Intro to Trigonometric Ratios (SOH CAH TOA) 1

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Question 1 of 5

1. Question
Name the following sides with respect to angle $θ$ $theta$ in the triangle below.

The sides can either be $A B$ $AB$ , $A C$ $AC$ or $B C$ $BC$ .
- $(i)$ $(i)$ Opposite: (BC, CB)
  
  $(i i)$ $(ii)$ Adjacent: (AB, BA)
  
  $(i i i)$ $(iii)$ Hypotenuse: (AC, CA)
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We can easily identify the hypotenuse as it is simply the side opposite the right angle.

Therefore, the hypotenuse is side $A C$ $AC$ .

Next, the opposite side is exactly the side opposite the angle $θ$ $theta$ .

Therefore, the opposite side is side $B C$ $BC$ .

Finally, the adjacent side is exactly the side next to the angle $θ$ $theta$ , but not the hypotenuse.

Therefore, the adjacent side is side $A B$ $AB$ .

$(i)$ $(i)$ Opposite: $B C$ $BC$

$(i i)$ $(ii)$ Adjacent: $A B$ $AB$

$(i i i)$ $(iii)$ Hypotenuse: $A C$ $AC$

Question 2 of 5

Find the following trigonometric ratios using the given triangle.

Write fractions in the format “a/b”

$(i) sin θ =$ $(i) sin theta=$ (5/13)

$(i i) cos θ =$ $(ii) cos theta=$ (12/13)

$(i i i) tan θ =$ $(iii) tan theta=$ (5/12)

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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}

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

First, label the values in the triangle.

opposite = 5

$\color{#004ec4}{\text{opposite}=5}$

adjacent = 12

$\color{#00880a}{\text{adjacent}=12}$

hypotenuse = 13

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

Now, solve each Trigonometric Ratio using the given formulas.

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

	$=$ $=$	$\frac{\color{#004ec4}{5}}{\color{#e85e00}{13}}$	Plug in the values

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

	$=$ $=$	$\frac{\color{#00880a}{12}}{\color{#e85e00}{13}}$	Plug in the values

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

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

(i) sin θ = \frac{5}{13}

$(i) sin theta=5/13$

(i i) cos θ = \frac{12}{13}

$(ii) cos theta=12/13$

(i i i) tan θ = \frac{5}{12}

$(iii) tan theta=5/12$

Question 3 of 5