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Using Bearings to Find Distance

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Using Bearings to Find Distance 1

Using Bearings to Find Distance 1

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Question 1 of 4

Find the distance the speedboat has traveled due south

(A B)

$(AB)$

Round your answer to two decimal places

(164.54) $km due South$ $\text(km due South)$

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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 $C A B$ $CAB$ .

Notice that angle

C A B

$CAB$ is part of the given bearing, which also consists of a straight angle from the North line clockwise to the South line.

To find the value of angle

C A B

$CAB$ , simply subtract the measure of the straight angle,

180 °

$180°$ , from the given bearing.

$∠ C A B$ $/_CAB$	$=$ $=$	$210 ° - 180 °$ $210°-180°$
	$=$ $=$	$30 °$ $30°$

To solve for line $A B$ $AB$ , we can use the known values of the hypotenuse and angle $C A B$ $CAB$ .

Use

cos

$cos$ to find the value of line

A B

$AB$ .

$cos$ $cos$ $30 °$ $30°$	$=$ $=$	$\frac{\color{#00880A}{AB}}{\color{#e85e00}{\text{hypotenuse}}}$

$cos$ $cos$ $30 °$ $30°$	$=$ $=$	$\frac{\color{#00880A}{AB}}{\color{#e85e00}{\text{190}}}$

$cos 30 °$ $cos30°$ $\times 190$ $xx190$	$=$ $=$	$(\frac{A B}{190})$ $((AB)/190)$ $\times 190$ $xx190$	Multiply both sides by $190$ $190$

$190 cos 30 °$ $190cos30°$	$=$ $=$	$A B$ $AB$

Using your calculator,

190 cos 30 ° = 164.54

$190cos30°=164.54$ .

Therefore, the speedboat is

164.54 km

$164.54 \text(km)$ due South.

164.54 km due South

$164.54 \text(km due South)$

Question 2 of 4

Find the distance (

d

$d$ ) the boat has traveled from the port (

P

$P$ )

Round your answer to the nearest kilometre

$d =$ $d=$ (25) $km$ $\text(km)$

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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.

To solve for line $d$ $d$ , we can use the known values of the line adjacent to the bearing.

Use

cos

$cos$ to find the value of line $d$ $d$ .

$cos 48 °$ $cos48°$	$=$ $=$	$\frac{\color{#00880A}{\text{adjacent}}}{\color{#e85e00}{d}}$

$cos 48 °$ $cos48°$	$=$ $=$	$\frac{\color{#00880A}{17}}{\color{#e85e00}{d}}$

$cos 48 °$ $cos48°$ $\times d$ $xxd$	$=$ $=$	$(\frac{17}{d})$ $((17)/d)$ $\times d$ $xxd$	Multiply both sides by $d$ $d$

$d cos 48 °$ $dcos48°$	$=$ $=$	$17$ $17$

$d cos 48 °$ $dcos48°$ $\div cos 48 °$ $divcos48°$	$=$ $=$	$17$ $17$ $\div cos 48 °$ $divcos48°$	Divide both sides by $cos 48 °$ $cos48°$

$d$ $d$	$=$ $=$	$\frac{17}{cos 48 °}$ $17/(cos48°)$	Divide both sides by $cos 48 °$ $cos48°$

Using your calculator,

\frac{17}{cos 48 °} = 25.46

$17/(cos48°)=25.46$ .

Rounding it to the nearest kilometre, the boat traveled

25 km

$25 \text(km)$ from the port.

25 km

$25 \text(km)$

Question 3 of 4

A ship has traveled

72 km

$72 \text(km)$ North West from the port (

O

$O$ ). How far North (

y_{n}

$y_n$ ) of the starting point has it traveled?

Round your answer to one decimal place

$y_{n} =$ $y_n=$ (50.9) $km$ $\text(km)$

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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.

To solve for the line $y_{n}$ $y_n$ , we can use the known values of the hypotenuse and the given angle

(45 °)

$(45°)$ .

Remember

N W

$NW$ (North West) means

45 °

$45°$ .

Use

cos

$cos$ to find the value of line $y_{n}$ $y_n$ .

$cos 45 °$ $cos45°$	$=$ $=$	$\frac{\color{#00880A}{{y_{n}}}}{\color{#e85e00}{\text{hypotenuse}}}$

$cos 45 °$ $cos45°$	$=$ $=$	$\frac{\color{#00880A}{{y_{n}}}}{\color{#e85e00}{72}}$

$cos 45 °$ $cos45°$ $\times 72$ $xx72$	$=$ $=$	$(\frac{y_{n}}{72})$ $((y_n)/72)$ $\times 72$ $xx72$	Multiply both sides by $72$ $72$

$72 cos 45 °$ $72cos45°$	$=$ $=$	$y_{n}$ $y_n$
$y_{n}$ $y_n$	$=$ $=$	$72 cos 45 °$ $72cos45°$

Using your calculator,

72 cos 45 ° = 50.9

$72cos45°=50.9$ .

Therefore, the boat traveled

50.9 km

$50.9 \text(km)$ to the North.

y_{n} = 50.9 km

$y_n=50.9 \text(km)$

Question 4 of 4

A ship has traveled

51 km

$51 \text(km)$ from the port (

P

$P$ ) with a bearing of

N 67 ° E

$N 67°E$ . How far east (

x_{e}

$x_e$ ) has it traveled?

Round your answer to one decimal place

$x_{e} =$ $x_e=$ (46.9) $km$ $\text(km)$

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Sin Ratio

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

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

Cos Ratio

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

Tan Ratio

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

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

To solve for $x_e$ , we can use the known values of the hypotenuse and the given angle,

$(67°)$ .

Use

$sin$ to find the value of line

$x_e$ .

$sin67°$	$=$	$\frac{\color{#004ec4}{x_e}}{\color{#e85e00}{\text{hypotenuse}}}$

$sin67°$	$=$	$\frac{\color{#004ec4}{x_e}}{\color{#e85e00}{51}}$

$sin67°$ $xx51$	$=$	$((x_e)/51)$ $xx51$	Multiply both sides by $51$

$51sin67°$	$=$	$x_e$
$x_e$	$=$	$51sin67°$

Using your calculator,

$51sin67°=46.9$ .

Therefore, the boat traveled

$46.9 \text(km)$ to the East.

$x_e=46.9 \text(km)$