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Trigonometry
Angles and Their Measure
Define:

Trigonometry:

Initial Side:

Terminal Side:

Positive Angle:

Negative Angle:

Standard Position:

Coterminal Angles:

Quadrantal Angle:

Reference Angle:
Angle Unit Measures:



Degree (º): 360º in a circle

1 Degree =
radians = 0.017 radians
180

Radian (π): 2π in circle
180
1 Radian =
degrees = 57.3 degrees

Examples: Conversions (D → R and R → D)


3
3 180
1. 30º = 30 *
=
2.
=
*
= 135 º
180
14
14

6
Practice: Conversions (D → R and R → D)
Convert the following degrees to radians.
3. 45º =
Convert the following radians to degrees.
7
4
5.
=
6.
=
8
3
4. 135º =
Coterminal Angles:



Angles differing in degree measure by multiples of 360 are equivalent.
If α is degree measure of angle, then all angles of form α + 360k are coterminal with α
If β is radian measure of angle, then all angles of form β + 2kπ are coterminal with β.
Examples: Coterminal Angles
Find 1 positive and 1 negative angle that is coterminal with the angle having the following measure:
11
a. 60º
b.
4
11
3
1 positive: 60º + 360º = 420º
1 positive:
- 2π =
4
4
11
 3
1 negative: 60º - 360º = -300º
1 negative:
- 4π =
4
4
1
Trigonometry
Practice: Coterminal Angles
Find 1 positive and 1 negative angle that is coterminal with the angle having the following measure:
13
c. 45º =
d.
=
4
Reference Angles:


If α >2π or α < 0, then associated with coterminal  of positive measure between 0 & 2π
Reference Angle Rule: reference angle defined by
o α when terminal side is in quadrant I
o π – α when terminal side is in quadrant II
o α - π when terminal side is in quadrant III
o 2π – α when terminal side is in quadrant IV
Examples: Reference Angles
Find the measure of the reference angle.
5
 13
1.
2.
4
3
5
 13
5
α=
(QIII)
α=
coterminal:
(QIV)
4
3
3
5

5

Reference =
-π=
Reference = 2π –
=
4
3
4
3
Practice: Reference Angles
Find the measure of the reference angle.
4. 142º =
6.
5. 320º =

=
4
7.
2
10
=
3
3. 510º
α = 510 coterminal: 150 (QII)
Reference = 180 – 150 = 30
Trigonometry
Central Angles and Arcs
Define:

Central Angle:

Arc:
Formula:
Length of Arc (s)
 rθ where r = radius and θ = radian measure

r
180
where r = radius and θ = degree measure
Examples: Central Angles & Arcs
1. Find the length of the arc, given r= 12 and central angle = 45º.
r
( 45)(12)
s=
=
= 24π
180
180
2. Find the central angle, given r = 9 and length of arc = 2π
r
180s 180(2 )
s=
θ=
=
= 40º
180
r
 (9)
3. Find the radius of the circle, given the length of arc = 6π and central angle = 45º
r
180s 180(6 )
s=
r=
=
= 24
180

 (45)
Practice: Central Angles & Arcs
4. Find the length of the arc, given radius = 8 and central angle = 42º
5. Find the central angle, given r = 10 and length = 5π
6. Find the radius of circle, given arc length of
25
5
and central angle =
3
3
3
Trigonometry
Unit Circle
Pythagorean Theorem



Pythagorean theorem: a2 + b2 = c2 (for right angles)
Used to find the length of a side of a right triangle when the lengths of the other 2 sides are known.
Pythagorean Triples:
 3-4-5
 5-12-13
 7-24-25
 8-15-17
 9-40-41
Special Right Triangles
45
√2
30
√3
1
2
45
1
60
1
Unit Circle:


x2 + y2 = 1 where Center (0,0), Radius = 1 and Vertices (1,0),(-1,0),(0,1),(0,-1)
Point of intersection
x y
o  ,  where r = x 2  y 2
r r
Examples: Finding points of intersection
1. Find the point of intersection of the unit circle and angle whose terminal sides contain
a. (3,4)
b. (5, -12)
x = 3, y = 4
x = 5, y = -12
r = 5 (Pythagorean triple)
r = 13 (Pythagorean Triple)
3 4
 5  12 
 , 
 ,

