Working with Radial Surveys 1

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Year 12

Radial Surveys

Working with Radial Surveys

Working with Radial Surveys 1

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Question 1 of 6

From the radial survey below, find the area of

△ A O B

$triangleAOB$ :

Round your answer to

3

$3$ decimal places

Area of $△ A O B =$ $triangleAOB=$ (23.909) $m^{2}$ $m^2$

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Area of a Non-Right Angled Triangle

A_{△} = \frac{1}{2}

$A_triangle=1/2$ $a$ $a$ $b$ $b$

\sin

$sin$ $C$ $C$

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 $C$ $C$

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.

First, identify the known values of the triangle

A O B

$AOB$ .

Now, substitute the known values to the formula and solve for the area.

a = 8 m

$a=8m$

b = 6 m

$b=6m$

C = 85 °

$C=85°$

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

	$=$ $=$	$\frac{1}{2} ($ $1/2($ $8$ $8$ $) ($ $)($ $6$ $6$ $) \sin$ $)sin$ $85 °$ $85°$	Substitute the values

	$=$ $=$	$\frac{1}{2} (48) \sin 85 °$ $1/2(48)sin85°$	Evaluate $\sin$ $sin$ $85$ $85$ on your calculator
	$=$ $=$	$24 \times 0.9961947$ $24times0.9961947$
	$=$ $=$	$23.909 m^{2}$ $23.909m^2$	Round off to $3$ $3$ decimal places

23.909 m^{2}

$23.909m^2$

Question 2 of 6

From the radial survey below, find the area of

△ B O C

$triangleBOC$ :

Round your answer to

3

$3$ decimal places

Area of $△ B O C =$ $triangleBOC=$ (12.586) $m^{2}$ $m^2$

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Area of a Non-Right Angled Triangle

A_{△} = \frac{1}{2}

$A_triangle=1/2$ $a$ $a$ $b$ $b$

\sin

$sin$ $C$ $C$

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 $C$ $C$

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.

First, identify the known values of the triangle

B O C

$BOC$ .

Now, substitute the known values to the formula and solve for the area.

a = 8 m

$a=8m$

b = 5 m

$b=5m$

C = 141 °

$C=141°$

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

	$=$ $=$	$\frac{1}{2} ($ $1/2($ $8$ $8$ $) ($ $)($ $5$ $5$ $) \sin$ $)sin$ $141 °$ $141°$	Substitute the values

	$=$ $=$	$\frac{1}{2} (40) \sin 141 °$ $1/2(40)sin141°$	Evaluate $\sin$ $sin$ $141$ $141$ on your calculator
	$=$ $=$	$20 \times 0.62932$ $20times0.62932$
	$=$ $=$	$12.586 m^{2}$ $12.586m^2$	Round off to $3$ $3$ decimal places

12.586 m^{2}

$12.586m^2$

Question 3 of 6

From the radial survey below, find the following:

$(i)$ $(i)$ Area of $△ A O C =$ $triangleAOC=$ (10.790) $m^{2} (3$ $m^2 (3$ decimal places $)$ $)$

$(i i)$ $(ii)$ Total Area of $△ A B C =$ $triangleABC=$ (47.3) $m^{2} (1$ $m^2 (1$ decimal place $)$ $)$

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Area of a Non-Right Angled Triangle

A_{△} = \frac{1}{2}

$A_triangle=1/2$ $a$ $a$ $b$ $b$

\sin

$sin$ $C$ $C$

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 $C$ $C$

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)

$(i)$ Area of

△ A O C

$triangleAOC$

First, identify the known values of the triangle

A O C

$AOC$ .

Now, substitute the known values to the formula and solve for the area.

a = 5 m

$a=5m$

b = 6 m

$b=6m$

C = 134 °

$C=134°$

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

	$=$ $=$	$\frac{1}{2} ($ $1/2($ $5$ $5$ $) ($ $)($ $6$ $6$ $) \sin$ $)sin$ $134 °$ $134°$	Substitute the values

	$=$ $=$	$\frac{1}{2} (30) \sin 134 °$ $1/2(30)sin134°$	Evaluate $\sin$ $sin$ $134$ $134$ on your calculator
	$=$ $=$	$15 \times 0.7193398$ $15times0.7193398$
	$=$ $=$	$10.790 m^{2}$ $10.790m^2$	Round off to $3$ $3$ decimal places

(i i)

$(ii)$ Total Area of

△ A B C

$triangleABC$

Finally, add the area of triangles

A O B

$AOB$ (from Question

1

$1$ ),

B O C

$BOC$ (from Question

2

$2$ ) and

A O C

$AOC$ .

