PYTHGORS THEOREM,TRIGONOMETRY,ERINGS ND THREE DIMENSIONL PROLEMS 1.1 PYTHGORS THEOREM: 1. The Pythgors Theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides in right - ngled tringle.. For right ngled Tringle C, h p + h p Where h is the hypotenuse. C The hypotenuse is the side opposite the right ngle nd is lso the longest side of tringle. EXMPLE1: In the right-ngled tringle C, D is point on C. Given tht D 15 m, D 5 m nd CD 6 m
Find: ) C ) 15 5 D 6 C SOLUTION: 15 5 D 6 C ) Using Pythgors Theorem on ΔDC, C + CD D C + 6 15 15-6 5-36
189 C 13.75 m Tke positive squre root on oth sides sine length nnot e negtive. ) Using Pythgors Theorem on ΔDC, C + CD D C + 6 15 15-6 5-36 189 C 13.75 m Tke positive squre root on oth sides sine length nnot e negtive. ) Using Pythgors Theorem on ΔC, + C C 11 + ( 13.75 ) 11 + 189 C 5 + 6 11 m C 13.75 m 310 EXMPLE: 17.60 m PQR is n isoseles tringle. Given tht PQ PR 19 m nd QR 3 m, find the re of tringle PQR.
SOLUTION: P 19 19 Q 16 T 16 R QT 3 16 m PT isets QR Using Pythgors Theorem on ΔPQT, PT + TQ D PT + 16 19 PT 19-16 361-56 105 PT 10. m Tke positive squre root on oth sides sine length nnot e negtive.
re of ΔPQR re of Δ X se X Height X QR X PT X 3 X 10. 163. m 3. The Converse of Pythgors Theorem is lso true. In tringle, if the squre of the longest side is equl to the squres of the other two sides, then the ngle opposite the longest side is right ngle. In tringle C, if +, then C is right - ngle tringle with the right ngle Opposite side. C EXMPLE3:
Tringle XYZ hs sides XY 9 m, YZ right - ngle tringle. m nd XZ 7 m. Show tht tringle XYZ is SOLUTION: YZ + XZ ( ) 7 81 9 XY ΔPQR right - ngle. (Converse of the Pythgors Theorem) 1. TRIGONOMETRIC RTIOS FOR CUTE NGLES: For right - ngle tringle C, sin (Hypotenuse) z (Opposite) y os tn x C (djent)
TIPS FOR STUDENTS: sin, os, tn re revition for sine, osine nd tngent. Hypotenuse side opposite the right ngle, the longest side. Opposite side opposite the ngle. djent side djent to ngle. To rememer the rtio, use TO CH SOH. T C S ( sin, os, tn ) PPLICTIONS OF TRIGONOMETRIC RTIOS TO RIGHT NGLE TRINGLES: We use trigonometri rtios to find the unknown sides or ngles in right - ngled tringle. EXMPLE: In the digrm, D 6 m, CD 1 m, D 45 0 nd D is perpendiulr to C
Clulte D ) DC, 1 6 ) D. 45 0 C SOLUTION: ) In ΔDC, D os DC os DC os -1 6 1 6 1 1 os -1 60 0 C ) In ΔD, D sin 45 0 sin D 6 6 0 sin 45 6 6 X 1 45 0
8.46 m (Corret to 3 sig.fig.) TIPS FOR STUDENTS: If the degree of ury is not stted in the question nd if the nswer is not ext, the nswer should e given orret to 3 signifint figures. This mens tht working should e performed orret to 4 signifint figures or more. nswer in degree should e given orret to 1 deiml ple. Chek tht your lultor is in degree mode. 1.3 TRIGONOMETRIC RTIOS FOR OTUSE NGLES: The trigonometri identities for n otuse ngle re: sin (180 0 - ) sin os (180 0 - ) - os