5 5
 13 13 
2. Find the measure of the angle if the terminal side contains (5,-5)
(5, -5) lies in Quadrant IV. Since both x and y are same value, dealing with 45-45-90 triangle
360 – 45 = 315º
3. Find the coordinates of point of intersection of unit circle and terminal ray of 210º
210 º lies in Quadrant III. Reference Angle: 210º - 180º = 30º
  3 1
, 
Dealing with 30-60-90 triangle. 
2 
 2
Practice: Finding points of intersection
4. Find the point of intersection of the unit circle and angle whose terminal sides contain
a. (-7,-24)
b. (-40, 9)
5. Find the measure of the angle if the terminal side contains (√3, -1)
6. Find the coordinates of point of intersection of unit circle and terminal ray of
a. -60º
b. 270º
c. 45º
4
Trigonometry
Circular Functions
Sine/Cosine

If terminal side of angle θ in standard position intersects unit circle at P(x,y) then cos θ = x and sin θ = y
Example: Find sin/cos
1. Find the value of sin 90º
Terminal side of 90º in standard position is positive y-axis intersecting unit circle at (0,1). y-coordinate
of this pair is sin 90º. Therefore, sin 90º = 1
2. Find the value of cos π.
Terminal side of angle in standard position measuring π radians is negative x-axis, which intersects unit
circle at (-1,0). X-coordinate of this pair is cos π. Therefore cos π = -1
Practice: Find sin/cos
3. Find the value of cos 90º.
4. Find the value of sin π.
Sine/Cosine of Angle in Standard Position

For an angle in standard position with measure θ, point P(x,y) on its terminal side and r =
y
x
o sin θ =
o cos θ =
r
r
x 2 y 2
Examples: Sin/Cos of Angle in Standard Position
5. Find values of sin and cos of angle in standard position with measure θ, point P(3,4) on its terminal side.
Step 1: find r
Step 2: write sin and cos ratios
5 (Pythagorean triple) or
y
4
o Sin θ =
=
2
2
2
2
5
r
r = x  y = 3 4 = 5
x 3
o Cos θ = =
r 5
5
& terminal side of θ in 1st quadrant.
13
Step 1: Identify the givens
Step 2: Find y.
12 (Pythagorean Triple) or
5
x
Cos θ =
= so x = 5, r = 13
132  5 2  y 2
13
r
y
12
Sin θ =
=
13
r
6. Find sin θ when cos θ =
Practice: Sin/Cos of Angle in Standard Position
7. Find values of sin and cos of angle in standard position with measure θ, point P(8,15) on its terminal
side.
8. Find sin θ when cos θ =
8
& terminal side of θ in 1st quadrant.
17
5
Trigonometry
Trig Functions of Angle in Standard Position

For angle in standard position with measure θ, point P(x,y) on its terminal side and r = x 2  y 2
y
x
y
opposite sin 
opposite
adjacent
sin θ =
=
cos θ =
=
tan θ =
=
=
r
r
x
adjacent cos 
hypotenuse
hypotenuse
csc θ =
1
r
hypotenuse
=
=
sin 
y
opposite
sec θ =
r
1
hypotenuse
=
=
x
cos 
adjacent
adjacent cos 
x
=
=
opposite sin 
y
1
=
tan 
cot θ =
Examples: Trig Functions of Angle in Standard Position
1. Terminal side of angle θ in standard position contains point (8,-15). Find tan θ, cot θ, sec θ and csc θ.
Step 1: Find r
r = 17 if you remember Pythagorean triples otherwise
r=
x 2 y 2 =
8 2 (15) 2 = 17
Step 2: Write ratios
y  15
17
r
tan θ = =
csc θ = =
x
8
y  15
sec θ =
r 17
=
x
8
cot θ =
8
x
=
y  15
2. If csc θ = -2 and θ lies in Quadrant III, find sin θ, cos θ, tan θ, cot θ and sec θ.
Step 1: find its reciprocal function
Since cscθ and sinθ are reciprocals, sin θ =
1
2
Step 2: Since we have y and r, find x
2 2  x 2 (1) 2 x = ±√3
Since terminal side of θ in Quandrant III, x = -√3
Step 3: Write ratios
 3
cos θ =
2
sec θ =
2 3
3
tan θ =
3
3
cot θ = √3
Practice: Trig Functions of Angle in Standard Position
3. Terminal side of angle θ in standard position contains point (-3,-4). Find tan θ, cot θ, sec θ and csc θ.
4. If sec θ = 2 and θ lies in Quadrant IV, find sin θ, cos θ, tan θ, cot θ, and csc θ.
6
Trigonometry
Trig Functions of Special Angles
Quandrantal Angles



