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AntonioBO ’’Cohen’’/Math/Doc/ MATHEMATICMY CONCISE NOTES about Trigonometry Sine, COSINEE andTANGENTSOME SOLVED PROBLEMS TO FIND UNKNOWN SIDES OF TRIANGLES, INVOLVING THE SINE, COSINE AND TANGENT RATIOS.SINE RATIO (SIN)THE FORMULA IS SINE θ (THETA) = QUOTE QUOTE There are three formulae that can help you to find the length of unknown side or angle of a right angle triangle. These are called THE SINE RATIO, CONSINE RATION AND TANGENT RATIO. B 9 m a A CAngle A = 48?Angle C = 90?Long side A-B (hypotenuse) = 9 metersUse the sine ratio if you know or need to know the opposite side or the hypotenuseExample 1Find the length of side a of triangle above, substitute the known values into the sine ratio.SOLUTION SIN θ = LENGTH OF OPPOSITE SIDELENGTH OF HYPOTENUSE Sin 48? = a9 ?Rearrange the formula to make a the subject and so it becomesa = 9 x sin 48?Use the sin button of calculator CASIO fx 992s or any scientific calculator to find the value of sin 48?, and then solve the equation:a = 9 x 0, 74314482547 a = 6.688 3034293 or 6.69 metersSo, the length of side a is 6.69 meters long (to 2 d.p.) or to 2 decimal proximately. COSINE RATIO (COS) THE FORMULA ISCOSINE θ (THETA) = QUOTE Use the cosine ratio if you know or need to know the adjacent side or the hypotenuse. θ=42° C b 10 m A BAngle C = 42°Angle B = 90°Side B-C = 10 metersExample 2Find the length of side b of this triangle; substitute the known values into the cosine ratio.SOLUTION COS θ = LENGTH OF ADJACENT SIDE LENGTH OF HYPOTENUSE COS 42? = 10b ?Rearrange the formula to make b the subject and so it becomes b = 10cos42°Use the “COS” button of calculator CASIO fx 992s or any scientific calculator to find the value of 42?, and then solve the equation. QUOTE 100.74314482547=13.4563272961 b = 100.74314482547=13.4563272961 So, the length of side b = 13.46 meters long (to 2 d.p.)TANGENT RATIO (TAN) THE FORMULA IS TANGENT θ (THETA) = LENGTH OF OPPOSITE SIDELENGTH OF ADJACENT SIDE QUOTE Error! Bookmark not defined.Use the tangent ratio if you know or need to know the opposite or adjacent side. B c A 5 meters CAngle A = 90Angle C = 50Side A-C = 5 metersExample 3Find the length of side c of this triangle; substitute the known values into the tangent ratio.SOLUTIONTAN θ = LENGTH OF OPPOSITE SIDE LENGTH OF ADJACENT SIDETan 50 = c5 ?Rearrange the formula to make c the subject and so it becomesc = 5 x tan 50Use the “TAN” button of calculator CASIO fx 992s or any scientific calculator to find the value of 50?, and then solve the equation.c = 5 x 1.19175359259c = 5.958767966297 ≈5.96 meter (to 2 d.p.)So the length of side c is 5.96 meter long (to 2 d.p.)SOME SOLVED PROBLEMS TO FIND UNKNOWN ANGLES OF TRIANGLES, INVOLVING THE SINE, COSINE AND TANGENT RATIOS.The SINE, COSINE AND TANGENT ratios can also be used to find the value of an unknown angle in a right angled triangle. Example 1The length of hypotenuse and adjacent sides on this triangle are known. Find the angle θ (theta), use the cosine ratio.SOLUTION: Cos θ= adj hyp= 612Cos θ= 12 = 0.5 ?Rearrange the formula to make θ the subject and so it becomesθ= cos 