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WARM-UP 1. What does SOH CAH TOA stand for? Sin θ = Cos θ = Tan θ = 2. Find the sine cosine and tangent of the acute angles. Round the decimal to four places. Challenge Question! Can the sine, cosine or tangent of a triangle be greater than 1? Why or Why not? The sine and the cosine cannot because this would mean the opposite or adjacent side is greater than the hypotenuse. The tangent can because one leg can be bigger than another leg.

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Warm-UP. What does SOH CAH TOA stand for? Sin θ = Cos θ = Tan θ = 2. Find the sine cosine and tangent of the acute angles. Round the decimal to four places. Challenge Question! Can the sine, cosine or tangent of a triangle be greater than 1? Why or Why not? - PowerPoint PPT Presentation

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Page 1: Warm-UP

WARM-UP1. What does SOH CAH TOA stand for?Sin θ = Cos θ = Tan θ = 2. Find the sine cosine and tangent of the acute

angles. Round the decimal to four places.

Challenge Question!Can the sine, cosine or tangent of a triangle be greater than 1? Why or Why not?

The sine and the cosine cannot because this would mean the opposite or adjacent side is greater than the hypotenuse. The tangent can because one leg can be bigger than another leg.

Page 2: Warm-UP

MORE TRIGONOMETRY

Page 3: Warm-UP

LEARNING OUTCOMESI will be able to use trigonometric ratios

to find missing side lengths of a right triangle

I will be able to use trigonometric ratios to find missing angle measures of a right triangle

Page 4: Warm-UP

USING TRIGONOMETRY RATIOSIn your homework you will be

given a problems like this.

Is there enough information to use the Pythagorean theorem to solve this problem?

28°

x

13

y

Page 5: Warm-UP

USING TRIGONOMETRIC RATIOSSince there’s not enough

information to solve this with the Pythagorean Theorem. How might our trig. ratios help us?

sin θ=

cos θ =

tan θ =

28°

x

13

y

Page 6: Warm-UP

USING TRIGONOMETRIC RATIOSLet’s solve this problem.

Step 1.) Label the triangle

Step 2.) Fill In your trig ratios

sin 28° = cos 28° = tan 28° =

Notice that the tangent has two unknowns. In this example the tangent does not help us.

28°

x

13

y

Opp.

Hyp.

Adj.

Page 7: Warm-UP

USING TRIGONOMETRIC RATIOSNow solve for x

sin 28° = x =13· sin 28 ≈ 6.1031

cos 28° = y = 13· cos 28 ≈ 11.4783

28°

x

13

y(13)(13)

(13) (13)

Do our answers make sense?

Page 8: Warm-UP

USING TRIGONOMETRIC RATIOSLet’s solve this problem.

Step 1.) Label the triangle

Step 2.) Fill In your trig ratios

sin 28° = cos 28° = tan 28° =

Notice that the cosine has two unknowns. In this example the cosine does not help us.

28°

13

x

y

Opp.

Hyp.

Adj.

Page 9: Warm-UP

USING TRIGONOMETRIC RATIOSNow solve for x

sin 28° = x · sin 28° = 13x = ≈ 27.6907tan 28° = y· tan 28° = 13y = ≈ 24.4494

28°

13

x

y(x)(x)

(y) (y)

Do our answers make sense?

Page 10: Warm-UP

LET’S PRACTICE!Solve for x.

x ≈ 11.8202 x ≈ 11.1809 x ≈ 17.9683

opp.

adj.

hyp.

SOH CAH TOA

Page 11: Warm-UP

HOW CAN I FIND ANGLES USING TRIG?• How can I use trig to find

the measure of angle A and angle C?

• Let’s set up a trig ratios for A using sine.

• sin A = • Now to solve for A we have

to use the inverse of sin.

3 5

4

A

B Csin‾¹ (sin A) =sin (‾¹ )A = sin (‾¹ )A ≈ 53.1301 °

Solve for A using the other two trig functions. Do you get the same answer?

Page 12: Warm-UP

HOW CAN I FIND ANGLES USING TRIG?• How can I use trig to find

the measure of angle A and angle C?

• Try to find the angle measure for C.

• sin C =

• cos C =

• tan C =

3 5

4

A

B C

C = sin (‾¹ )C ≈ 36.8699°C = cos (‾¹ )C ≈ 36.8699°C = tan (‾¹ )C ≈ 36.8699°

Page 13: Warm-UP

LET’S PRACTICE!Find the missing angles and side of the

right triangle.

m∠A ≈ 48.2 m∠B ≈ 41.8

BC = 4√5m∠R ≈ 54.0 m∠T ≈ 36.0

RT = √185

Page 14: Warm-UP

Solve for x and y.

Pg. 563: 28-36, 39,40 and Pg. 570: 14-33

EXIT TICKETHOMEWORK