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Intro to Trigonometric Ratios (SOH CAH TOA)

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Intro to Trigonometric Ratios (SOH CAH TOA) 2

Intro to Trigonometric Ratios (SOH CAH TOA) 2

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Question 1 of 4

1. Question
Find which angle in this triangle, $Q$ $Q$ , $P$ $P$ or $R$ $R$ , has the following trigonometric ratios:
- $(i) cos θ = \frac{21}{29} :$ $(i) cos theta=21/29:$ (R, r)
  
  $(i i) tan θ = \frac{21}{20} :$ $(ii) tan theta=21/20:$ (P, p)
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Trigonometric Ratios (SOHCAHTOA) for Right Angled Triangles

Sin Ratio (SOH)

$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio (CAH)

$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio (TOA)

$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

Label the triangle according to each given trigonometric ratio to find the angles.

$\cos\theta=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}=\frac{\color{#00880a}{21}}{\color{#e85e00}{29}}$

The angle adjacent to $21$ $21$ and has a hypotenuse of $29$ $29$ is $R$ $R$

$\tan\theta=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}=\frac{\color{#004ec4}{21}}{\color{#00880a}{20}}$

The angle opposite of $21$ $21$ and is adjacent to $20$ $20$ is $P$ $P$

$(i) cos θ = \frac{21}{29} : R$ $(i) costheta=21/29:R$

$(i i) tan θ = \frac{21}{20} : P$ $(ii) tantheta=21/20:P$

Question 2 of 4

If

tan θ = \frac{20}{21}

$tan theta=20/21$ , find the following trigonometric ratios.

Write fractions in the format “a/b”

$(i) sin θ =$ $(i) sin theta=$ (20/29)

$(i i) cos θ =$ $(ii) cos theta=$ (21/29)

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Trigonometric Ratios (SOHCAHTOA) for Right Angled Triangles

Sin Ratio (SOH)

sin = \frac{opposite}{hypotenuse}

$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio (CAH)

cos = \frac{adjacent}{hypotenuse}

$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio (TOA)

tan = \frac{opposite}{adjacent}

$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

Pythagoras’ Theorem

c^{2} = a^{2} + b^{2}

$\color{#e85e00}{c}^2=\color{#004ec4}{a}^2+\color{#00880a}{b}^2$

First, draw a random right triangle and use the

tan

$tan$ ratio to label it.

tan = \frac{opposite}{adjacent} = \frac{20}{21}

$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}=\frac{\color{#004ec4}{20}}{\color{#00880a}{21}}$

To find the missing side which is the hypotenuse, use Pythagoras’ Theorem.

a = 20

$a=20$

b = 21

$b=21$

$\color{#e85e00}{c}^2$	$=$ $=$	$\color{#004ec4}{a}^2+\color{#00880a}{b}^2$	Pythagoras’ Theorem
$\color{#e85e00}{c}^2$	$=$ $=$	$\color{#004ec4}{20}^2+\color{#00880a}{21}^2$	Plug in the values
$\color{#e85e00}{c}^2$	$=$ $=$	$400 + 441$ $400+441$
$\color{#e85e00}{c}^2$	$=$ $=$	$841$ $841$
$\sqrt{\color{#e85e00}{c}^2}$	$=$ $=$	$\sqrt{841}$ $sqrt841$	Take the square root of both sides
$c$ $c$	$=$ $=$	$29$ $29$

opposite = 20

$\color{#004ec4}{\text{opposite}=20}$

adjacent = 21

$\color{#00880a}{\text{adjacent}=21}$

hypotenuse = 29

$\color{#e85e00}{\text{hypotenuse}=29}$

Now, solve for the other Trigonometric Ratios using the given formulas.

$sin θ$ $sin theta$	$=$ $=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$	$sin$ $sin$ ratio

	$=$ $=$	$\frac{\color{#004ec4}{20}}{\color{#e85e00}{29}}$	Plug in the values

$cos θ$ $cos theta$	$=$ $=$	$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$	$cos$ $cos$ ratio

	$=$ $=$	$\frac{\color{#00880a}{21}}{\color{#e85e00}{29}}$	Plug in the values

(i) sin θ = \frac{20}{29}

$(i) sin theta=20/29$

(i i) cos θ = \frac{21}{29}

$(ii) cos theta=21/29$

Question 3 of 4

If

sin α = \frac{12}{13}

$sin alpha=12/13$ , find the following trigonometric ratios.

