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Using Bearings to Find Distance

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Using Bearings to Find Distance 2

Using Bearings to Find Distance 2

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Question 1 of 4

A ship has traveled

68 km

$68 \text(km)$ northwest from point

P

$P$ with a true bearing of

319 ° T

$319°T$ . How far west (

x_{w}

$x_w$ ) has it traveled?

Round your answer to one decimal place

$x_{w} =$ $x_w=$ (44.6) $km$ $\text(km)$

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Sin Ratio

s i n = \frac{opposite}{hypotenuse}

$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio

c o s = \frac{adjacent}{hypotenuse}

$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio

t a n = \frac{opposite}{adjacent}

$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

A true bearing is an angle measured clockwise from the True North around to the required direction.

First, find the value of

θ

$theta$ below.

Notice that it is part of the given bearing, which also consists of

3

$3$ quadrants from the North line moving clockwise.

To find the value of

θ

$theta$ , simply subtract the measure of the angles from the three quadrants,

270 °

$270°$ , from the given bearing.

$θ$ $theta$	$=$ $=$	$319 ° - 270 °$ $319°-270°$
	$=$ $=$	$49 °$ $49°$

To solve for $x_{w}$ $x_w$ , we can use the known values of the hypotenuse and

θ = 49 °

$theta=49°$ .

Use

\cos

$cos$ to find the value of $x_{w}$ $x_w$ .

$\cos 49 °$ $cos49°$	$=$ $=$	$\frac{\color{#00880A}{x_w}}{\color{#e85e00}{\text{hypotenuse}}}$

$\cos 49 °$ $cos49°$	$=$ $=$	$\frac{\color{#00880A}{x_w}}{\color{#e85e00}{68}}$

$\cos 49 °$ $cos49°$ $\times 68$ $xx68$	$=$ $=$	$(\frac{x_{w}}{68})$ $((x_w)/68)$ $\times 68$ $xx68$	Multiply both sides by $68$ $68$

$68 \cos 49 °$ $68cos49°$	$=$ $=$	$x_{w}$ $x_w$
$x_{w}$ $x_w$	$=$ $=$	$68 \cos 49 °$ $68cos49°$

Using your calculator,

68 \cos 49 ° = 44.6

$68cos49°=44.6$ .

Therefore, the speedboat is

44.6 km

$44.6 \text(km)$ to the West.

x_{w} = 44.6 km

$x_w=44.6 \text(km)$

Question 2 of 4

A marathon runner runs

25.3 km

$25.3 \text(km)$ at a true bearing of

129 ° T

$129°T$ . Find how far east (

x_{e}

$x_e$ ) the runner has traveled from the starting point (

S

$S$ ).

Round your answer to one decimal place

$x_{e} =$ $x_e=$ (19.7) $km$ $\text(km)$

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Sin Ratio

s i n = \frac{opposite}{hypotenuse}

$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio

c o s = \frac{adjacent}{hypotenuse}

$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio

t a n = \frac{opposite}{adjacent}

$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

A true bearing is an angle measured clockwise from the True North around to the required direction.

First, find the value of

θ

$theta$ below.

Notice that it is part of the given bearing, which also consists of a right angle from the North line moving to the East line.

To find the value of

θ

$theta$ , simply subtract the value of a right angle,

90 °

$90°$ , from the given bearing.

$θ$ $theta$	$=$ $=$	$129 ° - 90 °$ $129°-90°$
	$=$ $=$	$39 °$ $39°$

To solve for $x_{e}$ $x_e$ , we can use the known values of the hypotenuse and

θ = 39 °

$theta=39°$ .

Use

\cos

$cos$ to find the value of $x_{e}$ $x_e$ .

$\cos 39 °$ $cos39°$	$=$ $=$	$\frac{\color{#00880A}{x_e}}{\color{#e85e00}{\text{hypotenuse}}}$

$\cos 39 °$ $cos39°$	$=$ $=$	$\frac{\color{#00880A}{x_e}}{\color{#e85e00}{25.3}}$

$\cos 39 °$ $cos39°$ $\times 25.3$ $xx25.3$	$=$ $=$	$(\frac{x_{e}}{25.3})$ $((x_e)/25.3)$ $\times 25.3$ $xx25.3$	Multiply both sides by $25.3$ $25.3$

$25.3 \cos 39 °$ $25.3cos39°$	$=$ $=$	$x_{e}$ $x_e$
$x_{e}$ $x_e$	$=$ $=$	$25.3 \cos 39 °$ $25.3cos39°$

Using your calculator,

25.3 \cos 39 ° = 19.7

$25.3cos39°=19.7$ .

