n n n cute Tringe. right tringe. Ojective #3: ppying the Pythgoren Theore In right tringe, there is speci retionship etween the ength of the egs ( n ) n the hypotenuse (c). This is known s the Pythgoren Theore: Pythgoren Theore In right tringe, the squre of the hypotenuse (c 2 ) is equ to the su of the squres of the egs ( ) c c 2 = Keep in in tht the hypotenuse is the ongest sie of the right tringe. Fin the ength of the issing sies: Ex Ex ft 15 ft In this proe, we hve In this proe, we hve one eg the two egs of the tringe n the hypotenuse n we re n we re ooking for the ooking for the other eg: hypotenuse: c 2 = c 2 = (15) 2 = (6) c 2 = (7.8) 2 + (10.2) = (sove for 2 ) c 2 = = 36 c 2 = = 2

6 122 To fin c, tke the squre To fin, tke the squre root root* of : of 189: c = ± = ± = ± 189 = ± c ft * - The eqution c 2 = ctuy hs two soutions 12.8 n 12.8, ut the engths of tringes re positive so we ignore the negtive soution. Sove the foowing: Ex. 7 Mr. Henry Lee hs rectngur fower gren esures 50.4 feet y 14.7 feet. If he wnts to ui wkwy tht runs ong the igon of the gren, how ong (to the nerest tenth) wi the wkwy e? First, rw picture. Since the nges 14.7 feet of rectnge re right nges, then the two sies re the egs of right 50.4 feet tringe whie the igon is the hypotenuse of the right tringe. Using the Pythgoren Theore, we get: c 2 = c 2 = (14.7) 2 + (50.4) 2 c 2 = c 2 = To fin c, tke the squre root of : c = ± = ± So, the wkwy nees to e 52.5 feet in ength. Ojective #4: Working with siir figures. Siir Tringes Two tringes re siir if they hve the se shpe, ut not necessriy the se size. One figure is in proportion to the other figure. The nottion for writing tht tringe is siir to tringe HET is HET. The orering of the etters shows the corresponing vertices. In this cse, correspons to H, correspons to E, n correspons to T.

7 In siir figures, the corresponing nges re equ. Thus, if HET, then = H, = E, n = T. The corresponing sies re not equ. However, the rtios of corresponing sies re equ since one tringe is in proportion to the other tringe. Thus, if HET, then E = = HE ET HT We cn use proportions to then fin the issing sies in pir of siir tringes. T In setting up proportion, we H wys strt with pir of corresponing sies tht of which we know oth vues. Let s try soe expes. Fin the issing sies n nges of the foowing: 123 Ex. 8 W WT Y 8 ft 11 ft 15 ft ft T 112 Y The pir of corresponing sies we wi strt with re W n Y. We write the ength fro the ser tringe over the ength fro the igger tringe. So, our proportions re: W Y = WT 8 11 = WT 8 11 = WT 15 W n Y 15 n 8 11 = 5 11 Y = T Y. Now, cross utipy n sove: 8 11 = 5 Y 11WT = Y = WT = 120 8Y = WT = ft Y = ft. Since = Y n T =, then = 112 n T = so, W = = = 25.2.

8 Ex. 9 On, there is point on n point E on such tht E. If = 16 c, = 10 c, n E = 8 c, fin. To see how we wou use siir tringes to sove this proe, et's egin y rwing igr: Since E, then is trnsvers. This ens tht 16 c n re corresponing nges n therefore, 8 c =. Siiry, is E trnsvers. This ipies tht n E re corresponing nges n so, = E. 10 c Now, rerw the figure s two seprte tringes c 8 c E 10 c Note tht =. Since the esures of the corresponing nges re equ, then ~ E. We know the vues of n E n we so hve the vue for. Since is the issing sie n the tringes re siir, we cn set-up proportion to sove for. E = 10 8 = 16 (Sustitute) (ross Mutipy) 10 = = 128 (ivie y 10) = 12.8 c Suppose the question h ske for inste of. In tht cse, we wou work the proe excty the se wy n sti fin. Since =, then = 16 c 12.8 c = 3.2 c.

9 Ex. 10 tree csts 15-foot show t the se tie chi 3 feet t csts 4-foot show. How t is the tree? We egin y rwing picture: h 3 ft 4 ft 15 ft These two tringes re siir tringes so we wi tke the rtio the engths of the shows n set it equ to the rtio of the heights: Length of the chi s show = hi s height Length of the tree s show Tree s height 4 15 = 3 h (cross utipy) 4h = h = 45 (ivie y 4) h = ft The tree is feet t. 125 Ex. 11 If the foowing figures re siir, fin the issing iensions. 6.3 in 4.8 in Since we know the engths of oth figures, we wi use the to for our first rtio. Let's first fin the with of the ser rectnge rger figure ser figure : = 4.8 w 4.2 in (cross utipy) 1.5 in 6.3w = 4.2(4.8) 6.3w = (ivie y 6.3) w = 3.2 in The with of the ser figure is 3.2 in. Now, et's get the ieter of the rger circe: rger figure ser figure : = (cross utipy) (1.5) = = 4.2 (ivie y 4.2) = 2.25 in The ieter of the rger circe is 2.25 in. w

Mutipy by r sin RT P to get sin R r r R + T sin (sin T )+ P P = (7) ffi So we hve P P ffi = m (8) choose m re so tht P is sinusoi. If we put this in b

Mutipy by r sin RT P to get sin R r r R + T sin (sin T )+ P P = (7) ffi So we hve P P ffi = m (8) choose m re so tht P is sinusoi. If we put this in b Topic 4: Lpce Eqution in Spheric Co-orintes n Mutipoe Expnsion Reing Assignment: Jckson Chpter 3.-3.5. Lpce Eqution in Spheric Coorintes Review of spheric por coorintes: x = r sin cos ffi y = r sin sin