Terminal sides lie along axis.
Examples: 0, 90, 180, 270
Using unit circle, let (x,y) be coordinates of point of intersection of circle with terminal side of angle.
Example: Find trig functions of quandrantal angle
1. Find the 6 trig function values for 90º
Point is (0,1) therefore, sin 90º = 1 and cos 90º = 0
1
0
tan 90º = = undefined
cot 90º = = 0
csc 90º = 1
0
1
2. Find the 6 trig function values for π
Point is (-1,0) therefore, cos π = -1 and sin π = 0
0
1
tan π =
=0
cot π =
= undefined
1
0
sec 90º = undefined
csc π = undefined
sec π = -1
Practice: Find trig functions of quandrantal angle
3. Find the 6 trig function values for 2π
Use reference angle to find value of trig function

Given an angle, find the reference angle then determine the trig functions.
Example: Using reference angles
5
4. Find cos
6
Step 1: Find reference angle
5
α=
so terminal side is in QII.
6
5 
α’ = π - α = π = (30º)
6
6
Step 2: Find cos
5

cos
= -cos
(since cos < 0 in QII)
6
6
 3
=
2
Practice: Using reference angles
 11
5. Find tan
4
6. Find csc
29
3
7
Trigonometry
Using Calculators


Unless otherwise noted, make sure mode of calculator is in degrees not radians.
When trying to find the csc, sec and cot values, do not use sin-1, cos-1 and tan-1 function keys. Those are
1
1
1
used to find the angle value not the trig function value. Use the
tan, sin and cos functions.
x
x
x
Examples: Using Calculators – sin, cos, tan
Find each value to the nearest thousandth.
1. tan 45º
2. sin 80º
45 tan enter or tan 45 enter
80 sin enter or sin 80 enter
Answer: 1
Answer: 0.985
Examples: Using Calculators – csc, sec and tan
Find each value to the nearest thousandth.
4. cot -75º
5. csc 128º
1
tan (-75) Ans-1(x-1)
128 sin
x
Answer: -.2679
Answer: 1.269
Practice: Using Calculators
7. sec 90º
8. csc 125 º
3. cos 90º
90 cos enter or cos 90 enter
Answer: 0
6. sec 187º
1
x
Answer: -1.008
187 cos
9. cot(-40º)
10. sin 140º
11. cos (-82º)
Trig Identities
Pythagorean Identities

(cosθ)2 + (sin θ)2 = 1
 1 + (tan θ) 2 = (sec θ)2
 1 + (cot θ) 2 = (csc θ)2
Example: Writing 1 trig function in terms of others
1. Write cosθ in terms of tanθ
1
1
cosθ =
so (cosθ)2 =
sec 
(sec  ) 2
1
Using 1 + (tan θ) 2 = (sec θ)2, (cosθ)2 =
1 (tan  ) 2
cosθ = 
1
1 (tan  ) 2
Practice: Writing 1 trig function in terms of others
2. Use the statement (cosθ)2 + (sin θ)2 = 1 to prove each of the following
a. 1 + (tan θ) 2 = (sec θ)2
b. 1 + (cot θ) 2 = (csc θ)2
8
Trigonometry
Trig Relationships
Odd/Even Functions



For all angles θ, cos (-θ) = cos θ and sin (-θ) = -sin θ
For all angles θ (where cos θ ≠ 0), tan (-θ) = -tan θ and sec(-θ) = sec θ
For all angles θ (where sin θ ≠ 0), cot (-θ) = -cot θ and csc (-θ) = -csc θ
Supplementary Angles

For all angles θ, sin (π-θ) = sin θ and cos (π-θ) = -cos θ
Complementary Angles
 sin θ = cos (

2
 )
 sec θ = csc (
Examples: Simplify Expressions
1. Simplify the expression cos(-θ)tan(-θ)
sin(  )
cos(-θ) *
= sin (-θ) = -sin θ
cos(  )

