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Trigonometry Mixed Review: Part 1>
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Question 1 of 9
1. Question
Solve for dRound your answer to two decimal places- d= (12.86)
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Sin Ratio
sin=oppositehypotenuseCos Ratio
cos=adjacenthypotenuseTan Ratio
tan=oppositeadjacentFirst we need to identify which trig ratio to use.One of the known angles (37°42′) has d as an adjacent side and the other length (15.5) is the hypotenuseHence, we can use the cosratio to solve for dcosθ = adjacenthypotenuse cosratio cos(37°42′) = d15.5 Plug in the values Get d by itself to find its valuecos(37°42′) = d15.5 15.5×cos(65°) = d Multiply both sides by 15.5 15.5×0.829 = d Evaluate cos(37°42′) on the calculator 12.86 = d Round to two decimal places d = 12.86 d=12.86 -
Question 2 of 9
2. Question
Solve for θRound your answer to the nearest degree- θ= (24)°
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Sin Ratio
sin=oppositehypotenuseCos Ratio
cos=adjacenthypotenuseTan Ratio
tan=oppositeadjacentFirst we need to identify which trig ratio to use.One of the known lengths (22) is opposite to θ and the other length (55) is the hypotenuseHence, we can use the sinratio to solve for θsinθ = oppositehypotenuse sinratio sinθ = 2255 Plug in the values sinθ = 0.4 Use the inverse function for sin on your calculator to get θ by itselfθ = sin-1(0.4) The inverse of sin is sin-1 θ = 23.578° Use the shift sin function on your calculator θ = 24° Rounded to the nearest degree θ=24° -
Question 3 of 9
3. Question
Solve for aRound your answer to two decimal places- a= (19.37)
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Chapters- Chapters
Sin Ratio
sin=oppositehypotenuseCos Ratio
cos=adjacenthypotenuseTan Ratio
tan=oppositeadjacentFirst we need to identify which trig ratio to use.One of the known angles (61°15′) has a as an opposite side and 13.5 as an adjacent sideHence, we can use the tanratio to solve for atanθ = oppositeadjacent tanratio tan(61°15′) = a13.5 Plug in the values Now we need to have a on one side of the equationtan(61°15′) = a13.5 13.5×tan(61°15′) = a Multiply both sides by 13.5 13.5×1.43 = a Evaluate tan(61°15′) on the calculator 19.37 = a Round to two decimal places a = 19.37 a=19.37 -
Question 4 of 9
4. Question
Solve for θRound your answer to the nearest minute- 1.
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Chapters- Chapters
Sin Ratio
sin=oppositehypotenuseCos Ratio
cos=adjacenthypotenuseTan Ratio
tan=oppositeadjacentFirst we need to identify which trig ratio to use.One of the known lengths (14) is adjacent to θ and the other length (16.5) is opposite to θHence, we can use the tanratio to solve for θtanθ = oppositeadjacent tanratio tanθ = 16.514 Plug in the values tanθ = 1.1786 Use the inverse function for tan on your calculator to get θ by itselfθ = tan-1(1.1786) The inverse of tan is tan-1 θ = 49.686 Use the shift tan function on your calculator θ = 49°41’ Use the \text(degrees) function on your calculator theta = 49°41’ Rounded to the nearest minute theta=49°41’ -
Question 5 of 9
5. Question
Find the length of xRound your answer to one decimal place- x= (69.3) m
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The Angles of Elevation and Depression are the angles created by the upward or downward slope of the hypotenuse.First, we need to label the components of the triangle.To solve for x, we need to subtract the value of b from the value of aNext, we need to identify which trig ratio to use.The 28° angle has a as an \text(adjacent) side, the 54° angle has b as an \text(adjacent) side, and both angles have 60 m as their \text(opposite) side.Hence, we can use the tan \text(ratio) to solve for both a and bSolve for the value of a first:tan theta = \frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}} tan \text(ratio) tan28° = \frac{\color{#004ec4}{60}}{\color{#cc0000}{a}} Plug in the values axx tan28° = 60 Cross multiply a = 60/(tan28°) Divide tan28° from both sides to isolate a a = 112.8 m Rounded to one decimal place Next, use the tan \text(ratio) to solve for btan theta = \frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}} tan \text(ratio) tan54° = \frac{\color{#004ec4}{60}}{\color{#9e8600}{b}} Plug in the values bxx tan54° = 60 Cross multiply b = 60/(tan54°) Divide tan54° from both sides to isolate b b = 43.6 m Rounded to one decimal place Finally, subtract the value of b from the value of a to find xx = a-b x = 112.8-43.6 Plug in the values x = 69.3 m x=69.3 m -
Question 6 of 9
6. Question
Solve for xRound your answer to one decimal place- x = (11.7)
