Working with Radial Surveys 2

Years

Year 12

Radial Surveys

Working with Radial Surveys

Working with Radial Surveys 2

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Question 1 of 4

1. Question
From the radial survey below, find the following:
- $(i) ∠ B O C =$ $(i) angleBOC=$ (74) $°$ $°$
  
  $(i i) ∠ B O A =$ $(ii) angleBOA=$ (128) $°$ $°$
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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.

$(i)$ $(i)$ $∠ B O C$ $angleBOC$

Notice that $∠ B O C$ $angleBOC$ is the difference between the bearings of $B$ $B$ and $C$ $C$ .

Subtract the bearing of $B$ $B$ from the bearing of $C$ $C$ .

$∠ B O C$ $angleBOC$ $=$ $=$ $∠ C - ∠ B$ $angleC-angleB$

$=$ $=$ $139 ° - 65 °$ $139°-65°$ Substitute the values

$=$ $=$ $74 °$ $74°$

$(i i)$ $(ii)$ $∠ B O A$ $angleBOA$

Notice that $∠ B O A$ $angleBOA$ is the sum of the bearing of $B$ $B$ and $∠ N O A$ $angleNOA$ .

Add the bearing of $B$ $B$ to $∠ N O A$ $angleNOA$ .

$∠ B O A$ $angleBOA$ $=$ $=$ $∠ N O A + ∠ B$ $angleNOA+angleB$

$=$ $=$ $63 ° + 65 °$ $63°+65°$ Substitute the values

$=$ $=$ $128 °$ $128°$

$(i) ∠ B O C = 74 °$ $(i)/_BOC=74°$

$(i i) ∠ B O A = 128 °$ $(ii)/_BOA=128°$

Question 2 of 4

From the radial survey below, find the length of

B C

$BC$ .

Round your answer to

2

$2$ decimal places

$B C =$ $BC=$ (60.83) $m$ $m$

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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

B C

$BC$ 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 = 74 °

$A=74°$

b = 49 m

$b=49m$

c = 52 m

$c=52m$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{49}^2+\color{#9a00c7}{52}^2-2(\color{#00880A}{49})(\color{#9a00c7}{52})\cos\color{#007DDC}{74°}$	Substitute the values
$a^{2}$ $a^2$	$=$ $=$	$5105 - 5096 cos 74 °$ $5105-5096cos74°$	Evaluate $cos$ $cos$ $74$ $74$ on your calculator
$a^{2}$ $a^2$	$=$ $=$	$5105 - 5096 (0.275637)$ $5105-5096(0.275637)$
$a^{2}$ $a^2$	$=$ $=$	$5105 - 1404.64797$ $5105-1404.64797$
$a^{2}$ $a^2$	$=$ $=$	$3700.35203$ $3700.35203$
$\sqrt{a^{2}}$ $sqrt(a^2)$	$=$ $=$	$\sqrt{3700.35203}$ $sqrt3700.35203$	Take the square root of both sides
$a$ $a$	$=$ $=$	$60.8305 m$ $60.8305m$
$a$ $a$ or $B C$ $BC$	$=$ $=$	$60.83 m$ $60.83m$	Round off to $2$ $2$ decimal places

60.83 m

$60.83m$

Question 3 of 4

From the radial survey below, find the length of

B A

$BA$ .

Round your answer to

2

$2$ decimal places

$B A =$ $BA=$ (92.58) $m$ $m$

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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

B A

$BA$ or $c$ $c$ .

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

C, a

$C, a$ and

b

$b$ .

C = 128 °

$C=128°$

b = 51 m

$b=51m$

a = 52 m

$a=52m$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#9a00c7}{c}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#007DDC}{a}^2-2\color{#00880A}{b}\color{#007DDC}{a}\cos\color{#9a00c7}{C}$	Rewrite the formula according to the given values
$\color{#9a00c7}{c}^2$	$=$	$\color{#00880A}{51}^2+\color{#007DDC}{52}^2-2(\color{#00880A}{51})(\color{#007DDC}{52})\cos\color{#9a00c7}{128°}$	Substitute the values
$c^2$	$=$	$5305-5304cos128°$	Evaluate $cos$ $128$ on your calculator
$c^2$	$=$	$5305-5304(-0.615661475)$
$c^2$	$=$	$5305+3265.4685$
$c^2$	$=$	$8570.4685$
$sqrt(c^2)$	$=$	$sqrt8570.4685$	Take the square root of both sides
$c$	$=$	$92.5768m$
$c$ or $BA$	$=$	$92.58m$	Round off to $2$ decimal places

$92.58m$

Question 4 of 4

4. Question
Find the perimeter of the field shown by this radial survey.
- (253.41) $m$
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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.

To find the perimeter of the field, simply add the lengths of the sides of the field.

$AO=51m$

$OC=49m$

$BC=60.83m$

$BA=92.58m$

Perimeter $=$ $AO+OC+BC+BA$

$=$ $51+49+60.83+92.58$ Substitute the values

$=$ $253.41m$

$253.41m$

Working with Radial Surveys 2

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Quiz summary

Information

1. Question

Hint

Fantastic!

2. Question

Hint

Well Done!

Cosine Law

3. Question

Hint

Excellent!

Cosine Law

4. Question

Hint

Correct!

Quizzes

$∠ B O C$ $angleBOC$	$=$ $=$	$∠ C - ∠ B$ $angleC-angleB$
	$=$ $=$	$139 ° - 65 °$ $139°-65°$	Substitute the values
	$=$ $=$	$74 °$ $74°$

$∠ B O A$ $angleBOA$	$=$ $=$	$∠ N O A + ∠ B$ $angleNOA+angleB$
	$=$ $=$	$63 ° + 65 °$ $63°+65°$	Substitute the values
	$=$ $=$	$128 °$ $128°$

Perimeter	$=$	$AO+OC+BC+BA$
	$=$	$51+49+60.83+92.58$	Substitute the values
	$=$	$253.41m$