14right angle trigonometry

7/28/2019 14Right Angle Trigonometry

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I. Basic Facts and Definitions1. Right angle – angle measuring °90

2. Straight angle – angle measuring °180

3. Acute angle – angle measuring between °0 and °90

4. Complementary angles – two angles whose sum is °90

5. Supplementary angles – two angles whose sum is °180

6. Right triangle – triangle with a right angle

7. Isosceles triangle – a triangle with exactly two sides equal

8. Equilateral triangle – a triangle with all three sides equal

9. The sum of the angles of a triangle is °180 .

10. In general, capital letters refer to angles while small letters refer

to the sides of a triangle. For example, side a is opposite

angle A .

II. Right Triangle Facts and Examples

1. Hypotenuse – the side opposite the right angle, sidec .

2. Pythagorean Theorem - 222 cba =+

3. A∠ and B∠ are complementary.

Right AngleTrigonometry

B

A C

a

b

c




III. Examples:1. In a right triangle, the hypotenuse is 10 inches and one side is 8

inches. What is the length of the other side?

Solution:222

cba =+ 222 108 =+b

10064 2 =+b

362 =b

6=b

2. In a right triangle ABC Δ , if °=∠ 23 A , what is the measure of B∠ ?

Solution: The two acute angles in a right triangle arecomplementary.

°=∠+∠ 90 B A

°=∠+° 9023 B

°=∠ 67 B

IV. Similar Triangles:a. Two triangles are similar if the angles of one triangle are equal to

the corresponding angles of the other. In similar triangles, ratios of corresponding sides are equal.

B

A C

8=a

?=b

10=c

C

B

A

Conditions for Similar Triangles( EGF ABC ΔΔ ~ )

1. Corresponding angles in similar triangles areequal:

E A ∠=∠ F B ∠=∠ GC ∠=∠

2. Ratios of corresponding sides are equal:

FG

BC

EF

AB

EG

AC ==

F

G E




Example 1:

Example 2:

B C

A

F E

22

50

50

50= AE meters 22= EF m and 100= AB m

Find the len th of side BC .

Notice that ABC Δ and AEF Δ are similarsince corresponding angles are equal. (There

is a right angle at both F and C , A∠ is thesame in both triangles and B∠ equals theacute angle at E ∠ .)

Solution: BC

EF

AB

AE = so

BC

22

100

50=

By cross multiplying we get:

)100(22)(50 = BC

Therefore 44= BC meters.

50= AE meters 22= EF m and 100= AB m

Find the length of side BC .

a

a a2

45

45

a

a2 a3

60

30

All °−°−° 904545 triangles are similar to oneanother. Two sides are of equal length and the

hypotenuse is 2 times the length of each of the

equal sides.

All °−°−° 906030 triangles are similar to oneanother. The shortest side of length a is opposite

the smallest angle ( °30 ). The hypotenuse is twicethe length of the shortest side. The side opposite

the °60 has a length 3 times the shorter leg.




Problem: Find the lengths of the legs of a °−°−° 906030 triangle if thehypotenuse is 8 meters.

Solution: 1) If 82 =a , then 4=a meters and 2) 34)4(33 ==a meters.

V. The Six Trigonometric Ratios for Acute Angles

TRIG TRICK: A good way to remember the trig ratios is to use the mnemonicSOH CAH TOA!

sine = Ac

a

hypotenuse

opposite A ==sin cosecant = A

a

c

opposite

hypotenuse

A A ===

sin

1csc

cosine = Ac

b

hypotenuse

adjacent A ==cos secant = A

b

c

adjacent

hypotenuse

A A ===

cos

1sec

tangent = Ab

a

adjacent

opposite A ==tan cotangent = A

a

b

opposite

adjacent

A A ===

tan

1cot

c

a

bA

B

C

SOH CAH TOA ine

ppos

ite

y pot

enuse

osin

e

d jac

ent

y pot

enuse

ange

nt

ppos

ite

d jac

ent




Example 1:Find the six trigonometric ratios for the acute angle B .

Solution:

c

b

hypotenuse

opposite B ==sin . Using the above definitions, the rest are:

c

a B =cos ,

a

b B =tan ,

b

c B =csc ,

a

c B =sec ,

b

a B =cot

Example 2:In the right ABC Δ , 1=a and 3=b . Determine the six trigonometric ratios for B∠ .

Solution:Use Pythagorean Theorem:

222

bac += 222 31 +=c 22 10=c

10±=c

(Since length is positive, we will only use 10=c .)

c 3=b

1=a

A

C B

10

103

10

3sin ===

hyp

opp B

10

10

10

1cos ===

hyp

adj B 3

1

3tan ===

adj

opp B

310csc ==

opphyp B 10

110sec ===

adjhyp B

31cot ==

oppadj B




VI. Special Casesa. Trigonometric values of °30 and °60 (Use the °−°−° 906030

triangle from pg. 3.)

b. Trigonometric values of °45 (Use the °−°−° 904545 triangle frompg. 3.)

