Using Bearings to Find Distance 3

Topics

Trigonometry

Bearings

Using Bearings to Find Distance

Using Bearings to Find Distance 3

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

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) $km$ $\text(km)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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,

\frac{180 \times sin 32 °}{sin 101 °} = 97.17 km

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

97.17 km

$97.17 \text(km)$

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

Try VividMath Premium to unlock full access

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

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

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