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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 (43°) has x as an \text(adjacent) side and the other length (16) is the \text(hypotenuse)Hence, we can use the cos \text(ratio) to solve for xcos theta = \frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}} cos \text(ratio) cos (43°) = \frac{\color{#00880a}{x}}{\color{#e85e00}{16}} Plug in the values Get x by itself to find its valuecos (43°) = x/16 16 xx cos (43°) = x Multiply both sides by 16 16 xx 0.7313537016 = x Evaluate cos(43°) on the calculator 11.7 = x Round to one decimal place x = 11.7 x=11.7 -
Question 7 of 9
7. Question
Solve for hRound your answer to the nearest metre- h= (315, 316) m
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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 angles (32°15′) has h as an \text(opposite) side and 500 as an \text(adjacent) sideHence, we can use the tan \text(ratio) to solve for xtan theta = \frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}} tan \text(ratio) tan (32°15′) = \frac{\color{#004ec4}{h}}{\color{#00880a}{500}} Plug in the values Now we need to have x on one side of the equationtan (32°15′) = h/500 500 times tan (32°15′) = h Multiply both sides by 500 500 times 0.631 = h Evaluate cos(43°) on the calculator 315 = h Rounded to the nearest metre h = 315 m h=315 m -
Question 8 of 9
8. Question
Solve for thetaRound your answer to the nearest minute- theta= (69)° (20)'
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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 (6) is \text(adjacent) to theta and the other length (17) is the \text(hypotenuse)Hence, we can use the cos \text(ratio) to solve for thetacos theta = \frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}} cos \text(ratio) cos theta = \frac{\color{#00880a}{6}}{\color{#e85e00}{17}} Plug in the values cos theta = 0.353 Evaluate 6/17 to 3 decimal places Use the inverse function for cos on your calculator to get theta by itselftheta = cos^(-1) (0.353) The inverse of cos is cos^(-1) theta = 69.3327 Use the \text(shift) cos function on your calculator theta = 69°19’57” Use the \text(degrees) function on your calculator theta = 69°20’ Round up the minutes theta=69°20’ -
Question 9 of 9
9. Question
Solve for angle thetaRound your answer to the nearest minute-
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The Angles of Elevation and Depression are the angles created by the upward or downward slope of the hypotenuse.First, we need to label the components of the triangle.To solve for theta, we need to subtract the added value of gamma and beta from the total interior angle of a triangle, which is 180°In order to find gamma, we need to first solve for alpha.Angle alpha has 8.5 m as an \text(opposite) side and 10.2 m as an \text(adjacent) side.Hence, we can use the tan \text(ratio) to solve for alphatanalpha = \frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}} tan \text(ratio) tanalpha = \frac{\color{#004ec4}{8.5}}{\color{#00880a}{10.2}} Plug in the values tanalpha = 0.833 Evaluate Use the inverse function for tan on your calculator to get alpha by itselfalpha = tan^(-1) (0.833) The inverse of tan is tan^(-1) alpha = 39.794 Use the \text(shift) tan function on your calculator alpha = 39° 48′ 20.06″ Use the \text(degrees) function on your calculator alpha = 39° 48’ Rounded to the nearest minute Recall that a straight line has an angle of 180°Since we have the value of alpha, we can subtract that to 180° to find the value of gammagamma = 180°-alpha gamma = 180°-39°48’ Plug in the values gamma = 140° 12’ Next, identify which trig ratio to use for finding beta.Angle beta has 8.5 m as an \text(opposite) side and 17.2 m (10.2+7) as an \text(adjacent) side.Thus, use the tan \text(ratio) to solve for betatanbeta = \frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}} tan \text(ratio) tanbeta = \frac{\color{#004ec4}{8.5}}{\color{#00880a}{17.2}} Plug in the values tanbeta = 0.494 Evaluate Use the inverse function for tan on your calculator to get beta by itselfbeta = tan^(-1) (0.494) The inverse of tan is tan^(-1) beta = 26.298 Use the \text(shift) tan function on your calculator beta = 26° 17′ 53″ Use the \text(degrees) function on your calculator beta = 26° 18’ Rounded to the nearest minute Finally, we can subtract the added value of gamma and beta from 180° to find the value of thetatheta = 180°-(gamma+beta) theta = 180°-(140°12’+26°18’) Plug in the values theta = 13° 30’ theta=13° 30’ -
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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)