Using Bearings to Find Distance 3

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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

$B$ . Ship

$A$ traveled

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

$33°T$ and ship

$C$ traveled

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

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

$(AC)$ ?

Round your answer to the nearest kilometre

(222) $\text(km)$

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Pythagoras’ Theorem Formula

$a^2$

$=$ $b^2$

$+$ $c^2$

$a$ is the hypotenuse, and

$b$ and

$c$ are the two legs

First, find the value of the interior angle

$ABC$ by subtracting the two given bearings.

$/_ABC$	$=$	$123°-33°$
	$=$	$90°$

Since the triangle has an angle with a value of

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

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

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

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

$222 \text(km)$

Question 2 of 3

2. Question

Find the distance of point

$K$ to point

$M$

(10) $\text(km)$

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Pythagoras’ Theorem Formula

$a^2$

$=$ $b^2$

$+$ $c^2$

$a$ is the hypotenuse, and

$b$ and

$c$ are the two legs

First, find the value of the interior angle

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

$180°$ .

$/_KLM$	$=$	$180°-(35°+55°)$
	$=$	$180°-90°$
	$=$	$90°$

Since the triangle has an angle with a value of

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

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

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

$a^2$	$=$	$b^2$ $+$ $c^2$
$KM^2$	$=$	$6^2$ $+$ $8^2$	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