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Trigonometry Mixed Review: Part 1

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Trigonometry Mixed Review: Part 1 (3)

Trigonometry Mixed Review: Part 1 (3)

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Question 1 of 9

1. Question

Solve for

$h$

Round your answer to two decimal places

1. $16.92 m$
2. $7.95 m$
3. $6.81 m$
4. $12.17 m$

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Sin Ratio

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

Cos Ratio

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

Tan Ratio

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

First we need to identify which trig ratio to use.

One of the known angles $(29°35′)$ has

$h$ as an $\text(adjacent)$ side and the other length $(14)$ is the $\text(hypotenuse)$

Hence, we can use the

$cos \text(ratio)$ to solve for

$x$

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

$cos (29°35′)$	$=$	$\frac{\color{#00880a}{h}}{\color{#e85e00}{14}}$	Plug in the values

Now we need to have

$x$ on one side of the equation

$cos (29°35′)$	$=$	$h/14$

$14 times cos (29°35′)$	$=$	$h$	Multiply both sides by $14$
$14 times 0.8696385746$	$=$	$h$	Evaluate $cos (29°35′)$ on the calculator
$12.17$	$=$	$h$	Rounded to two decimal places
$h$	$=$	$12.17$

$h=12.17$

Question 2 of 9

2. Question
Find the area of the Triangle

The given measurements are in centimetres

Round your answer to the nearest whole number
- $\text(Area )=$ (14) $cm^2$
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Area of a Triangle Formula

$\text(Area )=1/2 xx$ $a$ $times$ $c$ $times sin$ $B$

Remember

Uppercase letters represent angles in the triangle

Lowercase letters represent the side lengths

Labelling the triangle

Solve for the area using the Area of a Triangle formula

$\text(Area)$ $=$ $1/2 xx$ $a$ $times$ $c$ $times sin$ $B$ Area of a Triangle formula

$=$ $1/2 xx$ $8$ $times$ $7$ $times sin$ $30°$ Plug in the known lengths

$=$ $14 cm^2$

The given measurements are in centimetres, so the area is measured as square centimetres

$\text(Area)=14 cm^2$

Question 3 of 9

3. Question

Solve for

$theta$

Round your answer to the nearest degree

$theta=$ (71) $°$

Correct

Fantastic!

Incorrect

Sin Ratio

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

Cos Ratio

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

Tan Ratio

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

First we need to identify which trig ratio to use.

One of the known lengths $(24)$ is $\text(adjacent)$ to

$theta$ and the other length $(70)$ is $\text(opposite)$ to

$theta$

Hence, we can use the

$tan \text(ratio)$ to solve for

$theta$

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

$tan theta$	$=$	$\frac{\color{#004ec4}{70}}{\color{#00880a}{24}}$	Plug in the values

$tan theta$	$=$	$2.9167$

Use the inverse function for

$tan$ on your calculator to get

$theta$ by itself

$theta$	$=$	$tan^(-1) (2.9167)$	The inverse of $tan$ is $tan^(-1)$
$theta$	$=$	$71.076°$	Use the $\text(shift) tan$ function on your calculator
$theta$	$=$	$71°$	Rounded to the nearest degree

$theta=71°$

Question 4 of 9

4. Question

Solve for

$x$

Round your answer to one decimal place

$x =$ (92.3)

Correct

Keep Going!

Incorrect

Sin Ratio

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

Cos Ratio

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

Tan Ratio

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

First we need to identify which trig ratio to use.

One of the known angles $(18°)$ has $30$ as an $\text(opposite)$ side and $x$ as an $\text(adjacent)$ side

Hence, we can use the

$tan \text(ratio)$ to solve for

$x$

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

$tan (18°)$	$=$	$\frac{\color{#004ec4}{30}}{\color{#00880a}{x}}$	Plug in the values

Now we need to have

$x$ on one side of the equation

$tan (18°)$	$=$	$30/x$

$x times tan (18°)$	$=$	$30$	Multiply both sides by $x$

$x$	$=$	$30/tan (18°)$	Divide both sides by $tan (18°)$

$x$	$=$	$30/0.3249196962$	Evaluate $tan (18°)$ on the calculator

$x$	$=$	$92.3$	Rounded to one decimal place

$x=92.3$

Question 5 of 9

5. Question

A wind turbine is to be built and

$4$ wires are needed to hold it to the ground to keep it stable. Using the information on the image below, how much wire should be prepared in total?

Round your answer to two decimal places

(117.18)m

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Sin Ratio

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

Cos Ratio

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

Tan Ratio

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

First we need to identify which trig ratio to use.

If we label the $(19°)$ angle as $theta$ , the $\text(adjacent)$ side would be $28$ m and the $\text(hypotenuse)$ would be $\text(w)$

Hence, we can use the

$cos \text(ratio)$ to solve for

$w$

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

$cos 19°$	$=$	$\frac{\color{#00880a}{28}}{\color{#e85e00}{w}}$	Plug in the values

Now we need to have

$w$ on one side of the equation

$cos 19°$	$=$	$28/w$

$cos 19°$ $times w$	$=$	$28/w$ $times w$	Multiply both sides by $w$

$w cos 19°$	$=$	$28$
$w cos 19°$ $divide cos 19°$	$=$	$28$ $divide cos 19°$	Divide both sides by $cos 19°$