1=b

2=c 3=a

60

30 2130sin =°

2360sin =°

2

330cos =°

2

160cos =°

3

3

3

130tan ==° 3

1

360tan ==°

21

230csc ==°

3

32

3

260csc ==°

332

3230sec ==° 2

1260sec ==°

31

330cot ==°

3

3

3

160cot ==°

1=b

1=a 2=c

45

45

30

60

22

2145sin ==° 2

1245csc ==°

2

2

2

145cos ==° 2

1

245sec ==°

11

145tan ==° 1

1

145cot ==°




VII. Converting Minutes and Seconds to Decimal Form

(Necessary for most calculator use in evaluating trig values)

1. To convert from seconds to a decimal part of a minute, divide the

number of seconds by 60.2. To convert from minutes to a decimal part of a degree, divide thenumber of minutes by 60.

Example 1: Convert 7464 ′° to degrees using decimals.

Solution:

°

⎟ ⎠

⎞⎜⎝

⎛ +°=′°

60

47647464

°=°+°=′° 783.64783.647464

Example 2: Convert 012115 ′′′° to degrees using decimals.

Solution: 012115012115 ′′+′°=′′′° °

⎟ ⎠

⎞⎜⎝

⎛ +′°=′′′°

60

102115012115

716.1215012115 ′°=′′′° °

⎟ ⎠

⎞⎜⎝

⎛ +°=′′′°

60

167.1215012115

°=°+°=′′′° 203.15203.15012115

VIII. Right Triangle Trigonometry Problems

To Solve Right Triangle Problems:There are six parts to any triangle; 3 sides and 3 angles. Each trig formula(ex: sin A = a/c) contains three parts; one acute angle and two sides. If youknow values for two of the three parts then you can solve for the thirdunknown part using the following method:

1. Draw a right triangle. Label the known parts with the given values andindicate the unknown part(s) with letters.

2. To find an unknown part, choose a trig formula which involves theunknown part and the two known parts.




Practice Problems:

1. In right triangle ABC Δ , if 39=c inches and 36=b inches, find a .

2.

Find the length of side AC . Note: This problem and diagram correspondsto finding the height of a street light pole (AC ) if a 6 ft. man ( EF ) casts ashadow ( BF ) of 15 ft. and the pole casts a shadow ( BC ) of 45 ft.

3. Evaluate:

a) sin E = _____________ b) tan E = _____________

c) cos F = _____________ d) sec F = _____________

4. Evaluate: (Draw reference °−°−° 906030 and °−°−° 904545 triangles)

a) sin 30 = _____________ b) tan 60 = _____________

c) sec 60 = _____________ d) tan 45 = _____________

e) csc 45 = _____________ f) cot 30 = _____________

B C

A

E

F

6

15 30

F

G E

13 12

5




5. Evaluate:

a) tan A = _____________ b) csc B = _____________

c) cot A = _____________ d) sec B = _____________

6. Convert to decimal notation using a calculator:a) 6076 ′°

b) 317245 ′′′°

Evaluate, using a calculator:c) 8452sin ′°

d) 2439cot ′°

7. Label the sides and remaining angles of right triangle ABC Δ , using A , B ,a , b and c . If 43=a and °= 37 A , find the values of the remaining parts.

B

C A

10

8

C




8. Given right triangle ABC Δ with 622.0=b and 0451 ′°= A , find c . Draw adiagram.

9. From a cliff 140 feet above the shore line, an observer notes that the angleof depression of a ship is 0321 ′° . Find the distance from the ship to a pointon the shore directly below the observer.

cliff ship




Answers to Right Triangle Trigonometry:

1. 15=a inches (use Pythagorean Theorem)

2.

BF

BC

EF

AC =

15

6

45

= AC

18= AC

3. a)13

12sin = E b)

5

12tan = E c)

13

12cos =F d)

12

13sec =F

4. (see part E of the handout)=

a)2

130sin =° b) 360tan =° c) 260sec =°

d) 145tan =° e) 245csc =° f) 330cot =°

5. 6=b (use Pythagorean Theorem)

a)3

4

6

8tan == A b)3

5csc = B c)4

3cot = A d)4

5sec = B

6. a) °1.76 b) °454.45 c) 7965. d) 2045.1

7. °=∠ 53 B 06.57≈b 45.71≈c

8. 1≈c (Usec

b A =cos or

b

c A =sec to solve for unknown c )

9. x

1400321tan =′°

41.355= x ft.

C

B

A b

a c

cliff ship

°5.21

140

°5.21 x

(Angle of depression)

14right angle trigonometry

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