AntonioBO ’’Cohen’’/Math/Doc/ MATHEMATIC

MY CONCISE NOTES about

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Trigonometry

Sine, COSINEE and

TAN

SOME SOLVED PROBLEMS TO FIND UNKNOWN SIDES OF TRIANGLES, INVOLVING THE SINE, COSINE AND TANGENT RATIOS.

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 C

Angle A = 48ᵒAngle C = 90ᵒLong side A-B (hypotenuse) = 9 meters

Use the sine ratio if you know or need to know the opposite side or the hypotenuse

Example 1Find the length of side a of triangle above, substitute the known values into the sine ratio.

SOLUTION

SIN θ = LENGTHOFOPPOSITE SIDELENGTHOF HYPOTENUSE

Sin 48˚ = a9 ⇨Rearrange the formula to make a the subject and so it becomes

a = 9 x sin 48˚

Use the sin button of calculator CASIO fx 992s or any scientific calculator to find the value of sin 48ᵒ, and then solve the equation:

a = 9 x 0, 74314482547 a = 6.688 3034293 or 6.69 meters

So, the length of side a is 6.69 meters long (to 2 d.p.) or to 2 decimal proximately.

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SINE RATIO (SIN)

THE FORMULA IS

SINE θ (THETA) =

COSINE RATIO (COS)

THE FORMULA IS

COSINE θ (THETA) =

Use the cosine ratio if you know or need to know the adjacent side or the hypotenuse. C

b 10 m

A BAngle C = 42°Angle B = 90°Side B-C = 10 meters

Example 2Find the length of side b of this triangle; substitute the known values into the cosine ratio.

SOLUTION COS θ = LENGTHOF ADJACENT SIDELENGTHOF HYPOTENUSE

COS 42ᵒ = 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 42ᵒ, and then solve the equation.

b = 100 .74314482547

=13 .4563272961

So, the length of side b = 13.46 meters long (to 2 d.p.)

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COSINE RATIO (COS)

THE FORMULA IS

COSINE θ (THETA) =

θ=42 °

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 = LENGTHOFOPPOSITE SIDELENGTHOF ADJACENT SIDE

Tan 50 =c5 ⇨Rearrange the formula to make c the subject and so it becomes

c = 5 x tan 50

Use the “TAN” button of calculator CASIO fx 992s or any scientific calculator to find the value of 50ᵒ, and then solve the equation.

c = 5 x 1.19175359259c = 5.958767966297 ≈5 .96meter (to 2 d.p.)

So the length of side c is 5.96 meter long (to 2 d.p.)

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TANGENT RATIO (TAN)

THE FORMULA IS

TANGENT θ (THETA) = LENGTHOFOPPOSITE SIDELENGTHOF ADJACENT SIDEQUOTEError ! Bookmark not defined .

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= 6

12

Cos θ=12 = 0.5 ⇨Rearrange the formula to make θthe subject and so it becomes

θ=cos 12=0.5 = 60

Use inverse or shift cos button of calculator CASIO fx 992s or CASIO fx 100AU or any scientific calculator to find the valueof θ. B

12 m

A CHOW TO USE CALCULATOR CASSIO fx 992s 6 M

TO FIND θ=cos 12=0.5 to find 60°

12=0.5⇨Press ⇨Press ⇨ Press , it will display

60⇨ Press ⇨ Press it will display 60

SO, THE ANGLEθ (THETA ) IS EQUAL¿60 °.

B

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ANGLE θ

90 °

SHIFT COS =

SHIFT ’ ”

12 m 9 m

A C

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

= 912

Sin θ=0.75 ⇨Rearrange the formula to make θthe subject and so it becomes

θ=sin 912

=0.75

Use inverse or shift sin button of calculator CASIO fx 992s or CASIO fx-100AU or any scientific calculator to find the valueof θ.

HOW TO USE CALCULATOR CASSIO fx 992s

TO FIND θ=sin 912

=0.75 to find 49

912

=0.75⇨Press ⇨ Press ⇨ Press , it will

display 48.590377890 ≈ 49 ⇨Press ⇨ Press it will

display 49° .

SO, THE ANGLE θ (THETA) IS EQUAL TO 49° .

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ANGLE θ

90 °

SFIFT SIN =

SHIFT ’ ”

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 SNE 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

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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

The 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 = 40º Angle B = 95º Side b = 12 meterFind side a =? b = 12 meter

SOLUTION:a

sin A= b

sinB= a

sin 40 °= 12

sin 95°

a0.6428

= 120.9968

a = 12

0.9968x 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.

I. An axial force is induced (is produced; is caused) in the mast by the two wire stays/girdle/cord as shown:

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25 18 A B 1000 N mast

1. Find the magnitude of the force acting along the mast.

2. What is the magnitude of the force acting on the stay B?

NB: Solve the problem analytic and graphically Giving:

o 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).

o B = unknown force, acting downwards to the right (sense) at 18° to the vertical (direction).

o Mast = unknown force =?

A) FIRST STEP, SOLVE THE PROBLEM ANAYTICALLY, USING SINES RULES

C

b =1 000 N

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ANGLE C

θ(THETA ) 25°

A a = 2 207 N

c = 1 368 N

B

SOLUTION:

1. asin A

= bsinB

= asin 137 °

=1 000Nsin18 °

a0.6820

=1 000N0.3090

a = 1000 N0.3090 x 0.6820

a = 2 207.1197411 ≈2 207N

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ANGLE Aα (ALPHA) 137 °=¿180

°−(25 °+18 ° ) ANGLE B β (BETHA)

18 °

So, the magnitude of Force acting along the mast is 2 207 N.