TIPS FOR STUDENTS: When is ute, it lies in the first qudrnt; sin nd os oth re positive. When is otuse, it lies the seond qudrnt ; sin is positive while os is negtive. y sin ( 180 0 - ) os ( 180 0 - ) sin - os r y 180 0 - y x -x O x EXMPLE1: In the digrm, C is right ngled tringle. is produed to P nd C is produed to Q. Given tht 15 m, C 8 m nd C 17 m, express s frtion
) tn C, P ) sin PC, ) os CQ. 15 17 C Q SOLUTION: ) tn C tn ) sin PC sin ( 180 0 - C ) sin C ) os CQ os ( 180 0 - C )
-os C - EXMPLE: In is n otuse ngle nd sin of os., find without the use of lultor, the vlue SOLUTION: sin ( Given ) Using Pythgors Theorem, y O + 4 5 5 O 5-4 4 49 (- 7) O x O 7 units os -
1.4 SOLUTIONS OF TRINGLES: THE SINE RULE,THE COSINE RULE ND RE OF TRINGLE: In ny tringle, ) + + C 180 0 ) The shortest side is opposite the smllest ngle nd the longest side is opposite the lrgest ngle, ) The sum of the length of the two shorter sides is lwys greter thn the length of the longest side. C THE SINE RULE:
1. In ny tringle C, the Sine Rule sttes tht: C. The lterntive form of the Sine Rule is lso useful : 3. We use the Sine Rule when we hve: ) ny ngles nd 1 side or ) side nd 1 ngle opposite one of those sides. THE COSINE RULE: 1. In ny tringle C, the Cosine Rule sttes tht:
+ os + os + os C C Use this to find the remining side when given sides nd n inluded ngle.. The lterntive from of Cosine Rule is shown elow. os os os C Use this ti find n ngle if ll three sides re known. TIPS FOR STUDENTS: We use trigonometri rtios ( sine, osine nd tngent ) of ute ngles to solve right ngled tringles. We use the Sine Rule nd the Cosine Rule to solve tringles tht re not right ngled.
RE OF TRINGLE: To find the re of tringle given sides nd the inluded ngle, use re of Δ C sin C sin sin C EXMPLE1: In the digrm, CD is stright line. C 100 0. C 5 0, C 6 m nd CD 8 m. Clulte ), 100 0 6 ) D, 5 0 C 8 D ) The re of tringle CD.
SOLUTION: 100 0 8 0 5 0 18 0 C 8 D 6 ) C 180 0 100 0 5 0 ( sum of Δ ) 8 0 Using the Sine Rule on Δ C, 6 0 0 sin 5 sin 8 Sine Rule 6 0 sin 8 X sin 5 0 C 10.1 (orret to 3 sig. fig.) ) CD 180 0 5 0 (dj. s on str. line) 18 0 Using the Cosine Rule on Δ CD, D 8 + 6 ( 8 ) ( 6 ) os 18 0 Cosine Rule 159.1 + os C
D. 1.6 m (orret to 3 sig. fig.) ) re of Δ CD X 8 X 6 X sin 18 0 18.9 m (orret to 3 sig.fig.) re of Tringle sin C EXMPLE: In the digrm, D 48 0, D 11. m, D 8.6 m, C 9.8 m nd CD 5.1 m. ) D, 11. D ) CD, 8.6 5.1 48 0 ) The re of tringle CD, d) The shortest distne from to CD. 9.8 C SOLUTION:
11. D 48 0 8.6 5.1 31.33 0 9.8 C ) Using the Sine Rule on Δ D, sin D 11. 0 sin 48 8.6 Sine Rule C sin D 0 sin 48 8.6 X 11. 0.9678 D 180 0 75.4 0 104.6 0 (orret to 1 d.p.) D is given s n otuse ngle in the digrm. ) Using the Cosine Rule on Δ CD, os CD 8.6 9.8 5.1 (8.6)(9.8) 0.854 CD 31.33 0 Cosine Rule os d C
31.3 0 (orret to 1 d.p.) ) re of Δ CD X 8.6 X 9.8 X sin 31.33 0 re Of Tringle 1.91 1.9 m ( orret to 3 sig. fig. ) sin C d) Let h m e the shortest distne from to CD. X CD X h re of Δ CD 8.6 5.1 X 5.1 X h 1.91 h D h.. 9.8 C 8.59 m ( orret to 3 sig. fig. ) The shortest distne from to CD is 8.59 m. TIP FOR STUDENTS: The shortest distne from to CD is the perpendiulr distne from to CD.
