25 solved25 solved Trigonometric ProblemsTrigonometric Problems

•Law of SineLaw of Sine

•Law of CosinesLaw of Cosines

•Law of TangentsLaw of Tangents

•Graphs of Sine and CosineGraphs of Sine and Cosine

•Graphs of Other FunctionsGraphs of Other Functions

Pampanga Agricultural CollegePampanga Agricultural CollegeMagalang, Pampanga, PhilippinesMagalang, Pampanga, Philippines

Solve Solve ∆ABC if A=39°, a=21inches and b=11inches. ∆ABC if A=39°, a=21inches and b=11inches. Find the remaining side and angles.Find the remaining side and angles.

SolutionSolutionUsing the Sine Law,Using the Sine Law,

• a a == b b sinA sinBsinA sinB

• 21 21 == 11 11 sin39sin39° sinB° sinB• sinB = sinB = 11sin3911sin39

2121• ≈≈0.3296440.329644• B ≈ 19,25°B ≈ 19,25°We can now say that,We can now say that,• C ≈ 180-(39+19.25)C ≈ 180-(39+19.25)• C ≈121.75C ≈121.75

To solve for the remaining side,To solve for the remaining side,

• c c == a a sinC sinAsinC sinA• c c == a a sin121.75sin121.75°° sin19.25 sin19.25°°• c=c=21sin121.75°21sin121.75° sin19.25°sin19.25°• c≈54.16inchesc≈54.16inches

Given a Given a ∆ABC, A=48°, b=16, c=32. Find a.∆ABC, A=48°, b=16, c=32. Find a.

Solution:Solution:

• a2 = b2 + c2 - 2·b ·c·Cos A a2 = b2 + c2 - 2·b ·c·Cos A

• a2 = 102 + 202 - 2·10 ·20·Cos 30a2 = 102 + 202 - 2·10 ·20·Cos 30

• a = ( 102 + 202 - 2·10 ·20·Cos 30 )0.5a = ( 102 + 202 - 2·10 ·20·Cos 30 )0.5

• a = 12.393137 a = 12.393137

Solve Solve ∆ABC shown at the right with A=65°, B=48°, and ∆ABC shown at the right with A=65°, B=48°, and c+76ft. Find the remaining sides and angle.c+76ft. Find the remaining sides and angle.

SolutionSolution• A+B+C=180A+B+C=180°°• C=180°-(B+C)C=180°-(B+C)• C=180°-(65+48)C=180°-(65+48)• C=180°-113C=180°-113• C=67°C=67°

To get the length of two sides, we use To get the length of two sides, we use the Law of Sines.the Law of Sines.

• sinAsinA==sinCsinC a ca c

• sin65sin65°°==sin67°sin67° a 76a 76• 76sin6576sin65°=asin67°°=asin67°

• 72sin6572sin65°=a°=a sin67°sin67°• 75=a75=a

• sinBsinB==sinCsinC b cb c

• sin48°sin48°==sin67°sin67° bb 76 76

• 76sin48°=bsin67°76sin48°=bsin67°• 76sin48°76sin48°=b=b sin67°sin67°• 67°=b67°=b

Two sides of Two sides of ∆ ABC measures 6cm and 8cm and ∆ ABC measures 6cm and 8cm and their included angle 40°. Find the third side. their included angle 40°. Find the third side.

Solution:Solution:

• cc²=a²+b²-2abcosC²=a²+b²-2abcosC

• c²=6²+8²-2(6)(8)cos40°c²=6²+8²-2(6)(8)cos40°

• c=5.144cmc=5.144cm

Given a triangle whose angles are A=40Given a triangle whose angles are A=40°, B=95° and °, B=95° and side b=30cm. Find the length of the bisector of angle C.side b=30cm. Find the length of the bisector of angle C.

Solution:Solution:

• Angle C=180°-40°-95°=45°Angle C=180°-40°-95°=45°

• In triangle ADC:In triangle ADC:

• ӨӨ=180=180°-40°-22.5°=117.5°°-40°-22.5°=117.5°

• c=c= x x == 30 30

• sin40sin40° sin117.5°° sin117.5°

• x=21.74cmx=21.74cm

Given Given ∆ ABC∆ ABC: C=100: C=100°, b=20. Find c.°, b=20. Find c.