2. bsinB

= csinC

¿ 1 000sin18 °

= csin 25 °

= 1 0000.3090

= c0.4226

c = 1 0000.3090

x 0.4226

c = 1 367.63754045 N ≈1368N

So, the magnitude of Force acting in the stay B is 1 368 N.

b) Second step, solve the problem graphically

A B

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1. FREE BODY DIAGRAM

25 18 Force A Force B 1 000 N

Mast

2. FORCE DIAGRAM ⇨ SCALE 1 CM=200N C

b

A c a

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PROJECTION LINE OF FORCE A

5 CM x 200 = 1 000 N

PROJECTION LINE OF FORCE B

6.8 CM x 200 = 1 360 N

PROJECTION LINE OF FORCE A AND B ACTING

ON MAST 11 CM x 200 = 2 200 N

B

CHECK THE ANGLES OF FORCE DIAGRAM, USING COSINES RULES OF NON-RIGHT TRIANGLES:

C

b =1 000 N

A a = 2 200 N1801800

c = 1 360 N

BSOLUTION:

Cos B =a2+c2−b2

2ac=2 200²+1360²−100 02

2x 2200 x1 360

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ANGLE θ(THETA )

24° 55'06.24 ' '

26.5615

ANGLE α (ALPHA) 137° 02’

42’’ = 180°−(24 ° 55' 06.2 4' '+18° 02' 49.56 ' ' )

ANGLE β (BETHA)18° 02 ' 49.56 ' '

18.0471

Cos B = 4 840 000+1 849 600−1000 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−c2

2ab =22002+1 0002−13602

2 x2200 x 1000

Cos C = 4 840 000+1 000 000−1849 6004 400000

Cos C = 3 990 4004 400 000 = 0.9069090909

Cos C = 24.9183544336 ≈24.9184=24 ° 55 '06.24 ' '⇨ θ(THETA)

Cos A = 180°−( cosB+cosC )=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 (βbeta) = 18.0471=18° 0 2'49.56 ' '; Cos C ¿theta) = 24.9184 = 24° 55 '06.24 ' ' ;

SOLUTION:

1. b=asinβsin α

=2200 x sin18 ° 02' 49.56(18.0471)sin 137 °0 2' 42(137. 0345)

b=2200 x0.309798705950.68155786016

b = 999.998962561 N ≈1 000N

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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.

2. c = asinθsinα = 2200 x24 °55' 06.24 ' '(24.9184)sin 137° 0 2' 42 ' ' (137. 0345)

c = 2200 x 0.42132708029

0.68155786016

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. FINAL CHECKIN ⇨

a² = (1000)²+ (1360)²- 2(1000) (1360) x cos137° 02'42' ' (137.0345 )

= 1000000 + 1849600 – 2720000 x (-0.73176422654) = 2 849 600 – (- 1 990 398.6962) = 4 839 998.6962

a = √4 839 998.6962=¿2 199.99970368 N ≈ 2 200 N

SO, 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.

II. 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

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a² = b²+ c²- 2bc cos α

y

40 kN

150° o x

SOLUTION GRAPHICALLY

FORCE DIAGRAMSCALE 1 CM = 5 kN

y

150°

o x

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.

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HORIZONTAL COMPONENT

6.9 CM x 5 kN = 34.5 kN

VERTICAL COMPONENT4 CM x 5 kN = 20 kN

40 kN

ANSWER

Horizontal component of force = 34.50 kN

Vertical component of force = 20 kN

SOLUTION ANALYTICALLY

FREE BODY DIAGRAM A y

40 kN 150° X B C

From the triangle above ABC

Sin 30° = ABAC=VERTICALCOMPONENTOF FORCE40kN

VERTICAL COMPONENT FORCE = 40 kN x sin 30° = 40 x 0.5 = 20 kN

Cos 30° = BCAC=HORIZONTALCOMPONENT OF FORCE40kN

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

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VERTICALCOMPONENT

HORIZONTAL COMPONENT 30°=180 °−150 °

AN EXAMPLE OF COMPONENT

Summing 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.

III. 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 METER

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SOLUTION:1. Draw a triangle from two chains and the top part of

the skip/hopper to find the angle B and C.2. 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 m

3. After 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 side

hypotenuse = 1.00m1.50m

= 0.6666 = 48.1975 ≈48=48 ° SO, COSINES θ 0R ANGLE D = 48° 48°D FTriangle GHI

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ANGLE θ?

H Cos θ=adjacent sidehypotenuse = 1.50m

1.00m

= 0.6666 = 48.1975 ≈48=¿ 48°

SO, COSINES θOR ANGLE I = 48°

48°

G I

4. Knowing 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

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ANGLEθ?

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/s²) 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 SOLUTION

FIND THE TENSINS ALONG THE TWO CHAINS

c = a sin θsinα

=24 500N x sin 42°sin 96 °

=24 500N x0.66910.9945

= 16 483.6098548 N ≈16 483.61

b = a sin βsin α

=24 500 N xsin 42 °sin 96°

=24500N x 0.66910.9945

= 16 483.6098548 N ≈16 483.61 N21

SO, THE TENSIONS ALONG THE TWO CHAINS ARE RESPECTIVELY 16 483.61 N OR 16.48361 kN.

Computed/calculated by: Antonio Barbosa de Oliveira AntonioBO ’’Cohen’’/Math/Doc/

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