12=0.5 = 60Use inverse or shift cos button of calculator CASIO fx 992s or CASIO fx 100AU or any scientific calculator to find the value of θ. B 12 mANGLE θ A CHOW TO USE CALCULATOR CASSIO fx 992s 6 M90 ° QUOTE TO FIND θ= cos 12=0.5 to find 60°SHIFT= COS 12=0.5 ? Press ?Press ? Press , it will display QUOTE ’ ”SHIFT60? Press ? Press it will display 60SO, THE ANGLE θ THETA IS EQUAL TO 60°. BANGLE θ 12 m 9 m A C90 ° QUOTE Example 2 The lengths of hypotenuse and opposite sides on this triangle are known. Find the angle θ (theta), use the sine ratio. SOLUTION: Sin θ= opphyp= 912Sin θ= 0.75 ?Rearrange the formula to make θ the subject and so it becomesθ=sin 912=0.75Use inverse or shift sin button of calculator CASIO fx 992s or CASIO fx-100AU or any scientific calculator to find the value of θ. HOW TO USE CALCULATOR CASSIO fx 992s TO FIND θ=sin 912=0.75 to find 49= SINSFIFT912=0.75 ?Press ? Press ? Press , it will QUOTE ’ ”SHIFTdisplay 48.590377890 ≈ 49 ?Press ? Press it will display 49°. SO, THE ANGLE θ (THETA) IS EQUAL TO 49°. NON-RIGHT-ANGLED TRIANGLES If a triangle does not contain a right angle, then the sine, cosine and tangent ratios cannot be used directly to calculate the sizes of sides and angles.Instead, other relationships, such as sine and cosine rules can be used. THE Sine RULE THE FORMULA a b c sin A sin B sin C This can be rearranged as: sin A sin B sin C a b c c??????THE SNE RULETHE FORMULA: a b c sin A sin B sin C This can be rearranged as: sin A sin B sin C a b cThe sine rule can be used to find a side when one side and any two angles are known.For example look at the triangle below, one side and two angles are known, so we use the sine rule to find side a. Angle A = 40? Angle B = 95? Side b = 12 meterFind side a =? b = 12 meterSOLUTION:asin A= bsin B= asin 40°=12sin 95°a0.6428=120.9968 a = 120.9968 x 0.6428=7.73836276083 a ≈ 7.74 meter →→ Side a is 7.74 meter SOME EXAMPLES OF ENGINEERING SOLVED PROBLEM ABOUT TENSION AND COMPONENT OF FORCES, INVOLVING SINE AND COSINE RATIOS AND RULES.An axial force is induced (is produced; is caused) in the mast by the two wire stays/girdle/cord as shown: 25 18 A B 1000 N mast Find the magnitude of the force acting along the mast.What is the magnitude of the force acting on the stay B?NB: Solve the problem analytic and graphically Giving:A = the magnitude of the force F 1000 N is equal to 102 kgf (kilogram force), because 1 N is equal to 0.102 kgf. The force acting downwards to the left (sense) at 25° to the vertical (direction).B = unknown force, acting downwards to the right (sense) at 18° to the vertical (direction).Mast = unknown force =? FIRST STEP, SOLVE THE PROBLEM ANAYTICALLY, USING SINES RULES C ANGLE C θ (THETA) 25° b =1 000 N AANGLE A α QUOTE (ALPHA) 137 °= 180 °-25°+18° a = 2 207 NANGLE B QUOTE β (BETHA)18 ° c = 1 368 N BSOLUTION: asin A= bsin B= asin 137° = 1 000 Nsin 18° a0.6820= 1 000 N0.3090 a = 1 000 N0.3090 x 0.6820 a = 2 207.1197411 ≈ 2 207N So, the magnitude of Force acting along the mast is 2 207 N. bsin B= csin C= 1 000sin 18°= csin 25° = 1 0000.3090= c0.4226c = 1 0000.3090 x 0.4226c = 1 367.63754045 N ≈1 368 NSo, the magnitude