1.5 ERINGS: 1. ering is n ngle tht tells the diretion of one ple from nother.. erings re lwys ) Mesured from the North. ) Mesured in Clokwise diretion nd ) Written s Three Digit numer (000 0 to 360 0 ) e.g.- N N N F 035 0 1150 335 0 The ering of The ering of D The ering of F From is 035 0. From C is 115 0. From E is 335 0. D EXMPLE1: In the digrm, C represents horizontl tringulr field. is 10 m from on ering of 048 0. C is 18 m from nd 0 m from.
Clulte N ) C, ) The ering of from C, 48 0 10 18 ) The re of tringulr field C. 0 C Given tht wter hydrnt, H, is due north of. The ering of from H is 07 0. d) Find the distne H. SOLUTION: N 48 0 86.18 0 10 18 0 C ) Using the Cosine Rule on Δ C,
os C 18 10 0 (18)(10) 0.06667 CD 86.18 0 86. 0 (orret to 1 d.p.) Cosine Rule os d C ) P 48 0 ( lt. s, M QC ) N P 86. 18 0 48 0 N 38.18 0 38.18 0 CQ 38.18 0 ( lt. s, P QC ) 10 48 0 48 0 P ering of from C 18 360 0 38.18 0 ( s, t point ) N 31.8 0 ( orret to 1 d.p. ) 0 38.18 0 Q C ) re of Δ C re of Tringle X 18 X 10 X sin 86.18 0 sin C
89.8 m 89.8 m (orret to 3 sig. fig.) The re of the tringulr field is 89.8 m d) H 180 0-7 0 ( di. s, on str. line ) 108 0 H 180 0-108 0-48 0 ( di. Δ sum of) 4 0 Using the Sine Rule on Δ H, H 10 0 0 sin 4 sin108 10 H 0 sin108 X sin 4 0 Sine Rule C 4.8 m (orret to 3 sig.fig.) EXMPLE :, nd C re three points on horizontl ground. C is 3 m from on ering of 18 0. C 110 0 nd 43 m.
) Clulte N i) C, 18 0 ii) The re of tringle C, 110 0 43 3 iii) The ering of from C, C ) oy wlks from point to point C until he rehes point X, where X is minimum. Find the length of X. SOLUTION: ) i) Using the Cosine Rule on Δ C, C 3 + 43 ( 3 ) ( 43 ) os 110 0 3814 D Cosine Rule + os C 61.76 61.8 m (orret to 3 sig. fig.) ii) re of Δ C re of Tringle X 3 X 43 X sin 110 0 646.5 sin C 647 m (orret to 3 sig. fig.)
iii) P 110 0 + 18 0-180 0 58 0 58 0 Q P (lt. s, Q P) 58 0 N Using the Sine Rule on Δ C, 18 0 sin C 3 0 sin110 61.76 Q 43 P 3 sin Q 0 sin110 61.76 3 58 0 9.14 0 0.4869 61.76 C C 9.14 0 QC 58 0 + 9.14 0 87.1 0 ( orret to 1 d.p. ) The ering of C from is 087.1 0 ) The oy losest to when X is to C. In Δ X, sin 9.140 X 43 X sin9.14 0 0.8 m (orret to 3 sig. fig.) lterntive method:
X C X X re of Δ C X 61.76 X X 646.5 X.. 0.9 m ( orret to 3 sig. fig. ) 1.6 NGLE OF ELEVTION ND NGLE OF DEPRESSION: LINE OF SIGHT OSERVER NGLE OF ELEVTION HORIZONTL LINE NGLE OF DEPRESSION LINE OF SIGHT
When you look t the irplne, the ngle etween the line of sight nd the horizontl line is lled the ngle of Elevtion. When you look down t the r, the ngle etween the horizontl line nd the line of sight is lled the ngle of Depression. NGLE OF ELEVTION ND NGLE OF DEPRESSION: Horizontl HORIZONTL O ngle of elevtion of from. ngle of depression of from. ( lt. s ) EXMPLE1: ) The ngle of elevtion of the foot of lighthouse from ot is 36 0. The lighthouse is loted on the edge of liff whih is 80 m high.