2
 )
 tan θ = cot (

2
 )
2. Rewrite sin (90- θ) in terms of cos θ
sin (90 – θ) = cos θ
Practice: Simplify Expressions in terms of sin or cos

3. sin (   )
4. cos (270 – θ)
5. sin (180 + θ)
2
Examples: True/False
Determine the validity of the following statements when θ = 125
7. cos (-θ) = cos θ
8. sin (-θ) = - sin θ
9. tan (-θ) = -tan θ
True
True
True
6. cos(

 )
2
10. sec (-θ) = -sec θ
False, should be sec θ
Examples: Finding values
11. Given terminal point (-2,11). Find the following
a. sin (-θ)
b. cos (-θ)
c. tan (-θ)
Step 1: Identify quadrant
Step 2: (x,y) are given, determine r
r = x 2  y 2 = (2) 2 (11) 2 = 5√5
Quadrant II
Step 3: Find the desired trig function
 11
2
sin(-θ) = -sinθ =
cos (-θ) = cos θ =
5 5
5 5
Practice: Finding values
12. Given terminal point (-8,-1). Find the following
a. sin (-θ)
b. cos (-θ)
9
tan (-θ) = -tanθ =
c. tan (-θ)
11
2
Trigonometry
Inverse Trig Functions
Inverses of Trig Functions




Sin func is set of all ordered pairs (x, sinx)
Arcsine func is set of all ordered pairs (sinx, x) with domain –1 ≤ x ≤ 1 & range set of all real #
Inverse of trig function gives angle that has the specified value of given function.
Note: none of inverses of trig functions are functions
Trig Function
y = sin x
y = cos x
y = tan x
Inverse Trig Function
x = sin-1 y or x = arcsin y
x = cos-1 y or x = arcos y
x = tan-1 y or x = arctan y
Example: Finding values of x for trig functions
1. Find all positive values of x for which cos x =
3
.
2
3
3
, then x is angle or real # whose cos =
.
2
2
3
 11
X = arccos
; Therefore, x = 30, 330, 390, 690, etc or ,
,etc
6
2
6
If cos x =
Practice: Finding values of x for trig functions
2. Find all positive values of x for which tan x = 1.
3. Find all positive values of x for which sin x =
3
2
Examples: Evaluate each expression.
1. sin (arcsin 0.4212)
let x = arcsin0.4212
sinx = 0.4212
Therefore, by substitution, sin(arcsin0.4212) = 0.421
5
2. tan (arcsin
)
13
5
5
let A = arcsin
then sin A =
. Since A > 0, A in either Quadrant I or II.
13
13
sin A
Since tan A =
, find cos A: x2 + y2 = r2; x = 12
cos A
12
Therefore, cos A =
.
13
5
5
sin A 13
5
tan (arcsin
) = tan A =
=
=
13
sin B 12 12
13
Practice: Evaluate each expression
5
3. tan (arctan 2.34)
4. cot(arctan )
5. sec(arccos 1)
8
10
Trigonometry
Principal Values of Inverse Trig Functions
Principal Values








Values in the domain of restricted domain trig functions.
Trig functions with restricted domains depicted with capital letters (Ex: Sin, Cos instead of sin, cos)
Y = Sin x iff y = sin x and –90 ≤ x ≤ 90
Y = Cos x iff y = cos x and 0 ≤ x ≤ 180
Y = Tan x iff y = tan x and –90 < x < 90
If Y = Sin x, inverse function is y = Sin-1 x or y = Arcsinx
If Y = Cos x, inverse function is y = Cos-1 x or y = Arccosx
If Y = Tan x, inverse function is y = Tan-1 x or y = Arctanx
Examples: Finding values of x for trig functions
Find each value.
1. Arccos ½
Let  = Arccos ½
Cos  = ½
(def of Arccos function)
 = 60
Therefore, Arccos ½ = 60

)
4

Let x = tan
x=1
2. Sin-1 (tan
4
Sin-1 (tan


) = Sin-1 (1) =
4
2
3. sin (Sin-1 1 – Cos-1 ½)
Let  = Sin-1 1 and  = Cos-1 ½
Sin  = 1
 = 90
Cos  = ½
 = 60
sin (Sin-1 1 – Cos-1 ½) = sin ( - ) = sin (90 - 60) = sin 30 = ½
Practice: Finding values of x for trig functions
5. Tan-1 (sin
4. Arcsin -½