Solution:Solution:

Using the Cosine Law:Using the Cosine Law:

• cc²=a²+b²-2abcosC²=a²+b²-2abcosC

• c²=15²+20²-2(15)(20)cos100°c²=15²+20²-2(15)(20)cos100°

• c=27c=27

Given ∆ABC, if A=40°°, B=20°, a=2, find C and b.Given ∆ABC, if A=40°°, B=20°, a=2, find C and b.

• A+B+C=180A+B+C=180°°

• C=180-(A+B)C=180-(A+B)

• C=180-(40+20)C=180-(40+20)

• C=180-60C=180-60

• C=120°C=120°

• sinAsinA==sinBsinB

a ba b

• sin40sin40==sin20sin20

2 b2 b

• bsin40=2sin20bsin40=2sin20• b=b=2sin202sin20

sin40sin40

• b=1.0642b=1.0642

Given a Given a ∆ABC, A=12°, C=45°, a=5m. Find c.∆ABC, A=12°, C=45°, a=5m. Find c.

Solution:Solution:

• a a == b b == c c

sinA sinB sinCsinA sinB sinC

• c=c= a(sinC) a(sinC) == 5(sin45) 5(sin45)

sinAsinA sin12 sin12

• c=17.005c=17.005

A small electric component is in the shape of a triangle with A small electric component is in the shape of a triangle with sides 6.23, 8.146 and 11.392 millimeter. Find the largest sides 6.23, 8.146 and 11.392 millimeter. Find the largest

angleangle

Solution:Solution:

• Let a = 6.23, b = 8.146 and c = 11.392 Let a = 6.23, b = 8.146 and c = 11.392

• Angle opposite to the largest side is the largest angle. Angle opposite to the largest side is the largest angle.

• cos C = a2 + b2 - c2 = (6.23)2 + (8.146)2 -(11.392)2 = cos C = a2 + b2 - c2 = (6.23)2 + (8.146)2 -(11.392)2 = -0.242-0.242

Given a Given a ∆ABC, A=36°, B=23°,b=11mm. Find a.∆ABC, A=36°, B=23°,b=11mm. Find a.

Solution:Solution:

• a a == b b == c c

sinA sinB sinCsinA sinB sinC

• b=b= b(sinA) b(sinA) == 11(sin35) 11(sin35)

sinBsinB sin23 sin23

• a=16.1475mma=16.1475mm

Given angle A=32Given angle A=32°, angle B=70°, and side °, angle B=70°, and side c=27 units. Solve for side a of the ∆.c=27 units. Solve for side a of the ∆.

SolutionSolution

Sine LawSine Law

A+B+C=180A+B+C=180°°

C=180°-32°-70°C=180°-32°-70°

C=78°C=78°

• a a == b b

sinA sinBsinA sinB

• a a == 27 27

sin32sin32° sin78 °° sin78 °

• a=14.63unitsa=14.63units

Solve for b in Solve for b in ∆ABC if a=4, c=7 and B=130°∆ABC if a=4, c=7 and B=130°

SolutionSolution

• bb²=a²+c²-2accosB²=a²+c²-2accosB

• 4²+7²-2(4)(7)cos130°4²+7²-2(4)(7)cos130°

• 16+49-56cos130°16+49-56cos130°

• 65+56(.6428)65+56(.6428)

• 100.9961100.9961

• b=10.05b=10.05

Given a ∆ABC, B=36°, C=24°, c=12cm. Find b.Given a ∆ABC, B=36°, C=24°, c=12cm. Find b.

Solution:Solution:

• a a == b b == c c

sinA sinB sinCsinA sinB sinC

• b= b= c(sinB) c(sinB) == 12(sin36) 12(sin36)

sinCsinC sin24 sin24

• b = 17.3415cmb = 17.3415cm

In a ∆ABC, A=45° and angle C=70°. The side opposite In a ∆ABC, A=45° and angle C=70°. The side opposite angle C is 40m long. What is the side of opposite angle A?angle C is 40m long. What is the side of opposite angle A?

SolutionSolution

• a a == b b

sinA sinBsinA sinB

• a a == b b

sin45° sin70sin45° sin70°°

• a=30.1ma=30.1m

Two sides of a Two sides of a ∆ABC are 50m and 60m long. The angle included ∆ABC are 50m and 60m long. The angle included between these sides is 30°. What is the interior angle opposite between these sides is 30°. What is the interior angle opposite

the longest side?the longest side?