i) Find the distne from the ot to the se of the liff. ii) Given tht the lighthouse is 45 m high, find the ngle of elevtion of the top of the lighthouse from the ot. ) n owl flies from the top of 10 m tll tree t n ngle of depression of 8 0 to th rt on the ground. Find the distne i) The owl flew to th the rt, ii) Of the rt from the se of the tree. SOLUTION: ) i) In ΔC, D tn 36 0 80 0 tn 36 Lighthouse 45 110.1 Cliff 110 m ( orret to 3 sig. fig. ) The distne from the est ot to the se of the liff is 110 m. 80 36 0 (110.1) ii) In ΔD,
tn D ( 80 45) 110.1 D 48.6 0 (orret to 1 d.p.) The ngle of elevtion of the top of the lighthouse from the ot is 48.6 0. ) i) sin 8 0 R 10 PR 0 sin 8 8 0 1.3 m ( orret to 3 sig. fig. ) 10 The owl flew 1.3 m to th the rt. 8 0 P Q ii) tn 8 0 10 PQ 0 sin 8 18.8 m ( orret to 3 sig. fig. ) The rt is 18.8 m from the se of the tree. EXMPLE:
In the digrm,, nd C re three points on horizontl ground. The ering of C from is 063 0 while the ering of from is 086 0. C 15 m nd 1 m. ) Find C. T ) vertil mst TC stnds on C. N C The ngle of elevtion of T from is 3 0. Find 15 i) TC, 86 0 ii) The ngle of depression of from T. 63 0 1 SOLUTION: C 86 0-63 0 T 3 0 N C 15 86 0 63 0 3 0 1
) i) Using the Cosine Rule on Δ C, Cosine Rule C 15 + 1 ( 15 ) ( 1 ) os 3 0 86.08 + os C C. 9.78 9.8 m (orret to 3 sig. fig.) ) i) tn 3 0 T TC 15 X tn 3 0 9.373 9.37 m ( orret to 3 sig. fig. ) 3 0 15 C ii) ngle of depression ngle of elevtion TC tn TC.. TC 45.3 0 ( orret to 1 d.p. ) The ngle of depression of from T is 45.3 0.
1.7 THREE DIMENSIONL PROLEMS: To solve given Three Dimensionl prolem: ) Redue in to prolem in plne, i.e. look for right ngled tringles in the relevnt plnes, nd then ) pply the Pythgors Theorem or trigonometri rtios (sin, os or tn) to these tringles to find the unknowns. EXMPLE1: The digrm shows wedge. EF is retngle nd CE nd DF re vertil lines. Given tht DC FE 4 m, E 18 m nd CE 1 m. Find ) E, D C ) E, 1 F E ) C, 18 d) CE. 4
SOLUTION: ) Using Pythgors Theorem on ΔDC, E E 4 + 18 900 18 E 30 m 4 ) In ΔE, tn E E 53.1 0 ( orret to 1 d.p. ) ) Using Pythgors Theorem on ΔEC, C C 30 + 1 1044 C 3.3 m ( orret to 3 sig. fig. ) 30 E d) In ΔEC, tn CE CE 1.8 0 (orret to 1 d.p.)
