Using Bearings to Find Distance 3

Years

Year 9

Bearings

Using Bearings to Find Distance

Using Bearings to Find Distance 3

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Question 1 of 3

1. Question

Two ships set sail straight away from point

B

$B$ . Ship

A

$A$ traveled

130 km

$130 \text(km)$ with a bearing of

33 ° T

$33°T$ and ship

C

$C$ traveled

180 km

$180 \text(km)$ with a bearing of

123 ° T

$123°T$ . How far away are the two ships

(A C)

$(AC)$ ?

Round your answer to the nearest kilometre

(222) $km$ $\text(km)$

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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

First, find the value of the interior angle

A B C

$ABC$ by subtracting the two given bearings.

$∠ A B C$ $/_ABC$	$=$ $=$	$123 ° - 33 °$ $123°-33°$
	$=$ $=$	$90 °$ $90°$

Since the triangle has an angle with a value of

90 °

$90°$ , that means it is a right triangle.

Use the Pythagoras’ Theorem to solve for the distance $A C$ $AC$ .

$a$ $a$	$=$ $=$	$A C$ $AC$
$b$ $b$	$=$ $=$	$A B$ $AB$	$=$ $=$	$130 km$ $130 \text(km)$
$c$ $c$	$=$ $=$	$B C$ $BC$	$=$ $=$	$180 km$ $180 \text(km)$

$a^{2}$ $a^2$	$=$ $=$	$b^{2}$ $b^2$ $+$ $+$ $c^{2}$ $c^2$
$A C^{2}$ $AC^2$	$=$ $=$	$130^{2}$ $130^2$ $+$ $+$ $180^{2}$ $180^2$	Substitute known values
$\sqrt{A C^{2}}$ $sqrt(AC^2)$	$=$ $=$	$\sqrt{16900 + 32400}$ $sqrt(16900+32400)$	Get the square root of both sides
$A C$ $AC$	$=$ $=$	$\sqrt{49300}$ $sqrt(49300)$
$A C$ $AC$	$=$ $=$	$222 km$ $222 \text(km)$	Rounded to the nearest kilometre

222 km

$222 \text(km)$

Question 2 of 3

2. Question

Find the distance of point

K

$K$ to point

M

$M$

(10) $km$ $\text(km)$

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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

First, find the value of the interior angle

K L M

$KLM$ by subtracting the sum of the two given bearings from the angle of a straight line, which is

180 °

$180°$ .

$∠ K L M$ $/_KLM$	$=$ $=$	$180 ° - (35 ° + 55 °)$ $180°-(35°+55°)$
	$=$ $=$	$180 ° - 90 °$ $180°-90°$
	$=$ $=$	$90 °$ $90°$

Since the triangle has an angle with a value of

90 °

$90°$ , that means it is a right triangle.

Use the Pythagoras’ Theorem to solve for the distance $K M$ $KM$ .

$a$ $a$	$=$ $=$	$K M$ $KM$
$b$ $b$	$=$ $=$	$K L$ $KL$	$=$ $=$	$6 km$ $6 \text(km)$
$c$	$=$	$LM$	$=$	$8 \text(km)$

$a^2$	$=$	$b^2$ $+$ $c^2$
$KM^2$	$=$	$6^2$ $+$ $8^2$	Substitute known values
$sqrt(KM^2)$	$=$	$sqrt(36+64)$	Get the square root of both sides
$KM$	$=$	$sqrt(100)$
$KM$	$=$	$10 \text(km)$

$10 \text(km)$

Question 3 of 3

3. Question

A yacht is located

$43°T$ from port

$X$ and

$302°T$ from port

$Z$ . Port

$Z$ is

$180 \text(km)$ directly east of port

$X$ . Find the distance of the yacht from port

$X$ .

Round your answer to two decimal places

(97.17) $\text(km)$

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Sine Rule

$x/(sinX)=y/(sinY)=z/(sinZ)$

Remember

Uppercase letters represent angles in the triangle
Lowercase letters represent the side lengths

First, find the value of $/_ X$ .

Notice that $/_ X$ and the bearing of

$Y$ from

$X$

$(43°)$ forms a complementary angle.

Since complementary angles are equal to

$90°$ , we can simply subtract

$43°$ from

$90°$ to find $/_ X$ .

$/_X$	$=$	$90°-43°$
	$=$	$47°$

Next, find the value of $/_ Z$ .

Notice that the true bearing of

$Y$ from

$Z$ includes $/_ Z$ along with

$3$ quadrants.

We can simply subtract the value of three quadrants, which is

$270°$ , from the given true bearing to find angle

$Z$ .

$/_Z$	$=$	$302°-270°$
	$=$	$32°$

Then, find the value of angle $/_ Y$ by subtracting the sum of $/_ X$ and $/_ Z$ from the total interior angle of a triangle, which is

$180°$ .

$/_Y$	$=$	$180°-($ $47°$ $+$ $32°$ $)$
	$=$	$180°-79°$
	$=$	$101°$

Since we know the values of angle

$Y$ , angle

$Z$ , and line

$XZ$ , we can use the sine law to solve for the distance of

$Y$ from

$X$ .

$Y$	$=$	$101°$
$y$	$=$	$180 \text(km)$
$Z$	$=$	$32°$

$y/(sinY)$	$=$	$z/(sinZ)$

$(180)/(sin101°)$	$=$	$z/(sin32°)$	Substitute known values

$zxxsin101°$	$=$	$180xxsin32°$	Cross-multiply
$zxxsin101°$ $divsin101°$	$=$	$180xxsin32°$ $divsin101°$	Divide both sides by $sin101°$

$z$	$=$	$(180xxsin32°)/(sin101°)$

Using a calculator,

$(180xxsin32°)/(sin101°)=97.17 \text(km)$ , rounded to two decimal places

$97.17 \text(km)$

Quizzes

Compass Bearings and True Bearings 1
Compass Bearings and True Bearings 2
Solving for Bearings
Bearings from Opposite Direction
Using Bearings to Find Distance 1
Using Bearings to Find Distance 2
Using Bearings to Find Distance 3
Using Bearings and Distances to Find Angles