Solution:Solution:

Solving for the third side by Solving for the third side by cosine law:cosine law:

• c²=a²+b²-2abcosCc²=a²+b²-2abcosC

• c=30.064c=30.064

Solving for the angle opposite the Solving for the angle opposite the 60m side by sine law:60m side by sine law:

• 60 60 == 30.064 30.064

sinsinӨӨ sin30 sin30°°

• sinsinӨӨ=0.9978=0.9978

• ӨӨ=86.26° and 93.74°=86.26° and 93.74°

In In ∆ABC, find the side c if angle C=100°, side b=20, ∆ABC, find the side c if angle C=100°, side b=20, and side a=15.and side a=15.

SolutionSolution

By Cosine Law:By Cosine Law:• cc²=a²+b²-2abcosC²=a²+b²-2abcosC

• cc²=(15) ²+(20) ²-2(15)(20)cos100°²=(15) ²+(20) ²-2(15)(20)cos100°

• C=27C=27

Find the difference between the largest and smallest Find the difference between the largest and smallest angles of a triangle if the lengths of sides are 10, 19 angles of a triangle if the lengths of sides are 10, 19

and 23.and 23. Solution:Solution:

The area of the triangle whose angles are 61The area of the triangle whose angles are 61°9’32”, 34°14’46”, and °9’32”, 34°14’46”, and 84°35’46” is 680.60. The length of the longest side is:84°35’46” is 680.60. The length of the longest side is:

Solution:Solution:

• A= ½ ac sinB=680.6 A= ½ ac sinB=680.6

• a a = = c c

• sin34°14’46” sin84°35’46” sin34°14’46” sin84°35’46”

• A=0.5653c1A=0.5653c1

• 680.6= ½ (0.5653 c) (c) sin61°9’32” 680.6= ½ (0.5653 c) (c) sin61°9’32”

• c=52.43unitsc=52.43units

Given a Given a ∆ABC, A=30°, b=10m, c=20m. Find a.∆ABC, A=30°, b=10m, c=20m. Find a.

Solution:Solution:

• a2 = b2 + c2 - 2·b ·c·Cos A a2 = b2 + c2 - 2·b ·c·Cos A

• a2 = 102 + 202 - 2·10 ·20·Cos 30a2 = 102 + 202 - 2·10 ·20·Cos 30

• a = ( 102 + 202 - 2·10 ·20·Cos 30 )0.5a = ( 102 + 202 - 2·10 ·20·Cos 30 )0.5

• a = 12.393137m a = 12.393137m

Graphs of Sine, Cosine and Graphs of Sine, Cosine and TangentsTangents

Some VocabularySome Vocabulary

Amplitude:Amplitude: height heightFrequency:Frequency: number of curves in a normal number of curves in a normal

periodperiodPeriod:Period: how long it takes to complete one how long it takes to complete one

curvecurveDomain:Domain: x values (the angle in radians) x values (the angle in radians)Range:Range: y values (the trig value) y values (the trig value)

Find the Amplitude and Period of each Find the Amplitude and Period of each function:function:

• y=-2cos 5/4 xy=-2cos 5/4 x

Amplitude=-2Amplitude=-2

Period=2Period=2π/5/4 = 8π/5π/5/4 = 8π/5

• y=2sin2xy=2sin2x

Amplitude=2Amplitude=2

Period=2π/2 = πPeriod=2π/2 = π

• Y= ¼ cos4xY= ¼ cos4xAmplitude= ¼Amplitude= ¼Period= 2π/4 = π/2Period= 2π/4 = π/2

• Y=3sin ¼ xY=3sin ¼ xAmplitude=3Amplitude=3Period=2π/ ¼ = 8πPeriod=2π/ ¼ = 8π

• Y= ½ cos ¼xY= ½ cos ¼xAmplitude= ½ Amplitude= ½

Period=2Period=2π/ ¼ = 8ππ/ ¼ = 8π

Prepared by:Prepared by:

Paula P. AlfonsoPaula P. Alfonso

Agricultural Science CurriculumAgricultural Science Curriculum

33rdrd year High School year High School

pau29_alfonso@yahoo.compau29_alfonso@yahoo.com