cos ⁡ θ = \cos\theta= cos θ =

a d j a c e n t h y p o t e n u s e \frac{adjacent}{hypotenuse} h y p o t e n u s e a d j a c e n t

o p p o s i t e h y p o t e n u s e \frac{opposite}{hypotenuse} h y p o t e n u s e o p p o s i t e

h y p o t e n u s e a d j a c e n t \frac{hypotenuse}{adjacent} a d j a c e n t h y p o t e n u s e

o p p o s i t e a d j a c e n t \frac{opposite}{adjacent} a d j a c e n t o p p o s i t e

tan ⁡ θ = \tan\theta= tan θ =

o p p o s i t e a d j a c e n t \frac{opposite}{adjacent} a d j a c e n t o p p o s i t e

o p p o s i t e h y p o t e n u s e \frac{opposite}{hypotenuse} h y p o t e n u s e o p p o s i t e

h y p o t e n u s e a d j a c e n t \frac{hypotenuse}{adjacent} a d j a c e n t h y p o t e n u s e

a d j a c e n t o p p o s i t e \frac{adjacent}{opposite} o p p o s i t e a d j a c e n t

sin ⁡ θ = \sin\theta= sin θ =

o p p o s i t e h y p o t e n u s e \frac{opposite}{hypotenuse} h y p o t e n u s e o p p o s i t e

h y p o t e n u s e o p p o s i t e \frac{hypotenuse}{opposite} o p p o s i t e h y p o t e n u s e

a d j a c e n t h y p o t e n u s e \frac{adjacent}{hypotenuse} h y p o t e n u s e a d j a c e n t

o p p o s i t e a d j a c e n t \frac{opposite}{adjacent} a d j a c e n t o p p o s i t e

What is the correct equation to solve for x?

sin ⁡ ( 45 ) = 13 x \sin(45)=\frac{13}{x} sin ( 4 5 ) = x 1 3

tan ⁡ ( 45 ) = x 13 \tan(45)=\frac{x}{13} tan ( 4 5 ) = 1 3 x

cos ⁡ ( 45 ) = x 13 \cos(45)=\frac{x}{13} cos ( 4 5 ) = 1 3 x

sin ⁡ ( 45 ) = x 13 \sin(45)=\frac{x}{13} sin ( 4 5 ) = 1 3 x

What is the correct equation to solve for x ?

tan ⁡ ( 46 ) = 17 x \tan(46)=\frac{17}{x} tan ( 4 6 ) = x 1 7

sin ⁡ ( 46 ) = x 17 \sin(46)=\frac{x}{17} sin ( 4 6 ) = 1 7 x

tan ⁡ ( 46 ) = x 17 \tan(46)=\frac{x}{17} tan ( 4 6 ) = 1 7 x

cos ⁡ ( 46 ) = x 17 \cos(46)=\frac{x}{17} cos ( 4 6 ) = 1 7 x

Trigonometry Basics

9th Grade

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Trigonometry Basics

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Mathematics

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9th Grade

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Practice Problem

•

Medium

•

CCSS

HSG.SRT.C.6, HSF.TF.A.2, HSG.SRT.C.7

Standards-aligned

Simon Garvey

Used 46+ times

FREE Resource

16 questions

Show all answers

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\cos\theta=$

$\frac{opposite}{hypotenuse}$

$\frac{hypotenuse}{adjacent}$

$\frac{adjacent}{hypotenuse}$

$\frac{opposite}{adjacent}$

Tags

CCSS.HSG.SRT.C.6

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\tan\theta=$

$\frac{opposite}{hypotenuse}$

$\frac{hypotenuse}{adjacent}$

$\frac{adjacent}{opposite}$

$\frac{opposite}{adjacent}$

Tags

CCSS.HSG.SRT.C.6

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\sin\theta=$

$\frac{opposite}{hypotenuse}$

$\frac{hypotenuse}{opposite}$

$\frac{adjacent}{hypotenuse}$

$\frac{opposite}{adjacent}$

Tags

CCSS.HSG.SRT.C.6

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

What is tan 45⁰?

0.5

0.75

Tags

CCSS.HSF.TF.A.2

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

What is cos 47⁰ ?

0.73

0.28

0.68

answer not shown

Tags

CCSS.HSG.SRT.C.7

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the length of the hypotenuse?

Tags

CCSS.8.G.B.7

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the length of the side opposite ∠X?

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

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Trigonometry Basics

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$\cos\theta=$

$\tan\theta=$

$\sin\theta=$

What is tan 45⁰?

What is cos 47⁰ ?

What is the length of the hypotenuse?

What is the length of the side opposite ∠X?

What is the length of the side adjacent to ∠Z?

Which trigonometric ratio should you use?

Which trigonometric ratio should you use?

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Trigonometry Basics

16 questions

cos⁡θ=\cos\theta=cosθ=

tan⁡θ=\tan\theta=tanθ=

sin⁡θ=\sin\theta=sinθ=

What is tan 45⁰?

What is cos 47⁰ ?

What is the length of the hypotenuse?

What is the length of the side opposite ∠X?

What is the length of the side adjacent to ∠Z?

Which trigonometric ratio should you use?

Which trigonometric ratio should you use?

Access all questions and much more by creating a free account

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$\cos\theta=$

$\tan\theta=$

$\sin\theta=$