EXMPLE: VCD is pyrmid with retngulr se. X is the point of intersetion of the digonls of the se nd V is vertilly ove X. Given tht 1 m, C 0 m nd VX 16 m, lulte ) C, V ) C, ) VX. 16 D C 0 X 1 ) Using Pythgors Theorem on ΔC, C C 1 + 16 C 56 0 C 16 m 1
) Using Pythgors Theorem on ΔVX, V C 30 + 1 1044 16 C 3.3 m (orret to 3 sig. fig.) 10 X ) In Δ VX, tn VX VX 3.0 0 (orret to 1 d.p.) 1.1 RELTION RETWEEN SIDES ND NGLES OF TRINGLE: 1. tringle onsists of three sides nd three ngles lled elements of the tringle. In ny tringle C,,, C denotes the ngles of the tringle t the verties. + + C 180 0. The sides of the tringle re denoted y,, opposite to the ngles, nd C respetively. C Fig (1)
3. + + s The perimeter of the tringle. 1. THE SINE RULE: In tringle C, prove tht R Where, R is the irum rdius of the tringle. PROOF: Let S e the irumentre of the tringle C. First prove tht R CSE (I) : Let S e n ute ngle. Let P e ny point on the irle. Join P, PC Whih pss through S. Join CP, so tht CP 90 O. C (ngles in the sme segment). P S FROM Δ PC, sin C C sin, R Fig ( )
CSE (II) : Let e right ngle, ie., C R 90 0 ( Fig 3 ), Then C is the dimeter. sin S C R Fig ( 3) CSE (III) : Let e n otuse ngle ( Fig 4 ). join P, pssing through S. Join CP, so tht P 90 0. Now C 180 0 ( C ) 180 0 ( Sine PC is yli qudrilterl ) From Δ PC, sin ( C ) S C Fig ( 4) i.e.- sin (180 0 ) sin R
R is true for ll vlues of. Similrly, we n prove, R, Thus, sinc. R. or R sin, R sin, R 1.3 THE COSINE RULE: In ny tringle C, prove tht + - os + - os + - os C C C D C D C ( Fig 5 ) ( Fig 6 ) ( Fig 7 )
PROOF: Cse ( I ) Let e n ute ngle ( Fig 5 ) Drw CD. From Δ DC C D + DC ( D ) + DC -. D + D + DC C -.D + C ( Sine D + DC C ).D + ut from Δ DC, Cos D os os..os + Or + os Cse ( II ) Let e right ngled, ie., 90 0 ( Fig 6 ) C + C ie., + ut + os + os 90 0 +, whih is true for right ngled tringle.
Cse ( III ) Let e otuse ngle, ie., > 90 0 ( Fig 7 ). Drw CD produed. From Δ DC C D + DC ( + D ) + DC C +. D + D + DC C +. D + C ( Sine, D + DC C ).C. D + ut from Δ DC, Cos ( D C ), ( D C 180 0 ) Cos ( 180 0 ) - os, D os + ( os ) + + os Similrly, we n prove, + - os + - os C Tips for Students: The ove formule n e written s: Cos, Cos, Cos C, These results re useful in finding the osines of the ngle when numeril vlues of the sides re given. Logrithmi omputtion is not pplile sine
the formule involve sum nd differene of terms. However, logrithmi method n e pplied t the end of simplifition to find ngle 1.4 THE PROJECTION RULE: In this rule, we show how, one side of tringle n e expressed in terms of other two sides. It is lled projetions rule. os C + os, os + os C, os + os. PROOF: Let C e n ute ngle Drw D C produed. In Fig ( i ) C D + DC [ NOTE : D is lled projetion of on C nd DC is the projetion of C ] D + DC (1) From Δ D, os D os os From Δ CD, os C CD C os C os C From (1) os + os C os C + os D C (Fig 8) CseII: When C is right ngle, ie., 90 0 ( Fig 9 ).
os, os () Sine 90 0, os os 90 0 0, We get, os 90 0 + os os (3) C ( Fig 9 ) Cse III : When is otuse ngle ( Fig (iii) ) From Δ D, C D - CD (4) From Δ D, os, D os From Δ CD Cos (180 0 C ) 180 0 C os, CD - os C From (4) C os ( - os C ) C D ie., os + os C os C + os ( Fig 10 ) Similrly, os + os C, os + os. 1.5 THE LW OF TNGENTS:
In ny Δ C, Prove tht : 1. tn tn,. tn tn C C, 3. tn tn C C PROOF: Using sine rule,.os os.sin os, os.tn tn tn tn 1 sin g ot u Similrly, other two results n e proved y hnging sides nd ngles in yle order.
