PREFACEThis project is for students who want to study further about

Plane and Spherical Trigonometry with deeper explanation and

examples. I made this project which contains a comprehensive

discussion on Plane and Spherical Trigonometry. Trigonometry is a

branch of mathematics that studies about triangles and the

relationships between the lengths of their sides and

the angles between those sides. Trigonometry defines

the trigonometric functions, which describe those relationships

and have applicability to cyclical phenomena, such as waves.

The exercises are arranged in such way as to allow the users

to warm up first with the very easy one before taking up more

complicated and challenging ones, thus enabling them to practice

and evaluate their knowledge and learning.

BEARINGI. Bearing is an angle, measured clockwise from the north

direction.

Direction: Draw the given bearing.

1. 45°N of E

45°N

2. 30° N

30°

3. S 35° E

35°

4. N 80° W

80°

5. S 34° E

34°

CONVERSIONII. Conversion - The act of turning or changing from one state or

condition to another or the state of being changed. It is very

essential in field of science and mathematics.

Direction: Convert Degrees to Radian.

1. 48° - Radian

45° πrad180°

2. 60.12°- Radian

= 0.7854

= 1.7199

= 28.6478

60.12° πrad180°

3. 57.65°- Radian

57.65°πrad180°

4. 18.2456°- Radian

18.2456°πrad180°

5. 98.54°- Radian

98.54° πrad180°

Direction: Convert Radian to Degrees

1. 0.5 – Degree

0.5 180°πrad

= 1.0493

= 1.0061

= 0.3184

=

=

=

= 43.1092

2. 1.98 – Degree

1.98 180°πrad

3. 12.5698- Degree

12.5698180°πrad

4. 0.2998- Degree

0.2998 180°πrad

5. 0.7524- Degree

0.7524 180°πrad

Direction: Convert Angle in Decimal to Degree, Minute and Seconds

69 27’

18 58’ 48”

127 5’

75 22’ 30”

1. 18.98°- Degree, Minute and Seconds

18 + 0.98(601 )

18 + 58’+ 0.8 (601)

2. 69.45°- Degree, Minute and Seconds

69 + 0.45 (601 )

3. 127.1°- Degree, Minute and Seconds

127 + 0.1 (601 )

4. 75.375°- Degree, Minute and Seconds

75 + 0.375(601 )

75 + 22’+ 0.5 (601)

5. 275.475°- Degree, Minute and Seconds

275 + 0.475(601 )

75 28’ 30”

181.8372

58.9403

275 + 28’+ 0.5 (601 )

Direction: Convert Degree, Minute and Seconds to Angle in Decimal

1. 181°49’ 74’’- Angle in Decimal

181 + 49(160) + 74 (

13600)

181 + .8167 + 0.0205

2. 58°45’ 685’’- Angle in Decimal

58 + 45(160) + 685 (

13600)

58 + 0.75 + 0.1903

3. 258°5’ 54’’- Angle in Decimal

258 + 5(160) + 54 (

13600)

258.0983

160.5206

143.2686

258 + 0.083 + 0.015

4. 160°31’ 14’’- Angle in Decimal

160 + 31(160) + 14 (

13600)

160 + 0.5167 + 0.0039

5. 143°15’ 67’’- Angle in Decimal

143 + 15(160) + 67 (

13600)

143 + 0.25 + 0.0186

Word ProblemsIII. Right Triangle, Angle of Elevation and Angle of Depression

Problem

Direction: Answer the following problems. Make an illustration and

show complete solution.

1. The angle of elevation from a point 49 meters from the base of a

flagpole to the top is 27 . Find the height of the flagpole.

Illustration

Solution:

tan27°= x49

49tan27°=x

49 (.5095)= x

X = 25 m

2. On the way to Tanauan, Rainier was told that when he stands 123 ft

from the base of a tree, the angle of elevation to the top is 26 40’.

If his eyes are 5.3 ft. above the ground, find the height of the tree.

Illustration

Solution:

tanA= side oppositeside adjacent

tan26°= a123

a=123tan26° 40'

A= 61.8 ft.

Since Rainier’s eyes are 5.3 above the ground, the height of the tree is

61.8 ft + 5.3 ft = 67.1 ft

3. A 15 ft ladder leans against a wall at Isaac at an angle of elevation of 60°

(a) How high up the wall does the ladder rest?

(b) How far from the wall is the base of the ladder?

Illustration:

Solution

sin 60°=h15

h=15sin60°

h=12.99 ft

cos 60 °=b15

b=15cos60°

b=7.5 ft

4. Professors of different schools at Faith will be having Shindig

2013. They will be needing a flat 12 foot plank rests with one end on

the ground and the other end upon a 4 foot ledge of the stage. How far

from the base of the ledge is the far end of the plank? What is the

angle of elevation?

