Using Bearings to Find Distance 3

Years

Year 11

Bearings

Using Bearings to Find Distance

Using Bearings to Find Distance 3

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Question 1 of 3

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) $km$ $\text(km)$

Question 2 of 3

Find the distance of point

K

$K$ to point

M

$M$

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

Question 3 of 3

A yacht is located

43 ° T

$43°T$ from port

X

$X$ and

302 ° T

$302°T$ from port

Z

$Z$ . Port

Z

$Z$ is

180 km

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

X

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

X

$X$ .

Round your answer to two decimal places

(97.17) $km$ $\text(km)$

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Sine Rule

\frac{x}{sin X} = \frac{y}{sin Y} = \frac{z}{sin Z}

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

Remember

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

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

Notice that $∠ X$ $/_ X$ and the bearing of

Y

$Y$ from

X

$X$

(43 °)

$(43°)$ forms a complementary angle.

Since complementary angles are equal to

90 °

$90°$ , we can simply subtract

43 °

$43°$ from

90 °

$90°$ to find $∠ X$ $/_ X$ .

$∠ X$ $/_X$	$=$ $=$	$90 ° - 43 °$ $90°-43°$
	$=$ $=$	$47 °$ $47°$

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

Notice that the true bearing of

Y

$Y$ from

Z

$Z$ includes $∠ Z$ $/_ Z$ along with

3

$3$ quadrants.

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

270 °

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

Z

$Z$ .

$∠ Z$ $/_Z$	$=$ $=$	$302 ° - 270 °$ $302°-270°$
	$=$ $=$	$32 °$ $32°$

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

180 °

$180°$ .

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

Since we know the values of angle

Y

$Y$ , angle

Z

$Z$ , and line

X Z

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

Y

$Y$ from

X

$X$ .

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

$\frac{y}{sin Y}$ $y/(sinY)$	$=$ $=$	$\frac{z}{sin Z}$ $z/(sinZ)$

$\frac{180}{sin 101 °}$ $(180)/(sin101°)$	$=$ $=$	$\frac{z}{sin 32 °}$ $z/(sin32°)$	Substitute known values

$z \times sin 101 °$ $zxxsin101°$	$=$ $=$	$180 \times sin 32 °$ $180xxsin32°$	Cross-multiply
$z \times sin 101 °$ $zxxsin101°$ $\div sin 101 °$ $divsin101°$	$=$ $=$	$180 \times sin 32 °$ $180xxsin32°$ $\div sin 101 °$ $divsin101°$	Divide both sides by $sin 101 °$ $sin101°$

$z$ $z$	$=$ $=$	$\frac{180 \times sin 32 °}{sin 101 °}$ $(180xxsin32°)/(sin101°)$

Using a calculator,

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

$97.17 \text(km)$

$a^{2}$ $a^2$	$=$ $=$	$b^{2}$ $b^2$ $+$ $+$ $c^{2}$ $c^2$
$A C^{2}$ $AC^2$	$=$ $=$	$130^{2}$ $130^2$ $+$ $+$ $180^{2}$ $180^2$	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

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

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

Using Bearings to Find Distance 3

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Quiz summary

Information

1. Question

Hint

Excellent!

Pythagoras’ Theorem Formula

2. Question

Hint

Exceptional!

Pythagoras’ Theorem Formula

3. Question

Hint

Correct!

Sine Rule

Remember

Quizzes

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

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