Unlocking the Secrets of Trigonometric Ratios of Acute Angles

Definition	Substitute
$sin θ = \frac{Length of opposite side to ∠ θ}{Hypotenuse}$	$sin θ = \frac{8}{1 7}$
$cos θ = \frac{Length of adjacent side to ∠ θ}{Hypotenuse}$	$cos θ = \frac{1 5}{1 7}$
$tan θ = \frac{Length of opposite side to ∠ θ}{Length of adjacent side to ∠ θ}$	$tan θ = \frac{8}{1 5}$

Definition

Substitute

sin θ = \frac{Length of opposite side to ∠ θ}{Hypotenuse}

sin θ = \frac{8}{1 7}

cos θ = \frac{Length of adjacent side to ∠ θ}{Hypotenuse}

cos θ = \frac{1 5}{1 7}

tan θ = \frac{Length of opposite side to ∠ θ}{Length of adjacent side to ∠ θ}

tan θ = \frac{8}{1 5}

a In the given diagram, it can be seen that the lengths of the opposite and adjacent sides to

∠ θ

are

35

and

12,

respectively. The trigonometric ratio that relates these two sides is the tangent ratio.

tan θ = \frac{length of opposite side to ∠ θ}{length of adjacent side to ∠ θ} ⇓ tan θ = \frac{3 5}{1 2}

To solve this equation, the inverse of the tangent function could be used.

tan θ = \frac{3 5}{1 2} ⇕ m ∠ θ = tan^{- 1} \frac{3 5}{1 2}

To find the value of

tan^{- 1} \frac{3 5}{1 2},

a calculator should to be used. First, the calculator must be set in degree mode. This is done by pushing

M O D E

and selecting Degree in the third row.

Next the value of $tan^{- 1} \frac{3 5}{1 2},$ can be calculated by pushing $2 N D,$ followed by $T A N,$ and $35 / 12 .$

Thereofre,

m ∠ θ \approx 7 1^{\circ} .

b In the given diagram, it is shown that the length of the adjacent side to

∠ θ

7

and that the hypotenuse of the right triangle is

25 .

The trigonometric ratio that relates these two sides is the cosine ratio.

cos θ = \frac{length of adjacent side to ∠ θ}{hypotenuse} ⇓ cos θ = \frac{7}{2 5}

To solve this equation, the inverse of the cosine function is needed.

cos θ = \frac{7}{2 5} ⇕ m ∠ θ = cos^{- 1} \frac{7}{2 5}

To find the value of

cos^{- 1} \frac{7}{2 5},

a calculator should be used. Just like before, the calculator must be set in degree mode. This is done by pushing

M O D E

and selecting Degree in the third row.

Next the value of $cos^{- 1} \frac{7}{2 5},$ is calculated by pushing $2 N D,$ followed by $C O S,$ and $7 / 25 .$

It was found that

m ∠ θ \approx 7 4^{\circ} .

c In the diagram, it can be seen that the hypotenuse of the right triangle is

29

and that the length of the opposite side to

∠ θ

20 .

The trigonometric ratio that relates these two sides is the sine ratio.

sin θ = \frac{length of opposite side to ∠ θ}{hypotenuse} ⇓ sin θ = \frac{2 0}{2 9}

To solve this equation, the inverse of the sine function can be used.

sin θ = \frac{2 0}{2 9} ⇕ m ∠ θ = sin^{- 1} \frac{2 0}{2 9}

To find the value of

sin^{- 1} \frac{2 0}{2 9},

a calculator should be used. Just like in Parts A and B, the calculator must be set in degree mode by pushing

M O D E

and selecting Degree in the third row.

Next the value of $sin^{- 1} \frac{2 0}{2 9},$ can be calculated by pushing $2 N D,$ followed by $S I N,$ and $20 / 29 .$

Thereofre,

m ∠ θ \approx 4 4^{\circ} .

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Challenge

Investigating a Side of a Right Triangle

Explore

Comparing Ratios of Sides in Similar Right Triangles

Discussion

Trigonometric Ratios

Example

Explaining Trigonometric Ratios in Right Triangles

Hint

Solution

Pop Quiz

Practice Finding Side Lengths Using Trigonometric Ratios

Discussion

Pythagorean Identities

Example

Using Trigonometry to Determine the Cosine of an Angle

Hint

Solution

Example

Calculating Angles of Right Triangles

Hint

Solution

Pop Quiz

Practice Finding Angles Using Trigonometric Ratios

Discussion

Reciprocal Trigonometric Ratios

Explore

Finding Reciprocal Identities

Hint

Solution

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