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Trig Ratios Word Problems: Solving for a Side

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Trig Ratios Word Problems: Solving for a Side

Trig Ratios Word Problems: Solving for a Side

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Question 1 of 7

A ship is anchored

$410 m$ from the base of a vertical cliff. The angle looking up to the top of the cliff is at an angle of

$59°24’$ . Calculate the height of the cliff (

$h$ ) to the nearest metre.

$h=$ (693) $m$

Incorrect

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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 function

DMS or $° ‘ ‘ ‘$

$=$ Degree/Minute/Second

$=$

$=$ Equal function

Notice that the scenario creates a triangle. Label it in reference to the given angle.

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

$\color{#00880a}{\text{adjacent}}=\color{#00880a}{410}$

Since we now have the opposite and adjacent values, we can use the

$tan$ ratio to find

$h$ .

$tan59°24’$	$=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

$tan59°24’$	$=$	$\frac{\color{#004ec4}{h}}{\color{#00880a}{410}}$

$410xx$ $tan59°24’$	$=$	$h/410$ $xx410$	Multiply both sides by $410$

$410tan59°24’$	$=$	$h$
$h$	$=$	$410tan59°24’$

Simplify this further by evaluating

$tan59°24’$ using the calculator:

$1.$ Press $tan$

$2.$ Press

$59$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$24$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$1.690908$

Continue solving for

$h$ .

$tan59°24’=1.690908$

$h$	$=$	$410tan59°24’$
	$=$	$410times1.690908$
	$=$	$693.27$
	$=$	$693 m$	Rounded off to the nearest metre

$693 m$

Question 2 of 7

A wire of length

$5.2 m$ is attached to the top of a flagpole. It is inclined to the ground at an angle of

$68°32’$ . Find the height of the flagpole (

$h$ ) rounded off to

$1$ decimal place.

$h=$ (4.8) $m$

Incorrect

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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 function

DMS or $° ‘ ‘ ‘$

$=$ Degree/Minute/Second

$=$

$=$ Equal function

Notice that the scenario creates a triangle. Label it in reference to the given angle.

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

$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{5.2}$

Since we now have the opposite and hypotenuse values, we can use the

$sin$ ratio to find

$h$ .

$sin68°32’$	$=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

$sin68°32’$	$=$	$\frac{\color{#004ec4}{h}}{\color{#e85e00}{5.2}}$

$5.2xx$ $sin68°32’$	$=$	$h/5.2$ $xx5.2$	Multiply both sides by $5.2$

$5.2sin68°32’$	$=$	$h$
$h$	$=$	$5.2sin68°32’$

Simplify this further by evaluating

$sin68°32’$ using the calculator:

$1.$ Press $sin$

$2.$ Press

$68$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$32$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$0.9306306$

Continue solving for

$h$ .

$sin68°32’=0.9306306$

$h$	$=$	$5.2sin68°32’$
	$=$	$5.2times0.9306306$
	$=$	$4.839$
	$=$	$4.8 m$	Rounded off to $1$ decimal place

$4.8 m$

Question 3 of 7

A

$9.7 m$ ladder leans against a wall and makes an angle of

$61°9’$ with the ground. How far is the foot of the ladder from the base of the wall (

$x$ )? (

$1$ decimal place).

$x=$ (4.7) $m$

Incorrect

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Mute

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 a triangle. Label it in reference to the given angle.

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

$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{9.7}$

Since we now have the adjacent and hypotenuse values, we can use the

$cos$ ratio to find

$x$ .

$cos61°9’$	$=$	$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

$cos61°9’$	$=$	$\frac{\color{#00880a}{x}}{\color{#e85e00}{9.7}}$

$9.7xx$ $cos61°9’$	$=$	$x/9.7$ $xx9.7$	Multiply both sides by $9.7$

$9.7cos61°9’$	$=$	$x$
$x$	$=$	$9.7cos61°9’$

Simplify this further by evaluating

$cos61°9’$ using the calculator:

$1.$ Press $cos$

$2.$ Press

$61$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$9$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$0.482518$

Continue solving for

$x$ .

$cos61°9’=0.482518$

$x$	$=$	$9.7cos61°9’$
	$=$	$9.7times0.482518$
	$=$	$4.68$
	$=$	$4.7 m$	Rounded off to $1$ decimal place

$4.7 m$

Question 4 of 7

A building casts a shadow

$40 m$ long on the ground when the sun has an altitude (elevation) from the ground of

$59°52’$ . Calculate the height of the building (

$h$ ) to the nearest metre.

$h=$ (69) $m$

Incorrect

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Mute

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 a triangle. Label it in reference to the given angle.

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

$\color{#00880a}{\text{adjacent}}=\color{#00880a}{40}$

Since we now have the opposite and adjacent values, we can use the

$tan$ ratio to find

$h$ .

$tan59°52’$	$=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

$tan59°52’$	$=$	$\frac{\color{#004ec4}{h}}{\color{#00880a}{40}}$

$40xx$ $tan59°52’$	$=$	$h/40$ $xx40$	Multiply both sides by $40$

$40tan59°52’$	$=$	$h$
$h$	$=$	$40tan59°52’$

Simplify this further by evaluating

$tan59°52’$ using the calculator:

$1.$ Press $tan$

$2.$ Press

$59$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$52$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$1.7227797$

Continue solving for

$h$ .