)
2
6. cos (Arctan 3 – Arcsin
Practice: Finding inverse trig functions using calculators
1. Find radian measure of angle  in Quadrant I with tangent of 1.3284
Arctan 1.3284 = 
Make sure calculator set to radian mode
1.3284 2nd TAN enter
Answer is 0.926
5
5
2. Find sin (Arcsin
– Arccot )
12
3
5 12 = 2nd SIN – (5  3) 1/x 2nd TAN = SIN
Answer is –0.1104185
3. Find radian measure of angle  in Quadrant I with sin of 0.4867
4. Find tan (Arccos
5
– Arcsin 0.3)
8
11
3
)
2
Trigonometry
Solving Right Triangles
Trig Functions in Right Triangle




SOH-CAH-TOA
a
opposite
Sin A =
=
hypotenuse c
b
adjacent
Cos A =
=
hypotenuse c
opposite a
Tan A =
=
adjacent b
Example: Find trig functions of right triangle
Find values of 6 trig functions of α.
5
12
5cm
13cm sin α =
cos α =
13
13
α
12cm
B
a
c
C
tan α =
b
5
12
csc α =
A
13
5
sec α =
13
12
cot α =
12
5
Practice: Find trig functions of right triangle
Given right triangle with sides of length 8,15, 17. Find values of 6 trig functions of α.
8cm
17cm
α
15cm
Finding value of x
When given a triangle and asked to find the value of x
 Identify the hypotenuse, leg opposite and leg adjacent to the given angle.
 Determine which trig ratio to use
 Set up ratio and solve for x.
Example: Finding value of x
Find the value of x to the nearest tenth.
Step 1: Identify the givens
73
42
73
Hypotenuse: 42
x
Leg opposite 73 = x
Step 3: Set up ratio and solve for x
x
sin 73 =
42
Step 2: Determine which trig ratio to use
Given hypotenuse & op leg; use sin
42 * sin 73 = x
42 * 0.956 = x
Practice: Finding value of x
Find the value of x to the nearest tenth.
x
x
17
13
32
36
12
40.2 = x
Trigonometry
Solve right triangle
 To solve a triangle means to find all of measures of its sides and angles. Usually given 2 measures.
A
b
c
C
a
B
Examples: Solve right triangle
1. Solve right ∆ABC. Round angle measures to nearest degree & side measure to nearest tenth. A =49º,
a=7
Step 1: Find B
49º + B = 90º (A & B are complementary)
B = 41º
Step 2: Find c
Step 3: Find b
7
c
7
0.7547 =
c
c = 9.3
7
b
7
1.1504 =
b
b = 6.1
Sin 49º =
tan 49º =
Practice: Solve right triangle
2. Solve right ∆ABC. Round angle measures to nearest degree & side measure to nearest tenth. A =52º,
a = 12
3.
Solve right ∆ABC. Round angle measures to nearest degree & side measure to nearest tenth. B =49º,
a=9
Finding measure of angle

When given 2 sides of triangle,  can be determined based on ratio of the 2 sides. This is easily done
with calculator.
Example: KC3: Finding measure of angle
1. In ∆ABC, c=14 and b =8, find the measure of B to nearest degree.
Step 1: Determine the ratio provided.
Step 2: Write ratio
Step 3: Use calculator to solve for B.
2nd sin(8 ÷ 14) enter
8
8 is opposite B
sin B =
14 is hypotenuse
14
Use sin
Practice: Finding measure of angle
2. In ∆ABC, c=12 and a =10, find the measure of B to nearest degree.
3.
In ∆ABC, b=10.4 and a =9, find the measure of B to nearest degree.
13
Trigonometry
Solving Triangles
Law of Sines



Sides of a triangle are proportional to sines of opposite s.
a
b
c
=
=
SinA SinB SinC
a SinA b SinB c SinC

, 
, 
b SinB c SinC a SinA
Law of Cosines




Square of any side is equal to sum of squares of other 2 sides minus twice product of these sides and
cosine of their included angle.
a2 = b2 + c2 – 2bc(cosA)
b2 = a2 + c2 – 2ac(cosB)
c2 = a2 + b2 – 2ab(cosC)
Types of Triangles
 If c is the length of the longest side of a triangle then
o If a2 + b2 > c2, triangle is acute
o If a2 + b2 = c2, triangle is right
o If a2 + b2 < c2, triangle is obtuse
Oblique Triangles
Case 1: given 1 side and 2 angles (use law of sines)
Case 2: given 2 sides &  opposite 1 of them (use law of sines)
Case 3: given 2 sides & included angle. (SAS) (use law of cos)
Case 4: given 3 sides. (SSS) (use law of cos)
Number of Solutions