Write fractions in the format “a/b”

$(i) cos α =$ $(i) cos alpha=$ (5/13)

$(i i) tan α =$ $(ii) tan alpha=$ (12/5)

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Trigonometric Ratios (SOHCAHTOA) for Right Angled Triangles

Sin Ratio (SOH)

sin = \frac{opposite}{hypotenuse}

$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio (CAH)

cos = \frac{adjacent}{hypotenuse}

$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio (TOA)

tan = \frac{opposite}{adjacent}

$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

Pythagoras’ Theorem

c^{2} = a^{2} + b^{2}

$\color{#e85e00}{c}^2=\color{#004ec4}{a}^2+\color{#00880a}{b}^2$

First, draw a random right triangle and use the

sin

$sin$ ratio to label it.

sin = \frac{opposite}{hypotenuse} = \frac{12}{13}

$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}=\frac{\color{#004ec4}{12}}{\color{#e85e00}{13}}$

To find the missing side which is the adjacent, use Pythagoras’ Theorem.

a = 12

$a=12$

c = 13

$c=13$

$\color{#e85e00}{c}^2$	$=$ $=$	$\color{#004ec4}{a}^2+\color{#00880a}{b}^2$	Pythagoras’ Theorem
$\color{#e85e00}{13}^2$	$=$ $=$	$\color{#004ec4}{12}^2+\color{#00880a}{b}^2$	Plug in the values
$169$ $169$	$=$ $=$	$144+\color{#00880a}{b}^2$
$169$ $169$ $- 144$ $-144$	$=$ $=$	$144+\color{#00880a}{b}^2\color{#CC0000}{-144}$	Subtract $144$ $144$ from both sides
$25$ $25$	$=$ $=$	$\color{#00880a}{b}^2$
$\sqrt{25}$ $sqrt25$	$=$ $=$	$\sqrt{\color{#00880a}{b}^2}$	Take the square root of both sides
$5$ $5$	$=$ $=$	$b$ $b$
$b$ $b$	$=$ $=$	$5$ $5$

opposite = 12

$\color{#004ec4}{\text{opposite}=12}$

adjacent = 5

$\color{#00880a}{\text{adjacent}=5}$

hypotenuse = 13

$\color{#e85e00}{\text{hypotenuse}=13}$

Now, solve for the other Trigonometric Ratios using the given formulas.

$cos α$ $cos alpha$	$=$ $=$	$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$	$cos$ $cos$ ratio

	$=$ $=$	$\frac{\color{#00880a}{5}}{\color{#e85e00}{13}}$	Plug in the values

$tan α$ $tan alpha$	$=$ $=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$	$tan$ $tan$ ratio

	$=$ $=$	$\frac{\color{#004ec4}{12}}{\color{#00880a}{5}}$	Plug in the values

(i) cos α = \frac{5}{13}

$(i) cos alpha=5/13$

(i i) tan α = \frac{12}{5}

$(ii) tan alpha=12/5$

Question 4 of 4

If

tan θ = \frac{1}{2}

$tan theta=1/2$ , find

x

$x$ .

$x =$ $x=$ (3)

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Trigonometric Ratios (SOHCAHTOA) for Right Angled Triangles

Sin Ratio (SOH)

sin = \frac{opposite}{hypotenuse}

$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio (CAH)

cos = \frac{adjacent}{hypotenuse}

$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio (TOA)

$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

First, label the given

$tan theta$ value.

$\tan\theta=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}=\frac{\color{#004ec4}{1}}{\color{#00880a}{2}}$

Also, find the opposite and adjacent values for the angle

$theta$ according to the given triangle.

$\tan\theta=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}=\frac{\color{#004ec4}{x}}{\color{#00880a}{6}}$

Finally, equate the two

$tan theta$ values and solve for

$x$ .

$\frac{\color{#004ec4}{1}}{\color{#00880a}{2}}$	$=$	$\frac{\color{#004ec4}{x}}{\color{#00880a}{6}}$

$2x$	$=$	$6(1)$	Cross multiply
$2x$	$=$	$6$
$2x$ $divide2$	$=$	$6$ $divide2$	Divide both sides by $2$
$x$	$=$	$3$

$x=3$