Trigonometry Word Problems 1
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Question 1 of 5
1. Question
Two poles are the same distance up a brick wall. The longer pole is `7.4 m` long. The shorter pole is `5.3 m` long and makes an angle of `47°` to the ground. What height `(h)` do the poles reach up the wall?
Round your answer to `1` decimal place- `h=` (3.9)`m`
Hint
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Great Work!
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Trigonometric Ratios (SOHCAHTOA)
Sin Ratio (SOH)
$$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio (CAH)
$$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio (TOA)
$$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$Calculator Buttons to Use
`sin` `=` Sine function`cos` `=` Cosine function`tan` `=` Tangent functionDMS or `° ‘ ‘ ‘` `=` Degree/Minute/Second`=` `=` Equal functionNotice that the scenario creates two right triangles. Focus on the smaller one and label it in reference to the given angle.
$$\color{#004ec4}{\text{opposite}}=\color{#004ec4}{h}$$$$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{5.3}$$Since we now have the opposite and hypotenuse values, we can use the `sin` ratio to find `h`.`sin47°` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$ `sin47°` `=` $$\frac{\color{#004ec4}{h}}{\color{#e85e00}{5.3}}$$ `5.3xx``sin47°` `=` `h/5.3``xx5.3` Multiply both sides by `5.3` `5.3sin47°` `=` `h` `h` `=` `5.3sin47°` Simplify this further by evaluating `5.3sin47°` using the calculator:`1.` Press `5.3``2.` Press `times``3.` Press `sin``4.` Press `47``5.` Press `=`The result will be `3.87617` or `3.9 m` when rounded off to `1` decimal place.
`3.9 m` -
Question 2 of 5
2. Question
Two poles are the same distance up a brick wall. The longer pole is `7.4 m` long. The shorter pole is `5.3 m` long and makes an angle of `47°` to the ground. What is the angle `(theta)` to the nearest degree that the longer pole reaches with the ground?
- (32)`°`
Hint
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Incorrect
Trigonometric Ratios (SOHCAHTOA)
Sin Ratio (SOH)
$$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio (CAH)
$$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio (TOA)
$$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$Calculator Buttons to Use
`sin` `=` Sine function`cos` `=` Cosine function`tan` `=` Tangent functionDMS or `° ‘ ‘ ‘` `=` Degree/Minute/SecondShift or 2nd F or INV `=` Inverse function`=` `=` Equal functionNotice that the scenario creates two right triangles. Let the missing angle be `theta` and label the sides of the larger triangle in reference to it.
$$\color{#004ec4}{\text{opposite}}=\color{#004ec4}{3.9}$$$$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{7.4}$$Since we now have the opposite and hypotenuse values, we can use the `sin` ratio to find `theta`.`sin theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$ `sin theta` `=` $$\frac{\color{#004ec4}{3.9}}{\color{#e85e00}{7.4}}$$ `theta` `=` `sin^(-1) (3.9/7.4)` Get the inverse of `sin` Simplify this further by evaluating `sin^(-1) (3.9/7.4)` using the calculator:`1.` Press Shift or 2nd F (depending on your calculator)`2.` Press `sin``3.` Press `3.9``4.` Press `divide``5.` Press `7.4``6.` Press `=`The result will be: `31.8048°`Finally, round off the answer to the nearest degree.`theta` `=` `31.8048°` `=` `31°48’` Press DMS on your calculator `=` `32°` Round up since the minutes is more than `30’` 
`32°` -
Question 3 of 5
3. Question
Two poles are the same distance up a brick wall. The longer pole is `7.4 m` long. The shorter pole is `5.3 m` long and makes an angle of `47°` to the ground. Find the distance between the poles on the ground.
Round your answer to `1` decimal place- (2.6)`m`
Hint
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Sine Rule
$$\frac{\color{#007DDC}{a}}{\sin\color{#007DDC}{A}}=\frac{\color{#00880A}{b}}{\sin\color{#00880A}{B}}=\frac{\color{#9a00c7}{c}}{\sin\color{#9a00c7}{C}}$$where:
`a` is the side opposite angle `A`
`b` is the side opposite angle `B`
`c` is the side opposite angle `C`When to use the Sine Rule
a) Given 2 sides and 1 angle to find the other angleorb) Given 2 angles 1 side to find the other sideNotice that the scenario creates an obtuse triangle. Let `x` be the distance of the two poles on the ground.Identify the two other angles in the triangle.