$tan59°52’=1.7227797$

$h$	$=$	$40tan59°52’$
	$=$	$40times1.7227797$
	$=$	$68.911$
	$=$	$69 m$	Rounded off to the nearest metre

$69 m$

Question 5 of 7

Jacob is flying a kite. He is holding the string

$1.5 m$ above ground. The string is

$32 m$ long and makes an angle of

$58°19’$ . How high is the kite above ground

$(h_t) ?$ (

$1$ decimal place)

$h_t =$ (18.3) $m$

Incorrect

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Mute

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 a triangle. Label it in reference to the given angle.

Let

$h$ be the height of the kite from where Jacob is holding the string.

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

$\color{#e85e00}{\text{hypotenuse}}=\color{#e85e00}{32}$

Since we now have the adjacent and hypotenuse values, we can use the

$cos$ ratio to find

$h$ .

$cos58°19’$	$=$	$\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

$cos58°19’$	$=$	$\frac{\color{#00880a}{h}}{\color{#e85e00}{32}}$

$32xx$ $cos58°19’$	$=$	$h/32$ $xx32$	Multiply both sides by $32$

$32cos58°19’$	$=$	$h$
$h$	$=$	$32cos58°19’$

Simplify this further by evaluating

$cos58°19’$ using the calculator:

$1.$ Press $cos$

$2.$ Press

$58$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$19$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$0.525224$

Continue solving for

$h$ .

$cos58°19’=0.525224$

$h$	$=$	$32cos58°19’$
	$=$	$32times0.525224$
	$=$	$16.80717$

Finally, solve for

$h_t$ or the total height of the kite above ground by adding

$1.5$ to

$h$ .

$h_t$	$=$	$h+1.5$
	$=$	$16.80717+1.5$	Substitute $h$
	$=$	$18.30717$
	$=$	$18.3 m$	Rounded off to $1$ decimal place

$18.3 m$

Question 6 of 7

Find the height,

$h cm$ , of the trapezium. (

$1$ decimal place)

$h=$ (3.3) $cm$

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Mute

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

First, form a triangle using the trapezium.

Identify the parts of the triangle using the given values.

Reference angle:

Notice that from the given obtuse angle, a right angle was formed as the triangle was drawn.

Subtract the right angle from the obtuse angle to find the reference angle of the triangle.

$151°7′-90°=61°7’$

Opposite Side:

Subtract the two bases of the trapezium to find the opposite side to the reference angle in the triangle.

$18 cm-12 cm=6 cm$

Adjacent Side:

The side

$h$ of the trapezium is equal to the adjacent side to the reference angle of the triangle.

Since we now have the opposite and adjacent values, we can use the

$tan$ ratio to find

$h$ .

$\color{#004ec4}{\text{opposite}}=\color{#004ec4}{6}$

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

$tan61°7’$	$=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

$tan61°7’$	$=$	$\frac{\color{#004ec4}{6}}{\color{#00880a}{h}}$

$h$	$=$	$6/(tan61°7′)$	Swap the value on the left side and the denominator from the right side

Simplify this further by evaluating

$tan61°7’$ using the calculator:

$1.$ Press $tan$

$2.$ Press

$61$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$7$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$1.812743$

Continue solving for

$h$ .

$tan61°7’=1.812743$

$h$	$=$	$6/(tan61°7′)$

	$=$	$6/1.812743$

	$=$	$3.3099$
	$=$	$3.3 cm$	Rounded off to $1$ decimal place

$3.3 cm$

Question 7 of 7

A wedge-shaped ramp is set up for a car stunt on a movie set. The ramp has an angle of inclination of

$19°48’$ and a vertical rise of

$2.6 m$ . Find the length of the ramp (

$x$ ) to the nearest metre.

$x=$ (8) $m$

Incorrect

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Mute

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 a triangle. Label it in reference to the given angle.

$\color{#004ec4}{\text{opposite}}=\color{#004ec4}{2.6}$

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

Since we now have the opposite and hypotenuse values, we can use the

$sin$ ratio to find

$x$ .

$sin19°48’$	$=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

$sin19°48’$	$=$	$\frac{\color{#004ec4}{2.6}}{\color{#e85e00}{x}}$

$x$	$=$	$2.6/(sin19°48′)$	Swap the value on the left side and the denominator from the right side

Simplify this further by evaluating

$sin19°48’$ using the calculator:

$1.$ Press $sin$

$2.$ Press

$19$ and DMS or $° ‘ ‘ ‘$

$3.$ Press

$48$ and DMS or $° ‘ ‘ ‘$ again

$4.$ Press $=$

The result will be:

$0.338738$

Continue solving for

$x$ .

$sin19°48’=0.338738$

$x$	$=$	$2.6/(sin19°48′)$

	$=$	$2.6/0.338738$

	$=$	$7.67$
	$=$	$8 m$	Rounded off to the nearest metre

$8 m$