of Force acting in the stay B is 1 368 N. b) Second step, solve the problem graphically A BFREE BODY DIAGRAM 25 18 Force A Force B 1 000 N MastFORCE DIAGRAM ? SCALE 1 CM=200N C PROJECTION LINE OF FORCE A 5 CM x 200 = 1 000 N b PROJECTION LINE OF FORCE A AND B ACTING ON MAST 11 CM x 200 = 2 200 N A c a PROJECTION LINE OF FORCE B6.8 CM x 200 = 1 360 N B CHECK THE ANGLES OF FORCE DIAGRAM, USING COSINES RULES OF NON-RIGHT TRIANGLES: C ANGLE θ (THETA)24° 55' 06.24''26.5615 b =1 000 N AANGLE α QUOTE (ALPHA) 137° 02’ 42’’ = 180°-(24° 55' 06.24''+ 18° 02' 49.56'') a = 2 200 N1801800ANGLE QUOTE β (BETHA)18° 02' 49.56''18.0471 c = 1 360 N BSOLUTION: Cos B =a2+c2-b22ac= 2 200?+1 360?-1 00022 x 2 200 x 1 360Cos B = 4 840 000+ 1 849 600 - 1 000 0005 984 000 Cos B = 5 689 6005 984 000= 0.95080213903 Cos B = 18.0471052702 ≈18.0471 =18° 02' 49.56''? β (BETA)Cos C = a2+b2-c22ab = 2 2002+ 1 0002- 1 36022 x 2 200 x 1 000 Cos C = 4 840 000+1 000 000-1 849 6004 400 000Cos C = 3 990 4004 400 000 = 0.9069090909Cos C = 24.9183544336 ≈ 24.9184=24° 55' 06.24''? θ (THETA)Cos A = 180°-cos B+cos C=180°-24.9184+18.0471=137.0345 = 137° 02’ 42’’ = ? α (ALPHA)CHECK ANALYTICALLY THE GRAPHICAL SOLUTION USING SINES RULES.GIVING:Cos A (α alpha) = 137.0345 = 137° 02' 42'';Cos B (β beta) = 18.0471=18° 02' 49.56'' ;Cos C (θ theta) = 24.9184 = 24° 55' 06.24'';SOLUTION: b= asinβsin α= 2 200 x sin18° 02'49.56 (18.0471)sin 137° 02' 42 (137. 0345)b= 2 200 x 0. 309798705950.68155786016 b = 999.998962561 N ≈1 000 N Analytically the force A projected in the line b is equal to 1 000 N, and it is same than the magnitude of force calculated using graphical solution.So, the solution is correct.c = asinθsinα = 2 200 x 24° 55' 06.24''(24.9184)sin 137° 02' 42'' (137. 0345) c = 2 200 x 0.421327080290.68155786016 c = 1 360.06585915 N≈ 1 360 N So, analytically the force B projected in the line c is equal to 1 360 N, and it is same than the magnitude of force calculated using graphical solution.So, the solution is correct.a? = b?+ c?- 2bc cos α FINAL CHECKIN ? a? = (1000)?+ (1360)?- 2(1000) (1360) x cos137° 02' 42''137.0345 = 1000000 + 1849600 – 2720000 x (-0.73176422654) = 2 849 600 – (- 1 990 398.6962) = 4 839 998.6962a = 4839 998.6962= 2 199.99970368 N ≈ 2 200 NSO, THE FINAL CHECKING WITH THE TRIGONOMETRY RULE, WE FIND THE MAGNITUDE OF FORCE OF 2 200 N ACTING ALONG THE MAST.THIS MAGNITUDE OF FORCE IS SAME THAN THE MAGNITUDE OF FORCE CALCULATED USING GRAPHICAL SOLUTION.CALCULATION IS CORRECT.Determine graphic and analytically the HORIZONTAL (x) and VERTICAL (y) COMPONENTS of a force of 40 kN acting to the left at 150° to the horizontal. FREE BODY DIAGRAM y 40 kN 150° o xSOLUTION GRAPHICALLY40 kNFORCE DIAGRAMSCALE 1 CM = 5 kNVERTICAL COMPONENT4 CM x 5 kN = 20 kN y 150° o x HORIZONTAL COMPONENT6.9 CM x 5 kN = 34.5 kN STEPS FOR GRAPHICAL SOLUTION:Draw the y axis perpendicular to the x axis.Draw the 40 kN force to the scale (1 cm = 5 kN ? 