First angle:Remember that a straight line measures `180°`. Subtract `47°` from `180°` to find the measure of the larger angle.`180-47` `=` `133°` 
Second angle:Remember that the sum of the interior angles in a triangle is `180°`. Subtract `32°` and `133°` from `180°` to find the measure of the smaller angle.`180-32-133` `=` `15°` 
Since `1` side and `2` angles are now known, use the Sine Rule to find the missing side.Side `1=x`Angle `1=15°`Side `2=7.4 m`Angle `2=133°`$$\frac{\color{#007DDC}{a}}{\sin\color{#007DDC}{A}}$$ `=` $$\frac{\color{#00880A}{b}}{\sin\color{#00880A}{B}}$$ $$\frac{\color{#007DDC}{x}}{\sin\color{#007DDC}{15°}}$$ `=` $$\frac{\color{#00880A}{7.4}}{\sin\color{#00880A}{133°}}$$ Substitute the values `xtimessin133°` `=` `7.4timessin15°` Cross multiply `xtimessin133°``dividesin133°` `=` `7.4timessin15°``dividesin133°` Divide both sides by `sin133°` `x` `=` `(7.4timessin15°)/(sin133°)` `x` `=` `1.9152609/0.7313537` Use the calculator to simplify `x` `=` `2.61879` `x` `=` `2.6 m` Rounded off to `1` decimal place 
`2.6 m` -
Question 4 of 5
4. Question
The triangle `ABD` has a right angle at `B`. If the length of side `DA` is `83 m`, find the length of `CA`.
Round your answer to `2` decimal places- `CA` = (65.07) `m`
Hint
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Sine Rule
$$\frac{\color{#007DDC}{a}}{\sin\color{#007DDC}{A}}=\frac{\color{#00880A}{b}}{\sin\color{#00880A}{B}}=\frac{\color{#9a00c7}{c}}{\sin\color{#9a00c7}{C}}$$where:
`a` is the side opposite angle `A`
`b` is the side opposite angle `B`
`c` is the side opposite angle `C`When to use the Sine Rule
a) Given 2 sides and 1 angle to find the other angleorb) Given 2 angles 1 side to find the other sideNotice that `CA` is the side opposite of angle `D`.
This means that we can use the Sine Rule to find the length of `CA`.Start by finding the value of angle `D`. Do this by subtracting the sum of angles `A` and `C` from `180°`.`D` `=` `180°-(8°+146°)` `D` `=` `180°-154°` `D` `=` `26°` Next, label the triangle according to the Sine Rule.
Substitute the three known values to the Sine Rule to find the fourth missing value.From labelling the triangle, we know that the known values are those with labels `d, D, c` and `C`.`d=CA``D=26°``c=83 m``C=146°`$$\frac{\color{#007DDC}{d}}{\sin\color{#007DDC}{D}}$$ `=` $$\frac{\color{#00880A}{c}}{\sin\color{#00880A}{C}}$$ $$\frac{\color{#007DDC}{CA}}{\sin\color{#007DDC}{26°}}$$ `=` $$\frac{\color{#00880A}{83}}{\sin\color{#00880A}{146°}}$$ Substitute the values `CAtimessin146°` `=` `83timessin26°` Cross multiply `CAtimessin146°``dividesin146°` `=` `83timessin26°``dividesin146°` Divide both sides by `sin146°` `CA` `=` `(83timessin26°)/(sin146°)` `CA` `=` `(36.384805)/(0.5591929)` Use the calculator to simplify `CA` `=` `65.0666` `CA` `=` `65.07 m` Rounded off to `2` decimal places 
`65.07 m` -
Question 5 of 5
5. Question
The triangle `ABD` has a right angle at `B` and a side `DA` that is `83 m`. The line `AC` goes through the triangle and has a length of `65.07 m`. Find the values of the following:`(i)` The value of `theta``(ii)` The length of `CB` rounded off to the nearest metre
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`(i)` `theta` = (34)`°``(ii)` `CB` = (54) `m`
Hint
Help VideoCorrect
Great Work!
Incorrect
Trigonometric Ratios (SOHCAHTOA)
Sin Ratio (SOH)
$$\sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio (CAH)
$$\cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio (TOA)
$$\tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$Calculator Buttons to Use
`sin` `=` Sine function`cos` `=` Cosine function`tan` `=` Tangent functionDMS or `° ‘ ‘ ‘` `=` Degree/Minute/Second`=` `=` Equal function`(i)` Find the value of `theta`Remember that a straight line measures `180°`. Subtract `146°` from `180°` to find the measurement of `theta`.`theta` `=` `180°-146°` `theta` `=` `34°` 
`(ii)` Find the length of `CB`Notice that `CB` is a side of the right triangle `ABC` and is adjacent to `theta`. The right triangle also has a hypotenuse `AC` with a length of `65.07 m`.
$$\color{#00880a}{\text{adjacent}}=\color{#00880a}{CB}$$$$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{65.07}$$ (AC)Since we now have the adjacent and hypotenuse values, we can use the `cos` ratio to find `CB`.`cos34°` `=` $$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$ `cos34°` `=` $$\frac{\color{#00880a}{CB}}{\color{#e85e00}{65.07}}$$ `65.07xx``cos34°` `=` `(CB)/65.07``xx65.07` Multiply both sides by `65.07` `65.07cos34°` `=` `CB` `CB` `=` `65.07cos34°` Simplify this further by evaluating `cos34°` using the calculator:`1.` Press `cos``2.` Press `34``3.` Press `=`The result will be: `0.8290376`Continue solving for `CB`.`cos34°=0.8290376``CB` `=` `65.07cos34°` `=` `65.07times0.8290376` `=` `53.9454749` `=` `54 m` Rounded off to the nearest metre 
`(i) theta=34°``(ii) CB=54 m` -
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
- 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)