Therefore, the runner runs

19.7 km

$19.7 \text(km)$ to the East.

x_{e} = 19.7 km

$x_e=19.7 \text(km)$

Question 3 of 4

A ship sails on a bearing of

200 ° T

$200° T$ towards

P

$P$ . If

P

$P$ is

80

$80$ nautical miles west from

O

$O$ , find how far the ship has sailed

(x)

$(x)$ .

Round your answer to one decimal place

(233.9) $nautical miles$ $\text(nautical miles)$

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>

Sin Ratio

s i n = \frac{opposite}{hypotenuse}

$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio

c o s = \frac{adjacent}{hypotenuse}

$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio

t a n = \frac{opposite}{adjacent}

$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

A true bearing is an angle measured clockwise from the True North around to the required direction.

First, find the value of angle

Q O P

$QOP$ .

Notice that adding angle

Q O P

$QOP$ to the given bearing

(200 °)

$(200°)$ will cover

3

$3$ quadrants, which is equal to

270 °

$270°$

To find the value of angle $Q O P$ $QOP$ , simply subtract the value of the bearing from

270 °

$270°$ .

$∠ Q O P$ $/_QOP$	$=$ $=$	$270 ° - 200 °$ $270°-200°$
	$=$ $=$	$70 °$ $70°$

To solve for $x$ $x$ , we can use the known values of the line adjacent to angle $Q O P$ $QOP$ .

Use

\cos

$cos$ to find the value of $x$ $x$ .

$\cos 70 °$ $cos70°$	$=$ $=$	$\frac{\color{#00880A}{\text{adjacent}}}{\color{#e85e00}{x}}$

$\cos 70 °$ $cos70°$	$=$ $=$	$\frac{\color{#00880A}{80}}{\color{#e85e00}{x}}$

$\cos 70 °$ $cos70°$ $\times x$ $xxx$	$=$ $=$	$(\frac{80}{x})$ $((80)/x)$ $\times x$ $xxx$	Multiply both sides by $x$ $x$

$x \cos 70 °$ $xcos70°$	$=$ $=$	$80$ $80$

$x \cos 70 °$ $xcos70°$ $\div \cos 70 °$ $divcos70°$	$=$ $=$	$80$ $80$ $\div \cos 70 °$ $divcos70°$	Divide both sides by $\cos 70 °$ $cos70°$
$x$ $x$	$=$ $=$	$\frac{80}{\cos 70 °}$ $(80)/(cos70°)$

Using your calculator,

\frac{80}{\cos 70 °} = 233.9

$(80)/(cos70°)=233.9$ .

Therefore, the boat traveled

233.9 nautical miles

$233.9 \text(nautical miles)$ in total.

233.9 nautical miles

$233.9 \text(nautical miles)$

Question 4 of 4

Bianca leaves her home and cycles due North for

12

$12$ km, then

7

$7$ km due West to go to the gym. How far is the gym from her home

(x)

$(x)$ ?

Round your answer to one decimal place

(13.9) $km$ $\text(km)$

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Pythagoras’ Theorem Formula

$a^{2}$ $a^2$

=

$=$ $b^{2}$ $b^2$

+

$+$ $c^{2}$ $c^2$

a

$a$ is the hypotenuse, and

b

$b$ and

c

$c$ are the two legs

Use the Pythagoras’ Theorem to solve for the distance

(

$($ $x$ $x$

)

$)$ .

$a$ $a$	$=$ $=$	$x$ $x$
$b$ $b$	$=$ $=$	$12 km$ $12 \text(km)$
$c$ $c$	$=$ $=$	$7 km$ $7 \text(km)$

$a^{2}$ $a^2$	$=$ $=$	$b^{2}$ $b^2$ $+$ $+$ $c^{2}$ $c^2$
$x^{2}$ $x^2$	$=$ $=$	$12^{2}$ $12^2$ $+$ $+$ $7^{2}$ $7^2$	Substitute known values
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{144 + 49}$ $sqrt(144+49)$	Get the square root of both sides
$x$ $x$	$=$ $=$	$\sqrt{193}$ $sqrt(193)$
$x$ $x$	$=$ $=$	$13.9 km$ $13.9 \text(km)$	Rounded to one decimal place

13.9 km

$13.9 \text(km)$