Trigonometry Word Problems 2 | VividMath

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Year 11

Trigonometry

Trigonometry Word Problems

Trigonometry Word Problems 2

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Question 1 of 4

1. Question

Noah and Amelia are standing `91 m` apart and observe a drone at an angle of elevation of `63°` and `42°` respectively. Find the distance `(ND)` between Noah `(N)` and the drone `(D)` to the nearest metre.

`\text(ND) =` (170)`m`

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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 angle

b) Given 2 angles 1 side to find the other side

Notice that the scenario creates an obtuse triangle. Let `ND` 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 `63°` from `180°` to find the measure of the larger angle.

`180-63`

`=`

`117°`

Second angle:

Remember that the sum of the interior angles in a triangle is `180°`. Subtract `42°` and `117°` from `180°` to find the measure of the smaller angle.

`180-42-117`

`=`

`21°`

Since `1` side and `2` angles are now known, use the Sine Rule to find the missing side.

Side `1=ND`

Angle `1=42°`

Side `2=91 m`

Angle `2=21°`

$$\frac{\color{#007DDC}{a}}{\sin\color{#007DDC}{A}}$$	`=`	$$\frac{\color{#00880A}{b}}{\sin\color{#00880A}{B}}$$

$$\frac{\color{#007DDC}{ND}}{\sin\color{#007DDC}{42°}}$$	`=`	$$\frac{\color{#00880A}{91}}{\sin\color{#00880A}{21°}}$$	Substitute the values

`NDtimessin21°`	`=`	`91timessin42°`	Cross multiply
`NDtimessin21°``dividesin21°`	`=`	`91timessin42°``dividesin21°`	Divide both sides by `sin21°`

`ND`	`=`	`(91timessin42°)/(sin21°)`

`ND`	`=`	`(91times0.6691306)/0.3583679`	Use the calculator to simplify

`ND`	`=`	`169.991164`
`ND`	`=`	`170 m`	Rounded off to the nearest metre

`170 m`

Question 2 of 4

2. Question

Noah and Amelia are standing `91 m` apart and observe a drone at an angle of elevation of `63°` and `42°` respectively. Find the height `(h)` of the drone to the nearest metre.

`h=` (151)`m`

Hint

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Correct

Excellent!

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 function

DMS or `° ‘ ‘ ‘` `=` Degree/Minute/Second

`=` `=` Equal function

Notice that the scenario creates two right triangles. Focus on the smaller one and label it in reference to the given angle.

$$\color{#004ec4}{\text{opposite}}=\color{#004ec4}{h}$$

$$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{170}$$

Since we now have the opposite and hypotenuse values, we can use the `sin` ratio to find `h`.

`sin63°`	`=`	$$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$

`sin63°`	`=`	$$\frac{\color{#004ec4}{h}}{\color{#e85e00}{170}}$$

`170xx``sin63°`	`=`	`h/170``xx170`	Multiply both sides by `170`

`170sin63°`	`=`	`h`
`h`	`=`	`170sin63°`

Simplify this further by evaluating `sin63°` using the calculator:

`1.` Press `sin`

`2.` Press `63`

`3.` Press `=`

The result will be: `0.89100652`

Continue solving for `h`.

`sin63°=0.89100652`

`h`	`=`	`170sin63°`
	`=`	`170times0.89100652`
	`=`	`151.47`
	`=`	`151 m`	Rounded off to the nearest metre

`151 m`

Question 3 of 4

3. Question

Two support wires, one `13 m` long and the other `12 m` long, are attached to a pole from a common point `A`. The longer wire is attached to the top of the pole. The shorter wire is attached `4 m` below the longer wire. Find the measurement of `/_ABD`.

Round your answer to the nearest minute

`/_ABD` = (66)`°` (47)`'`

Hint

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Correct

Fantastic!

Incorrect

Cosine Rule

$$\cos\color{#007DDC}{A}=\frac{\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-\color{#007DDC}{a}^2}{2\color{#00880A}{b}\color{#9a00c7}{c}}$$

where:
`a` is the side opposite angle `A`
`b` is the side opposite angle `B`
`c` is the side opposite angle `C`

Calculator Buttons to Use

`sin` `=` Sine function

`cos` `=` Cosine function

`tan` `=` Tangent function

DMS or `° ‘ ‘ ‘` `=` Degree/Minute/Second

Shift or 2nd F or INV `=` Inverse function

`=` `=` Equal function

Since `3` sides of the triangle `ABC` are given, we can use the Cosine Rule.

First, label the triangle `ABC` according to the Cosine Rule.

Substitute the three known values to the Cosine Rule to find `B` or `/_ABD`.

From labelling the triangle, we know that the known values are those with labels `a, b` and `c`.

`B=/_ABD`

`a=15 cm`

`b=11 cm`

`c=6 cm`

$$\cos\color{#007DDC}{B}$$	`=`	$$\frac{\color{#00880A}{a}^2+\color{#9a00c7}{c}^2-\color{#007DDC}{b}^2}{2\color{#00880A}{a}\color{#9a00c7}{c}}$$

$$\cos\color{#007DDC}{B}$$	`=`	$$\frac{\color{#00880A}{13}^2+\color{#9a00c7}{4}^2-\color{#007DDC}{12}^2}{2(\color{#00880A}{13})(\color{#9a00c7}{4})}$$	Substitute the values

`cos B`	`=`	`(169+16-144)/(104)`	Simplify

`cos B`	`=`	`41/104`

`B`	`=`	`cos^(-1) (41/104)`	Get the inverse of the cosine

Simplify this further by evaluating `cos^(-1) (41/104)` using the calculator:

`1.` Press Shift or 2nd F (depending on your calculator)

`2.` Press `cos`

`3.` Press `41`

`4.` Press `divide`

`3.` Press `104`

`4.` Press `=`

The result will be: `66.78199226°`

Proceed with solving for `/_ABD`.

`cos^(-1) (41/104)=66.78199226°`

`B`	`=`	`cos^(-1) (41/104)`
`B`	`=`	`66.78199226°`
`B`	`=`	`66°46’55”`	Press DMS on the calculator
`B` or `/_ABD`	`=`	`66°47’`	Round off to the nearest minute

`66°47’`

Question 4 of 4

4. Question

Round your answer to `1` decimal place

`BD` = (5.1)`m`

Hint

Help Video

Correct

Nice Job!

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 function

DMS or `° ‘ ‘ ‘` `=` Degree/Minute/Second

`=` `=` Equal function

Notice that the scenario creates two right triangles. Focus on the bigger one and label it in reference to the given angle.

$$\color{#00880a}{\text{adjacent}}=\color{#00880a}{BD}$$

$$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{13}$$

Since we now have the adjacent and hypotenuse values, we can use the `cos` ratio to find `BD`.

`cos66°47’`	`=`	$$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$

`cos66°47’`	`=`	$$\frac{\color{#00880a}{BD}}{\color{#e85e00}{13}}$$

`13xx``sin66°47’`	`=`	`(BD)/13``xx13`	Multiply both sides by `13`

`13cos66°47’`	`=`	`BD`
`BD`	`=`	`13cos66°47’`

Simplify this further by evaluating `13cos66°47’` using the calculator:

`1.` Press `13`

`2.` Press `times`

`3.` Press `cos`

`4.` Press `66`

`5.` Press DMS

`6.` Press `47`

`7.` Press DMS

`8.` Press `=`

The result will be `5.1247` or `5.1 m` when rounded off to `1` decimal place.

`5.1 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
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)