Illustration:

Solution:

42 + b2=122

b= √122+ 42

b= √144-16

b= √128

b = 11.3 ft

What is the Angle of Elevation?

sinA= 412

A=sin-1(412 )A=19.47°

5. Benjamin is 5’ ft. tall. Find the length of his shadow if the angle

of elevation of the sun is 30.2°

Illustration:

Solution:

tan30.2°= (5 )s

s= 5tan30.2

s=8.5908 ft

IdentitiesIV. Trigonometric Identities. I want the topic of proving

trigonometric identities because I enjoyed using formulas we

learned to substitute other formulas in order to prove

equality. This was a bit difficult for me and sometimes I had

to step away from the problem and try to see it from a

different point of view because there are so many different

ways to prove an identity

Direction: Prove the following Trigonometric Identities.

1.sinθ cotθ=cosθ

sinθ∙ cosθsinθ

=cosθ

cosθ=cosθ

2. tanθ+cotθ=secθcscθ¿.

sinθcosθ

+cosθsinθ

=(1cosθ )(1sinθ )sinθcosθ

∙sinθsinθ

+cosθsinθ

∙cosθcosθ

=1cosθ

∙1sinθ

sin2θ+cos2θsinθcosθ =

1cosθ∙sinθ

1sinθcosθ

=1sinθcosθ

3. tan Acot A+ 1=sec2A

sinAcosAcosAsinA

+ 1= sec2A

sinAcosA

∙sinAcosA

+1=sec2A

sin2Acos2A

+1=sec2A

tan2A+1= sec2A

sec2A=sec2A

4. cosθ sinθ+cosθcos2θ

=1

cosθ (sinθ+1)cos2θ

=1

(sinθ+1)cosθ

=1

cosθcosθ

=1

1=1

5.sinx+cosx=secx+cscxtanx+cotx

sinx+cosx=

1cosx

+ 1sinx

sinxcosx+

cosxsinx

sinx+cosx=

sinx+cosxsinxcosx

sin2x+cos2xsinxcosx

sinx+cosx=¿sinx+cosxsinxcosx

∙ sinxcosx1

¿

sinx+cosx=¿sinxcosx ¿

Trigonometric EquationDirection: Answer the following trigonometric identities usingquadratic equation.

1.6 (tanθ−3 )=5tanθ−3

6tanθ−18=5tanθ−3

6tanθ−5tanθ=18−3

[tanθ=15]sin−1

86.1859° and 266.1859°

2. 10cosθ+6=9cosθ+5

10cosθ−9cosθ=5−6

[cosθ=−1 ]−1

180°∧360°

3.tan2θ−3tanθ=5

.tan2θ−3tanθ−5=0

Let tanθbex

x2−3x−2=0

x=−b +¿−¿√b2−4ac2a

¿

3 +¿−¿√32−4 (1 )(2)

2(1)¿

3 +¿−¿√9−82

¿

3 +¿−¿12

¿

3+12

[tan=2]tan−1

63.4349°, 243.4349°

3−12

[tan=1]tan−1

45°, 225°

4. 4sin2x+12sinx=9

4sin2x+12sinx−9=0

Let sinbex

4x2+12x−9=0

x=−b +¿−¿√b2−4ac2a

¿

x=−10¿¿

x=−10¿¿

x=−10¿¿

x=−10 +¿−¿08

¿

x=−108

[sin=−108 ]

sin−1

−0.9490

0.9490°, 180.9490.

5. sinx+2=3

sinx=3−2

sinx=1

[sin=1]sin−1

90°, 180

Sine Law

Sine Law. In trigonometry, the law of sines, sine law, sine formula,or sine rule is an equation relating the lengths of the sides of anarbitrary triangle to the sines of its angles.

Direction: Answer the missing sides and angle of each given.

1. a=5, B=84.3 A=33.5

asinA

= bsinB

5sin33.5

= bsin84.3

b=5sin84.3sin33.5

b= 2.9036

C= 180 – (A+B)

C= 180 – (33.5+84.3)

C= 180 – (117.8)

C= 62.2

asinA

=c

sinC

5sin33.5

=c

sin62.2

b=5sin62.2sin33.5

b= 3.7759

2. a=10, C=30° A=75°

asinA

=c

sinC

10sin75

= csin30

c=10sin30sin75

c= 25.4791

B= 180 – (A+C)

B= 180 – (30+75)

B= 180 – (105)

B= 75

asinA

= bsinB

6sin75

= bsin75

b=6sin70sin75

b= 6

3. B=71 C=34 c=12 cmA= 180 – (B+C)

A= 180 – (71+34)

A= 180 – (105)

A= 75

csinC

=b

sinB

12sin34

=b

sin71

b=12sin71sin34

b= 21.5706 cm

asinA

=c

sinC

12sin75

=c

sin24

c=12sin24sin75

c= 28.0233cm

4. B=81 C=64 c=15 ftA= 180 – (B+C)

A= 180 – (81+64)