1.6 EXPRESSIONS FOR HLF NGLES IN TERMS OF,,: In ny tringle C, prove tht 1. sin S S C,. os S S, 3. tn S S SS PROOF: (1) We know tht sin 1 os sin 1 - ( Using osine rule for ) sin
S S Sine + + s + s - + s - sin s s (Divide y ) sin ± S S If is ute, then sin is lwys positive. sin S S. sin 1 + os 1 + (Using osine rule for ) sin sin s s Using + + s + s
Dividing y, we get os s s ± S S Sine is ute, os is lwys positive nd therefore, os S S 3. tn sin os s s ss s s ss Similrly, we n show tht sin s s, os s s, tn s s ss sin s s, os s s, tn s s ss WORKED EXMPLES
1. If 3, 4, 5, in tringle C, find the vlue of.) sin C.) sin 4C + os 4C SOLUTION: Sine, + is stisfied y the given sides, they form right ngled tringle. 5 4 + 3 C 90 0, sin C 90 0 sin sin 45 0 1 nd, sin 4C + os 4C sin( 4 X 90 0 ) + os( 4 X 90 0 ) sin 360 0 + os 360 0 0 + 1 1. Prove tht sin ( C ) + sin ( C ) + C sin ( ) 0. SOLUTION: Now, sin( C ) R sin. sin( C ) ( Sine, R sin ) R sin.sin( C ) Sine + + C 180 0, + C 180 0 sin( + C ) sin
R sin ( + C ) sin ( C ) R sin [ sin sin C ] Similrly, sin ( C ) R [ sin C sin ] C sin ( ) R [ sin sin ] L. H. S. sin ( C ) + sin ( C ) + C sin ( ) R [ sin sin C ] + R [ sin C sin ] + R [ sin sin ] R [ sin sin C + sin C sin + sin sin ] 3. Prove tht, in ΔC, SOLUTION: sin C sin C L. H. S. sin C sin C X sin C sin C sin sin sin C Sine sin ( + C ) sin ( - C ) sin sin C sin sin sin C sin( + ) sin C in ΔC
C 4R 4R 4R 4R 4R ( using sine rule ) R. H. S. 4. Prove tht ( os C os ) - SOLUTION: L. H. S. ( os C os ) os C os - (Using osine rule) - - R. H. S. 5. Prove tht sin + sin + sinc 0 SOLUTION:
Now sin X sin os (Sine, sin sin os ) X R X (Using sine rule nd osine rule) R. 4 Similrly, sin R. 4 sinc R. 4 L. H. S R. 4 + R. 4 + R. 4 4R. 1 [ 4 4 ( ) + 4 4 ( ) + 4 4 ( ) ] 4R. 1 [ 0 ] 0 R. H. S 6. Find the gretest side of the tringle, whose sides re x + x + 1, x + 1, x 1.
SOLUTION: Let x + x + 1, x + 1, x 1 Then, is the gretest side. Therefore is the gretest ngle. os 1 1 1 1 1 x x x x x x 1 1 1 1 4 4 3 3 4 4 x x x x x x x x x x x x os 1 1 3 3 x x x x x x 1 - os60 0 os( 180 0 60 0 ) 1 os10 0 10 0 Therefore, the gretest ngle is 10 0 7. If sin + sin sin C in Δ C, Prove tht either 90 0 or 90 0. SOLUTION: sin + sin sin C sin.os sin C Using sin C + sin D sin C D. os C D
sin ( + ). os ( ) sin C os C sin ( + ) Sin( 180 0 C ) sin C sin C os ( ) sin C os C Dividing y sin C oth sides, we get, Cos( ) os C lso, Cos( ) os -C [ Sine, os( -C ) os C ] ± C When C, + C ut, + + C 180 0, gives - 180 0, i.e., 180 0, 90 0 When - C, + C + + C 180 0, gives Therefore, tringle is right ngled tringle. + 180 0, i.e., 180 0, 90 0 os os 1 8. Prove tht - - 1 SOLUTION:
os os L. H. S - (Using os 1 sin ) 1 sin 1 sin - 1 - sin 1 - sin + 1-1 - sin + sin 1-1 - 1 R + 1 R Sine, sin 1 R sin 1 R 1-1 R. H. S SUMMRY ND KEY POINTS 1.The Pythgors Theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides in right - ngled tringle.. For right ngled Tringle C, h p + h p
Where h is the hypotenuse. C The hypotenuse is the side opposite the right ngle nd is lso the longest side of tringle. 3. The Converse of Pythgors Theorem is lso true. In tringle, if the squre of the longest side is equl to the squres of the other two sides, then the ngle opposite the longest side is right ngle. In tringle C, if +, then C is right - ngle tringle with the right ngle Opposite side. 4. TRIGONOMETRIC RTIOS FOR CUTE NGLES: C For right - ngle tringle C, sin os (Hypotenuse) z y (Opposite) tn (djent) C sin, os, tn re revition for sine, osine nd tngent.
