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Trigonometry Mixed Review: Part 1>
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 placesHint
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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 TriangleThe given measurements are in centimetresRound your answer to the nearest whole number `\text(Area )=` (14) `cm^2`
Hint
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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)` sideHence, 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
Hint
Help VideoCorrect
Correct!
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.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 TriangleRound your answer to two decimal placesHint
Help VideoCorrect
Excellent!
Incorrect
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 TriangleThe given measurements are in unitsRound 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 minuteHint
Help VideoCorrect
Nice Job!
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 `(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 NonRight Angled Triangles 1
 Area of NonRight Angled Triangles 2
 Law of Sines: Solving for a Side
 Law of Sines: Solving for an Angle
 Law of Cosines: Solving for a Side
 Law of Cosines: 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)