40 kN÷ 5 kN = 8 cm) at 150° to the horizontal (x axis).From the terminal point of the 40 kN force draw a line parallel to the x axis and a line parallel to the y axis.The intersection of these two lines with the x and y axes give the magnitude, to scale, of the required horizontal and vertical components.ANSWER Horizontal component of force = 34.50 kN Vertical component of force = 20 kN SOLUTION ANALYTICALLYFREE BODY DIAGRAM VERTICALCOMPONENT A y 40 kN 150° X B C HORIZONTAL COMPONENT 30°=180°-150°From the triangle above ABCSin 30° = ABAC= VERTICAL COMPONENT OF FORCE 40 kN VERTICAL COMPONENT FORCE = 40 kN x sin 30° = 40 x 0.5 = 20 kNCos 30° = BCAC= HORIZONTAL COMPONENT OF FORCE40 kN HORIZONTAL COMPONNENT FORCE = 40 kN x cos 30° = 40 x 0.8660 = 34.64 kNANSWER Horizontal component of force = 34.64 kN Vertical component of force = 20 kN AN EXAMPLE OF COMPONENTSumming A + B COMPONENT B A B + = COMPONENT A RESULTANT R = A + B NOTE: here R is the resultant of A + B, and A and B are called the components of R.A concrete skip or hopper at the BCP Factory in Morebank is attached to a crane hook by two chains each 1.50 meter long (1500 millimeter). Find the tension in the chains if the loaded skip’s or hoper’s mass is 2.50 tonnes. 1.50 METER CHAINS EACH 1.50 METER LONG SKIP/HOPPER FULLOF CONCRETE (2.50 TONES) 2.00 METERSOLUTION:Draw a triangle from two chains and the top part of the skip/hopper to find the angle B and C.Divide the triangle in the middle into two right angled triangles to make ease the calculation using COSINE RATIO. A E H 1.50 m B C D F G I 2.00 m 1.00 m 1.00 mAfter dividing the triangle ABC into two right angled triangles DEF and GHI as shown above, then find their respective angles using COSINES RATIOS.Triangle DEF: E Cos θ= adjcent sidehypotenuse = 1.00 m1.50 m ANGLE θ? = 0.6666 = 48.1975 ≈48= 48° SO, COSINES θ 0R ANGLE D = 48° 48°D FTriangle GHI H Cos θ= adjacent sidehypotenuse = 1.50 m1.00 mANGLE θ? = 0.6666 = 48.1975 ≈48= 48° SO, COSINES θ OR ANGLE I = 48° 48° G IKnowing the angles of each chain draw a new non-right-angled triangle according to each angle of 48°. K I = 48° L D’= 48° D = 48° J Angle D is equal to angle D’ because they are alternate angles, which state that angles formed on alternate sides of a transversal between parallel line are equal. Transform the non-right-angled triangle JKL above in a new non-right-angled triangle A’B’C’ to find the tension of each chain using SINE RULES. B’ c A’ a b C’GIVING:ANGLE A’ (α) = 48° x 2 = 96°. ANGLE B’ (β) AND ANGLE C’ (θ) =180°-96°2=42°.MASS OF CONRETE = 2.50 TONNES.TRANSFORM UNIT OF MASS TO UNIT OF FORCES W = mg i.e. W (N) = m (kg) x 9.8 (m/s?) gravity. It is the relationship between mass m, weight W and acceleration due to gravity g, of a body.SO, IF MAS OF 2,50 TONES TO BE TRANSFORMED INTO NEWTONS IS 2.50 TONNES TIMES 1 000 KG TIMES 9.8 N IS EQUAL TO 24 500 N.a = 24 500 N FINAL SOLUTIONFIND THE TENSINS ALONG THE TWO CHAINSc = a sin θsin α=24 500 N x sin 42°sin 96°= 24 500 N x 0.66910.9945 = 16 483.6098548 N ≈ 16 483.61b = a sin βsin α= 24 500 N x sin 42°sin 96° = 24 500 N x 0.66910.9945 = 16 483.6098548 N ≈ 16 483.61 NSO, THE TENSIONS ALONG THE TWO CHAINS ARE RESPECTIVELY 16 483.61 N OR 16.48361 puted/calculated by: Antonio Barbosa de Oliveira AntonioBO ’’Cohen’’/Math/Doc/ ................
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