Hypotenuse side opposite the right ngle, the longest side. Opposite side opposite the ngle. djent side djent to ngle. To rememer the rtio, use TO CH SOH. T C S (sin, os, tn ) We use trigonometri rtios to find the unknown sides or ngles in right - ngled tringle. 5. Importnt Note: If the degree of ury is not stted in the question nd if the nswer is not ext, the nswer should e given orret to 3 signifint figures. This mens tht working should e performed orret to 4 signifint figures or more. nswer in degree should e given orret to 1 deiml ple. Chek tht your lultor is in degree mode. 6. TRIGONOMETRIC RTIOS FOR OTUSE NGLES: The trigonometri identities for n otuse ngle re: sin(180 0 - ) sin os(180 0 - ) - os When is ute, it lies in the first qudrnt; sin nd os oth re positive. When is otuse, it lies the seond qudrnt ; sin is positive while os is negtive. 7. SOLUTIONS OF TRINGLES: THE SINE RULE,THE COSINE RULE: In ny Tringle, Sum of ll three ngles ) + + C 180 0
) The shortest side is opposite the smllest ngle nd the longest side is opposite the lrgest ngle, ) The sum of the length of the two shorter sides is lwys greter thn the length of the longest side. C ) THE SINE RULE: i) In tringle C, the Sine Rule sttes tht: C ii) The lterntive form of the Sine Rule is lso useful:
iii) We use the Sine Rule when we hve: ) ny ngles nd 1 side or ) side nd 1 ngle opposite one of those sides. In tringle C, R Where, R is the irum rdius of the tringle. The lterntive form of Sine Rule s shown elow: Rsin, Rsin, RsinC ) THE COSINE RULE: i)in ny tringle C, the Cosine Rule sttes tht: + os + os + os C C Use this to find the remining side when given sides nd n inluded ngle.
ii) The lterntive form of Cosine Rule is shown elow. os os os C Use this ti find n ngle if ll three sides re known. Importnt points: We use trigonometri rtios ( sine, osine nd tngent ) of ute ngles to solve right ngled tringles. We use the Sine Rule nd the Cosine Rule to solve tringles tht re not right ngled. 8. RE OF TRINGLE: To find the re of tringle given sides nd the inluded ngle, use re of Δ C sin C sin sin
C 9. THE ERINGS: 1. ering is n ngle tht tells the diretion of one ple from nother.. erings re lwys ) Mesured from the North. ) Mesured in Clokwise diretion nd ) Written s Three Digit numer (000 0 to 360 0 ) e.g.- N N N F 035 0 1150 335 0 The ering of The ering of D The ering of F From is 035 0. From C is 115 0. From E is 335 0. D 10. THE THREE DIMENSIONL PROLEMS: To solve given Three Dimensionl prolem: ) Redue in to prolem in plne, i.e. look for right ngled tringles in the relevnt plnes, nd then ) pply the Pythgors Theorem or trigonometri rtios ( sin, os or tn )
to these tringles to find the unknowns. 11. RELTION RETWEEN SIDES ND NGLES OF TRINGLE: ) tringle onsists of three sides nd three ngles lled elements of the tringle. In ny tringle C,,, C denotes the ngles of the tringle t the verties. + + C 180 0 ) The sides of the tringle re denoted y,, opposite to the ngles, nd C respetively. C ) + + s The perimeter of the tringle. 1. THE PROJECTION RULE: In this rule, we show how, one side of tringle n e expressed in terms of other two sides. It is lled projetions rule. os C + os, os + os C, os + os. 13. THE LW OF TNGENTS: In ny Δ C, Prove tht :
.) tn tn,.) tn tn C C,.) tn tn C C 14. EXPRESSIONS FOR HLF NGLES IN TERMS OF,,: In ny tringle C, prove tht.) sin C S S,.) os S S,.) tn S S S S Similrly, we n show tht sin s s, os s s, tn s s s s sin s s, os s s, tn s s s s