$w$	$=$	$28/(cos 19°)$

$w$	$=$	$29.29$	Evaluate using the calculator

Finally, multiply the number to

$4$ since

$4$ wires are to be built to hold the wind turbine

$29.29xx4$

$=$

$117.18$

$117.18$ m of wire would be used for the wind turbine

$117.18$ m

Question 6 of 9

6. Question

Find the area of the Triangle

Round your answer to two decimal places

1. $8.86 m^2$
2. $5.64 m^2$
3. $15.98 m^2$
4. $18.27 m^2$

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Area of a Triangle Formula

$\text(Area )=1/2 xx$ $a$

$times$ $b$

$times sin$ $C$

Remember

Uppercase letters represent angles in the triangle
Lowercase letters represent the side lengths

Labelling the triangle

Solve for the area using the Area of a Triangle formula

$\text(Area)$	$=$	$1/2 xx$ $a$ $times$ $b$ $times sin$ $C$	Area of a Triangle formula

	$=$	$1/2 xx$ $8.7$ $times$ $4.2$ $times sin$ $61°$	Plug in the known lengths

	$=$	$15.98 m^2$	Rounded to two decimal places

The given measurements are in metres, so the area is measured as square metres

$\text(Area)=15.98 m^2$

Question 7 of 9

7. Question

Find the area of the Triangle

The given measurements are in units

Round your answer to one decimal place

$\text(Area )=$ (14.7) $units^2$

Correct

Well Done!

Incorrect

Area of a Triangle Formula

$\text(Area )=1/2 xx$ $b$

$times$ $c$

$times sin$ $A$

Remember

Uppercase letters represent angles in the triangle
Lowercase letters represent the side lengths

Labelling the triangle

Solve for the area using the Area of a Triangle formula

$\text(Area)$	$=$	$1/2 xx$ $b$ $times$ $c$ $times sin$ $A$	Area of a Triangle formula

	$=$	$1/2 xx$ $9$ $times$ $4$ $times sin$ $55°$	Plug in the known lengths

	$=$	$14.7 units^2$	Rounded to one decimal place

The given measurements are in units, so the area is measured as square units

$\text(Area)=14.7 units^2$

Question 8 of 9

8. Question

Solve for

$x$

Round your answer to one decimal place

$x =$ (31.2)

Correct

Great Work!

Incorrect

Sin Ratio

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

Cos Ratio

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

Tan Ratio

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

First we need to identify which trig ratio to use.

One of the known angles $(33°)$ has $17$ as an $\text(opposite)$ side and $(x)$ is the $\text(hypotenuse)$

Hence, we can use the

$sin \text(ratio)$ to solve for

$x$

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

$sin (33°)$	$=$	$\frac{\color{#004ec4}{17}}{\color{#e85e00}{x}}$	Plug in the values

Now we need to have

$x$ on one side of the equation

$sin (33°)$	$=$	$17/x$

$x times sin (33°)$	$=$	$17$	Multiply both sides by $x$

$x$	$=$	$17/sin (33°)$	Divide both sides by $sin (33°)$

$x$	$=$	$17/0.544639035$	Evaluate $sin (33°)$ on the calculator

$x$	$=$	$31.2$	Rounded to one decimal place

$x=31.2$

Question 9 of 9

9. Question

Solve for

$theta$

Round your answer to the nearest minute

1. $10˚59'$
2. $27˚19'$
3. $27˚49'$
4. $27˚49' 5''$

Hint

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Chapters

Chapters

Sin Ratio

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

Cos Ratio

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

Tan Ratio

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

First we need to identify which trig ratio to use.

One of the known lengths $(22)$ is $\text(opposite)$ to

$theta$ and the other length $(55)$ is the $\text(hypotenuse)$

Hence, we can use the

$sin \text(ratio)$ to solve for

$theta$

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

$sin theta$	$=$	$\frac{\color{#004ec4}{7}}{\color{#e85e00}{15}}$	Plug in the values

$sin theta$	$=$	$0.466…$
$sin theta$	$=$	$0.4dot 6$

Use the inverse function for

$sin$ on your calculator to get

$theta$ by itself

$theta$	$=$	$sin^(-1) (0.4dot 6)$	The inverse of $sin$ is $sin^(-1)$
$theta$	$=$	$27.818$	Use the $\text(shift) sin$ function on your calculator
$theta$	$=$	$27°49’5.3”$	Use the $\text(degrees)$ function on your calculator
$theta$	$=$	$27°49’$	Rounded to the nearest minute

$theta=27°49’$

Quizzes

Intro to Trigonometric Ratios (SOH CAH TOA) 1
Intro to Trigonometric Ratios (SOH CAH TOA) 2
Round Angles (Degrees, Minutes, Seconds)
Evaluate Trig Expressions using a Calculator 1
Evaluate Trig Expressions using a Calculator 2
Trig Ratios: Solving for a Side 1
Trig Ratios: Solving for a Side 2
Trig Ratios: Solving for an Angle
Angles of Elevation and Depression
Trig Ratios Word Problems: Solving for a Side
Trig Ratios Word Problems: Solving for an Angle
Area of Non-Right Angled Triangles 1
Area of Non-Right Angled Triangles 2
Sine Rule: Solving for a Side
Sine Rule: Solving for an Angle
Cosine Rule: Solving for a Side
Cosine Rule: Solving for an Angle
Trigonometry Word Problems 1
Trigonometry Word Problems 2
Trigonometry Mixed Review: Part 1 (1)
Trigonometry Mixed Review: Part 1 (2)
Trigonometry Mixed Review: Part 1 (3)
Trigonometry Mixed Review: Part 1 (4)
Trigonometry Mixed Review: Part 2 (1)
Trigonometry Mixed Review: Part 2 (2)
Trigonometry Mixed Review: Part 2 (3)

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