Working with Radial Surveys 4

Years

Year 11

Bearings

Working with Radial Surveys

Working with Radial Surveys 4

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Question 1 of 4

1. Question
From the radial survey below, find $∠ P O Q$ $anglePOQ$ :
- $∠ P O Q =$ $anglePOQ=$ (137) $°$ $°$
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A radial survey is a tool used for land and seafloor mapping. Each corner of the area being measured is connected to a central point.

Since $∠ P O Q$ $anglePOQ$ is the sum of $∠ P O N$ $anglePON$ and $∠ N O Q$ $angleNOQ$ , start with solving for $∠ P O N$ $anglePON$ .

Knowing that a full revolution from $N$ $N$ measures $360 °$ $360°$ , subtract the bearing of $P$ $P$ to get $∠ P O N$ $anglePON$

$∠ P O N$ $anglePON$ $=$ $=$ $360 ° - ∠ P$ $360°-angleP$

$=$ $=$ $360 ° - 275 °$ $360°-275°$ Substitute the values

$=$ $=$ $85 °$ $85°$

Finally, proceed with adding $∠ P O N$ $anglePON$ and $∠ N O Q$ $angleNOQ$ .

$∠ P O Q$ $anglePOQ$ $=$ $=$ $∠ P O N + ∠ N O Q$ $anglePON+angleNOQ$

$=$ $=$ $85 ° + 52 °$ $85°+52°$ Substitute the values

$=$ $=$ $137 °$ $137°$

$137 °$ $137°$

Question 2 of 4

From the radial survey below, find the area of

△ P O Q

$trianglePOQ$ to the nearest square metre.

Area of $△ P O Q =$ $trianglePOQ=$ (1590) $m^{2}$ $m^2$

Question 3 of 4

3. Question
From the radial survey below, find $∠ S O R$ $angleSOR$ :
- $∠ S O R =$ $angleSOR=$ (68) $°$ $°$
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A radial survey is a tool used for land and seafloor mapping. Each corner of the area being measured is connected to a central point.

Notice that $∠ S O R$ $angleSOR$ is the difference between the bearings of $S$ $S$ and $R$ $R$ .

Subtract the bearing of $R$ $R$ from the bearing of $S$ $S$ .

$∠ S O R$ $angleSOR$ $=$ $=$ $∠ S - ∠ R$ $angleS-angleR$

$=$ $=$ $201 ° - 133 °$ $201°-133°$ Substitute the values

$=$ $=$ $68 °$ $68°$

$68 °$ $68°$

Question 4 of 4

From the radial survey below, find the length of

S R

$SR$ to the nearest metre.

$S R =$ $SR=$ (67) $m$ $m$

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Cosine Law

a^{2} = b^{2} + c^{2} - 2 b c cos A

$\color{#007DDC}{a}^2=\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$

where:
$a$ $a$ is the side opposite angle $A$ $A$
$b$ $b$ is the side opposite angle $B$ $B$
$c$ $c$ is the side opposite angle $C$ $C$

Since

2

$2$ sides are given together with an angle between them, use the Cosine Law.

First, label the triangle according to the Cosine Law.

Substitute the three known values to the Cosine Law to find the length of side

S R

$SR$ or $a$ $a$ .

From labelling the triangle, we know that the known values are those with labels

A, b

$A, b$ and

c

$c$ .

A = 68 °

$A=68°$

b = 61 m

$b=61m$

c = 58 m

$c=58m$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{61}^2+\color{#9a00c7}{58}^2-2(\color{#00880A}{61})(\color{#9a00c7}{58})\cos\color{#007DDC}{68°}$	Substitute the values
$a^{2}$ $a^2$	$=$ $=$	$3721 + 3364 - 7076 cos 68 °$ $3721+3364-7076cos68°$	Evaluate $cos$ $cos$ $68$ $68$ on your calculator
$a^{2}$ $a^2$	$=$ $=$	$7085 - 7076 (0.37460659)$ $7085-7076(0.37460659)$
$a^{2}$ $a^2$	$=$ $=$	$7085 - 2650.7162$ $7085-2650.7162$
$a^{2}$ $a^2$	$=$ $=$	$4434.283769$ $4434.283769$
$\sqrt{a^{2}}$ $sqrt(a^2)$	$=$ $=$	$\sqrt{4434.283769}$ $sqrt4434.283769$	Take the square root of both sides
$a$ $a$	$=$ $=$	$66.59 m$ $66.59m$
$a$ $a$ or $B C$ $BC$	$=$ $=$	$67 m$ $67m$	Round off to the nearest metre

67 m

$67m$

$A_{△}$ $A_triangle$	$=$ $=$	$\frac{1}{2}$ $1/2$ $a$ $a$ $b$ $b$ $sin$ $sin$ $C$ $C$

	$=$ $=$	$\frac{1}{2} ($ $1/2($ $63$ $63$ $) ($ $)($ $74$ $74$ $) sin$ $)sin$ $137 °$ $137°$	Evaluate $sin$ $sin$ $137$ $137$ on your calculator

	$=$ $=$	$2331 \times 0.68199836 °$ $2331times0.68199836°$	Simplify

	$=$ $=$	$1589.74$ $1589.74$

	$=$ $=$	$1590 m^{2}$ $1590m^2$	Round off to the nearest metre

Working with Radial Surveys 4

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Quiz summary

Information

1. Question

Hint

Correct!

2. Question

Hint

Exceptional!

Area of a Non-Right Angled Triangle

3. Question

Hint

Keep Going!

4. Question

Hint

Fantastic!

Cosine Law

Quizzes

$∠ P O N$ $anglePON$	$=$ $=$	$360 ° - ∠ P$ $360°-angleP$
	$=$ $=$	$360 ° - 275 °$ $360°-275°$	Substitute the values
	$=$ $=$	$85 °$ $85°$

$∠ P O Q$ $anglePOQ$	$=$ $=$	$∠ P O N + ∠ N O Q$ $anglePON+angleNOQ$
	$=$ $=$	$85 ° + 52 °$ $85°+52°$	Substitute the values
	$=$ $=$	$137 °$ $137°$

$∠ S O R$ $angleSOR$	$=$ $=$	$∠ S - ∠ R$ $angleS-angleR$
	$=$ $=$	$201 ° - 133 °$ $201°-133°$	Substitute the values
	$=$ $=$	$68 °$ $68°$