A= 180 – (145)

A= 35

csinC

=b

sinB

15sin64

=b

sin81

b=15sin81sin64

b= 16.4836 ft

asinA

= csinC

asin35

= 15sin64

a=15sin35sin64

a= 9.5724cm

5. a=50, B=80 A=18

asinA

=b

sinB

50sin18

=b

sin80

b=50sin80sin18

b= 159.3452 ft

C= 180 – (A+B)

C= 180 – (80+50)

C= 180 – (130)

C= 50

asinA

= csinC

50sin18

=c

sin50

b=50sin50sin18

b= 123.9486ft

Cosine LawCosine Law. In trigonometry, the law of cosines (also known asthe cosine formula or cosine rule) relates the lengths of the sides ofa triangle to the cosine of one of its angles. 

Direction: Answer the missing sides and angle of each given.

1. a= 3 b=4 c=5

cosA=b2+c2−a2

2bc

cosA=42+52−32

2 (4)(5)

cosA=3240

[cosA=3240

]cos−1

A=36.8699°

cosB=a2+c2−b2

2bc

cosB=32+52−42

2 (3 )(5)

cosB=35

[cosB=35

]cos−1

B=53.1301°

C=180-(A+B)

C=180-(36.8699°+57.7690 )

C=180-(90.0000 )

C=90.0000

2. a= 13 b=24 c=32

cosA=b2+c2−a2

2bc

cosA=242+322−132

2 (24 )(32)

cosA=477512

[cosA=477512

]cos−1

A=21.3080°

cosB=a2+c2−b2

2bc

cosB=132+322−242

2 (13)(32)

cosB=617832

[cosB=6230

]cos−1

B=42.1333°

C=180-(A+B)

C=180-(21.3080 +42.1333°)

C=180-(63.4413°)

C=116.5587°

3. a= 10 b=12 c=18

cosA=b2+c2−a2

2bc

cosA=122+182−102

2 (12) (18)

cosA=2327

[cosA=2327

]cos−1

A=31.5863°

cosB=a2+c2−b2

2bc

cosB=102+182−122

2 (10)(18)

cosB=79

[cosB=79

]cos−1

B=38.9424°

C=180-(A+B)

C=180-(31.5863°+38.9424°)

C=180-(70.5287°)

C=109.4713°

4. a= 12 b=16 c=21

cosA=b2+c2−a2

2bc

cosA=162+212−122

2 (16) (21)

cosA=7996

[cosA=7996

]cos−1

A=34.6221°

cosB=a2+c2−b2

2bc

cosB=122+212−162

2 (12)(21)

cosB=4772

[cosB=4772 ]

cos−1

B=49.2486°

C=180-(A+B)

C=180-(34.6221°+49.2486°)

C=180-(83.8707°)

C=96.1293°

5. a= 6 b=8 c=5

cosA=b2+c2−a2

2bc

cosA=82+52−62

2 (8) (5)

cosA=5380

[cosA=5380 ]

cos−1

A=48.5092°

cosB=a2+c2−b2

2bc

cosB=62+52−82

2 (6 )(5)

cosB=−120

[cosB=−120

]cos−1

B=92.8660°

C=180-(A+B)

C=180-(48.5092°+92.8660°)

C=180-(141.3752°)

C=38.6248°

Spherical TrigonometrySpherical Trigonometry. The last part of our topic. One of theeasiest topic we’ve discussed. It is just a summary of sine law andcosine law.

Direction: Give what is asked.

1. A=100°B=30°a=69°findb

sin asin A

=sin bsin B

sin 69sin 100

=sin bsin 30

sin b=sin 69 sin 30sin 100

b=[sin69sin30sin100

] sin−1

b=28.2937°

2. a=50° b=82° c=150° find c

cos c= cos a cos b +sin a sin b cosC

cos c= cos50cos82+sin50sin82cos150

c = [cos50cos82+sin50sin82cos150]cos−1

c = 124.5760°

3. Find the area of ∆ ABCwhose sides measures 100°, 96° and 82° respectively and its radius is 29 cm.

E= A + B+ C -180

E=100° + 96° +82° -180

E=278° -180

E=98°

A= πr2E

180

A= (3.1416 )292(98)

180

A=1438.4688cm2

4. Find the area of ∆ ABCwhose sides measures 98°, 76° and 39° respectively and its radius is 13 ft.

E= A + B+ C -180

E=98° + 76° +39° -180

E=213° -180

E=33°

A= πr2E

180

A= (3.1416 )132(33)

180

A=97.3372ft2

5. Given ∆ ABC , ¿Ais59°and side b and side c is 50°∧68°

cos a= cos b cos c +sin b sin c cos A

cos a= cos50cos68+sin50sin68cos59

a = [cos50cos68+sin50sin68cos59cos−1

c = 52.6556°