2 Solutions: if A is acute and value of a lies between b and b sinA
No Solution: if A is acute and a < b sinA or if A is obtuse & a < b or a = b
1 Solution: in all other cases
Case 1 Example: 1 side, 2 angles
Given B=30º, C=105º and a = 7.07 find the remaining parts.
Step 1: Find A, using 180 – (B+C) = 180 – (30+105) = 180 – 135 = 45º.
b SinB
c SinC
Step 2: Find b using law of sines: 
Step 3: Find c using law of sines: 
a SinA
a SinA
bSinB
aSinC
b=
c=
SinA
SinA
7.07(sin 30)
7.07(sin( 180  105)) 7.07(sin 75)
b=
c=
=
sin 45
sin 45
sin 45
7.07(0.5)
7.07(0.966)
b=
c=
.707
.707
b=5
c= 9.66
14
Trigonometry
Case 1 Practice
Given A=65º, B=40º and a = 50, find C, b, and c.
Case 2 Example: 2 sides, angle opposite 1 of them
Given a = 40, b = 30 and A=75º. Find the remaining parts.
Step 1: Since a > b & A is acute, only 1 solution.
Step 3: Find C. C = 180 – (A + B) = 180 - 121º 25’ =
58º 35’
a
b
=
SinA SinB
bSinA 30(0.9659)
Sin B=
=
= 0.7244
a
40
B = 46º 25’
Step 2: Find B, using law of sines:
c SinC

a SinA
aSinC 40(0.8535)
c=
=
= 35.3
sin A
.9659
Step 4: Find c using law of sines:
Case 2 Practice: 2 sides, angle opposite 1 of them
Given a = 40, b = 30 and A=75º. Find the remaining parts.
Case 3 Example: 2 sides, included angle
Given a = 132, b = 224 and C=28º40’. Find the remaining parts.
Step 1: Find c, using law of cosines: c2 = a2 + b2 – 2ab(cosC)
= (132)2 + (224)2 – 2(132)(224)(cos28º40’)
= (132)2 + (224)2 – 2(132)(224)(0.8774)
= 15,714
c = 125
aSinC 132(sin 2840' ) 132(0.4797)
Step 2: Find A using law of sines: sinA =
=
=
c
125
125
= 0.5066
so A = 30º30’
bSinC 224(sin 2840' ) 224(0.4797 )
Step 3: Find B using law of sines: sinB =
=
=
c
125
125
= 0.8596
so B = 120º40’
Case 3 Practice: 2 sides, included angle
Given a = 30, b = 54 and C=46º. Find the remaining parts.
15
Trigonometry
Case 4 Example: 3 sides
Given a = 30.3, b = 40.4 and c=62.6. Find the remaining parts.
b 2 c 2  a 2
2bc
40.4 2 62.6 2  30.32
=
2(40.4)(62.6)
= 0.9519
Step 1: Find A, using law of cosines: cos A =
A = 23º40’
a 2 c 2  b 2
2ac
30.32 62.6 2  40.4 2
=
2(30.3)(62.6)
= 0.8448
Step 2: Find B, using law of cosines: cos B =
B = 32º20’
a 2 b 2  c 2
2ab
30.32 40.4 2  62.6 2
=
2(30.3)( 40.4)
= -0.5590
Step 3: Find C, using law of cosines: cos C =
C = 124º
Step 4: Check A + B + C = 180º
23º40’ + 32º20’ + 124º = 180º
Case 4 Practice: 3 sides
Given a = 24.5, b = 18.6 and c=26.4. Find the remaining parts.
More Practice: Law of Sine/Cosine
1. a = 125, A = 54º40’, B=65º10’
2.
b = 321, A = 75º20’, C= 38º30’
3.
b = 215, c = 150, B = 42º40’
4.
a = 512, b = 426, A = 48º50’
5.
b = 120, c = 270, A = 118º40’
6.
a = 6.34, b = 7.30, c = 9.98
16