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UNIT 3: RIGHT TRIANGLE TRIGONOMETRY

Vocabulary Builder

Similar Triangles Square RootHypotenuse

legSimplest radical

form Variable

Pythagorean Theorem

45-45-90 Triangles 30-60-90 Triangles

UNIT 3: RIGHT TRIANGLE TRIGONOMETRY3.1 Trigonometric Ratios

Square rootPerfect Square

Radical expressionradicand

Perfect Square

5 8 14

30 18 23

𝟔 𝟐 2 𝟔 3 𝟏𝟏

10 𝟑 3 𝟏𝟎 𝟓 𝟑

right triangle

equals

hypotenuse

Side lengths

42 + 82 = 𝑥2

16 + 64 = 𝑥2

80 = 𝑥2

4 5 = 𝑥

8.9 = 𝑥

62 + 32 = 𝑥2

36 + 9 = 𝑥2

45 = 𝑥2

3 5 = 𝑥

6.7 = 𝑥

𝑥2 + 3.92 = 112

𝑥2 + 15.21 = 121

𝑥2 = 95.79

𝑥 = 9.75

32 + 102 = 𝑥2

9 + 100 = 𝑥2

109 = 𝑥2

10.44 = 𝑥

72 + 22 = 𝑥2

49 + 4 = 𝑥2

53 = 𝑥2

7.3 = 𝑥

𝑥2 + 3.62 = 7.42

𝑥2 + 12.96 = 54.76

𝑥2 = 41.8

𝑥 = 6.47

hypotenuse

adjacent

opposite

𝑌𝑍

𝑋𝑌

𝑋𝑍

𝐻𝑈

𝑈𝐼

𝐻𝐼

trigonometric ratio

SOH CAH TOA

lengthright acute

SOH CAH TOA

𝑂𝑃𝑃

𝐴𝐷𝐽=

20

15=

4

3

𝑂𝑃𝑃

𝐴𝐷𝐽=

16

30=

8

15

𝑂𝑃𝑃

𝐴𝐷𝐽=

14

48=

7

24

𝑂𝑃𝑃

𝐻𝑌𝑃=

15

25=

3

5

𝑂𝑃𝑃

𝐻𝑌𝑃=

30

34=

15

17

𝐴𝐷𝐽

𝐻𝑌𝑃=

48

50=

24

25

=11

29=

11

29=

25

29

=25

11=

18

7= 1


3.2 Finding Sides and Angles

of Right Triangles

function

inputangle

output

x

sin 19° =𝑥

11

11 ∙ sin 19° = 𝑥

3.58 = 𝑥

x

cos 46° =19.2

𝑥

𝑥 =19.2

cos 46°

𝑥 = 27.64

sin 𝑅 =𝑂𝑃𝑃

𝐻𝑌𝑃 cos 𝐶 =𝐴𝐷𝐽

𝐻𝑌𝑃

x

sin 34° =11

𝑥

𝑥 =11

sin 34°

𝑥 = 19.67

sin 𝐻 =𝑂𝑃𝑃

𝐻𝑌𝑃

14.63 = 𝑥

x

𝑥 = 15.49

tan 26° =𝑂𝑃𝑃

𝐴𝐷𝐽tan 50° =

𝐴𝐷𝐽

𝐻𝑌𝑃

𝑥 = 6.92

tan 20° =𝑂𝑃𝑃

𝐴𝐷𝐽

sin-1 cos-1 tan-1

tan−1 𝑥 = 1

tan−1 1 = 𝑥

45° = 𝑥

tan−1 𝑥 =5

5

tan−15

12= 𝑥

22.6° = 𝑥

tan−1 𝑥 =5

12

cos−120

21= 𝑥

17.8° = 𝑥

cos−1 𝑥 =40

42

cos−115

28= 𝑥

57.6° = 𝑥

sin−119

23= 𝑥

55.7° = 𝑥cos−1

35

43= 𝑥

35.5° = 𝑥


3.3 Sine and Cosine of

Complementary Angles

•G.SRT.7 Explain and use

the relationship between the sine and cosine of complementary angles

What am I learning today?

•Find the complement of sine and cosine

What am I going to do today?

•5 Question Daily Assessment

•Daily Ticket out the Door

How will I show you that I’ve learned it?

