Trigonometry Mixed Review: Part 1 (1)

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Question 1 of 9

1. Question
Which of the following are labelled correctly?

There can be more than one answer
- 1.
- 2.
- 3.
- 4.
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Check each triangle and identify if the labels are correct

The side opposite of the right angle is labelled as hyp

The side opposite of $θ$ $theta$ is labelled as opp

The side adjacent to $θ$ $theta$ is labelled as adj

The triangle is labelled correctly

The side opposite of the right angle is labelled as hyp

The side opposite of $θ$ $theta$ is labelled as opp

The side adjacent to $θ$ $theta$ is labelled as adj

The triangle is labelled correctly

The side opposite of the right angle is labelled as hyp

The side opposite of $θ$ $theta$ is labelled as opp

The side adjacent to $θ$ $theta$ is labelled as adj

The triangle is labelled correctly

The side opposite of the right angle is labelled as “opp”, but it should be hyp

The side opposite of $θ$ $theta$ is labelled as “hyp”, but it should be opp

The side adjacent to $θ$ $theta$ is labelled as adj

The triangle is labelled incorrectly

Question 2 of 9

Solve for:

(i) sin θ

$(i) \text(sin)theta$

(i i) cos (90 - θ)

$(ii) \text(cos)(90-theta)$

Enter fractions as: x/y

$sin θ =$ $\text(sin)theta =$ (3/5)
$cos (90 - θ) =$ $\text(cos)(90 - theta) =$ (3/5)

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Sin Ratio

s i n = \frac{opposite}{hypotenuse}

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

Cos Ratio

c o s = \frac{adjacent}{hypotenuse}

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

Tan Ratio

t a n = \frac{opposite}{adjacent}

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

Solving for (i) sin θ

$\text(Solving for) (i) sin theta$

First we need to identify which trig ratio to use.

One of the known lengths $(3)$ $(3)$ is $opposite$ $\text(opposite)$ to

θ

$theta$ and the other length $(5)$ $(5)$ is the $hypotenuse$ $\text(hypotenuse)$

Hence, we can use the

sin ratio

$sin \text(ratio)$ to solve for

sin θ

$sintheta$

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

$cos θ$ $cos theta$	$=$ $=$	$\frac{\color{#004ec4}{3}}{\color{#e85e00}{5}}$	Plug in the values

Solving for (i i) cos (90 - θ)

$\text(Solving for) (ii) cos (90-theta)$

First we need to understand which angle is

(90 - θ)

$(90-theta)$ .
Notice that the two known angles of the triangle is the right angle

(90 °)

$(90°)$ and

θ

$theta$ .
Hence, the other angle left is

(90 - θ)

$(90-theta)$ .

Now, one of the known lengths $(3)$ $(3)$ is $adjacent$ $\text(adjacent)$ to

(90 - θ)

$(90-theta)$ and the other length $(5)$ $(5)$ is the $hypotenuse$ $\text(hypotenuse)$

Hence, we can use the

cos ratio

$cos \text(ratio)$ to solve for

cos (90 - θ)

$cos(90-theta)$

$cos (90 - θ)$ $cos (90-theta)$	$=$ $=$	$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$	$cos ratio$ $cos \text(ratio)$

$cos (90 - θ)$ $cos (90-theta)$	$=$ $=$	$\frac{\color{#00880a}{3}}{\color{#e85e00}{5}}$	Plug in the values

sin θ = \frac{3}{5}

$sintheta=3/5$

cos (90 - θ) = \frac{3}{5}

$cos (90-theta)=3/5$

Question 3 of 9

Solve for

x

$x$ if:

θ = 30

$theta=30$

$x =$ $x=$ (60) $°$ $°$

Question 4 of 9

Solve for

θ

$theta$

Round your answer to the nearest degree

$θ =$ $theta=$ (37) $°$ $°$

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