180˚

90˚

complementary

Opposite angle

cosine









= 𝐜𝐨𝐬 𝟐𝟔° = 𝐬𝐢𝐧 𝟔° = 𝐬𝐢𝐧 𝟓𝟐°

= 𝐜𝐨𝐬 𝟔𝟔° = 𝐬𝐢𝐧 𝟏𝟖° = 𝐜𝐨𝐬 𝟒𝟓°

= 𝐜𝐨𝐬 𝟔𝟖° = 𝐬𝐢𝐧 𝟐𝟓° = 𝐬𝐢𝐧 𝟒𝟔°

= 𝐜𝐨𝐬 𝟓𝟖° = 𝐬𝐢𝐧 𝟔𝟓° = 𝐜𝐨𝐬 𝟕𝟓°









𝟒

𝟓

𝟒

𝟓

𝟑

𝟓

𝟑

𝟓

Sin A and Cos B are the same which means The complement of sine and cosine are equal.









𝟎. 𝟓

𝟎. 𝟓𝟎. 𝟖𝟔𝟔

𝟎. 𝟖𝟔𝟔

Then mQ = 60˚

The complement of sine and cosine are equal.









102 + 242 = 𝑥2

100 + 576 = 𝑥2

676 = 𝑥2

26 = 𝑥

26

𝟐𝟒

𝟐𝟔

𝟏𝟎

𝟐𝟔=

𝟓

𝟏𝟑

The complement of sine and cosine are equal.

=𝟏𝟐

𝟏𝟑

𝟐𝟒

𝟐𝟔=

𝟏𝟐

𝟏𝟑

𝟏𝟎

𝟐𝟔=

𝟓

𝟏𝟑 𝟎. 𝟕𝟎𝟕 𝟎. 𝟕𝟎𝟕

𝟎. 𝟕𝟎𝟕 𝟎. 𝟕𝟎𝟕

Then mT = 45˚









𝐜𝐨𝐬 𝟓𝟐° ≈ 𝟎. 𝟔𝟏𝟔

𝐬𝐢𝐧 𝟑𝟎° ≈ 𝟎. 𝟓









5

C B

A

12

35

H T

A

12

52 + 122 = 𝑥2

25 + 144 = 𝑥2

169 = 𝑥2

13 = 𝑥

13

13

𝟓

𝟏𝟑𝟓

𝟏𝟑

122 + 352 = 𝑥2

144 + 1225 = 𝑥2

1369 = 𝑥2

37 = 𝑥

37

37𝟏𝟐

𝟑𝟕𝟏𝟐

𝟑𝟕









8

Y Z

X

6

82 + 62 = 𝑥2

64 + 36 = 𝑥2

100 = 𝑥2

10 = 𝑥

10

10

𝟒

𝟓𝟒

𝟓

3

I M

X

4

32 + 𝑥2 = 52

9 + 𝑥2 = 25

𝑥2 = 16

𝑥 = 4

10

5

𝟒

𝟓𝟒

𝟓

UNIT 3: RIGHT TRIANGLE TRIGONOMETRY3.4 Solve Right Triangles

•G.SRT.8 Use trigonometric

ratios and the Pythagorean Theorem to solve triangles in applied problems


•Use the Pythagorean Theorem to Solve Right Triangles

•Use Trigonometric Ratios to Solve Right Triangles


• 5 Question Daily Assessment



solve

Side lengths

angle

The length of two sides

The length of 1 side and 1 acute angle

The length of 1 side and 1 trigonometric ratio










3.12 + 8.92 = 𝑥2

9.61 + 79.21 = 𝑥2

88.82 = 𝑥2

9.4 = 𝑥9.4

19.2˚

70.8˚

9.4tan−1

3.1

8.9= ∠𝑃

19.2° = ∠𝑃

R = 90 – 19.2 = 70.8˚










𝑥2 + 3.92 = 7.42

𝑥2 + 15.21 = 54.76𝑥2 = 39.55

𝑥 = 6.3

3.9

6.3˚

58˚

3.9

C = 90 – 32 = 58˚

sin 32° =𝐵𝐶

7.4

3.58 = 𝑥

6.3










Angle of Elevation

Angle of Depression










618.75 ft2










991.86 ft2

•G.SRT.8 Use trigonometric ratios and the Pythagorean Theorem to solve triangles in applied problems








unit 3: right triangle trigonometry - wordpress.com · unit 3: right triangle trigonometry 3.2...

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