Total Area	$=$ $=$	$△ A O B + △ B O C + △ A O C$ $triangleAOB+triangleBOC+triangleAOC$
	$=$ $=$	$23.909 + 12.586 + 10.790$ $23.909+12.586+10.790$	Substitute the values
	$=$ $=$	$47.285$ $47.285$	Use the calculator
	$=$ $=$	$47.3 m^{2}$ $47.3m^2$	Rounded off to $1$ $1$ decimal place

(i) 10.790 m^{2}

$(i) 10.790m^2$

(i i) 47.3 m^{2}

$(ii) 47.3m^2$

Question 4 of 6

4. Question
From the radial survey below, find $∠ B O C$ $angleBOC$ :
- $∠ B O C =$ $angleBOC=$ (106) $°$ $°$
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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 $∠ B O C$ $angleBOC$ is the difference between the bearings of $B$ $B$ and $C$ $C$ .

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

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

$=$ $=$ $258 ° - 152 °$ $258°-152°$ Substitute the values

$=$ $=$ $106 °$ $106°$

$106 °$ $106°$

Question 5 of 6

From the radial survey below, find the area of

△ B O C

$triangleBOC$ to the nearest square metre.

Area of $△ B O C =$ $triangleBOC=$ (943) $m^{2}$ $m^2$

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Area of a Non-Right Angled Triangle

A_{△} = \frac{1}{2}

$A_triangle=1/2$ $a$ $a$ $b$ $b$

\sin

$sin$ $C$ $C$

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 $C$ $C$

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.

First, identify the known values of the triangle

B O C

$BOC$ .

Now, substitute the known values to the formula and solve for the area.

a = 37 m

$a=37m$

b = 53 m

$b=53m$

C = 106 °

$C=106°$

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

	$=$ $=$	$\frac{1}{2} ($ $1/2($ $37$ $37$ $) ($ $)($ $53$ $53$ $) \sin$ $)sin$ $106 °$ $106°$	Evaluate $\sin$ $sin$ $106$ $106$ on your calculator

	$=$ $=$	$980.5 \times 0.9612617 °$ $980.5times0.9612617°$	Simplify
	$=$ $=$	$942.517$ $942.517$
	$=$ $=$	$943 m^{2}$ $943m^2$	Round off to the nearest square metre

943 m^{2}

$943m^2$

Question 6 of 6

From the radial survey below, find the length of

B C

$BC$ to the nearest metre.

$B C =$ $BC=$ (73) $m$ $m$

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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 = 106 °

$A=106°$

b = 37 m

$b=37m$

c = 53 m

$c=53m$

$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{b}^2+\color{#9a00c7}{c}^2-2\color{#00880A}{b}\color{#9a00c7}{c}\cos\color{#007DDC}{A}$
$\color{#007DDC}{a}^2$	$=$ $=$	$\color{#00880A}{37}^2+\color{#9a00c7}{53}^2-2(\color{#00880A}{37})(\color{#9a00c7}{53})\cos\color{#007DDC}{106°}$	Evaluate $\cos$ $cos$ $106$ $106$ on your calculator
$a^{2}$ $a^2$	$=$ $=$	$1369 + 2809 - 3922 (- 0.275637)$ $1369+2809-3922(-0.275637)$	Simplify
$a^{2}$ $a^2$	$=$ $=$	$4178 + 1081.0499$ $4178+1081.0499$
$a^{2}$ $a^2$	$=$ $=$	$5259.0499$ $5259.0499$
$\sqrt{a^{2}}$ $sqrt(a^2)$	$=$ $=$	$\sqrt{5259.0499}$ $sqrt5259.0499$	Take the square root of both sides
$a$ $a$	$=$ $=$	$72.519 m$ $72.519m$
$a$ $a$ or $B C$ $BC$	$=$ $=$	$73 m$ $73m$	Round off to the nearest metre

73 m

$73m$

Working with Radial Surveys 1

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Quiz summary

Information

1. Question

Hint

Great Work!

Area of a Non-Right Angled Triangle

2. Question

Hint

Correct!

Area of a Non-Right Angled Triangle

3. Question

Hint

Well Done!

Area of a Non-Right Angled Triangle

4. Question

Hint

Good Job!

5. Question

Hint

Great Work!

Area of a Non-Right Angled Triangle

6. Question

Hint

Keep Going!

Cosine Law

Quizzes

$∠ B O C$ $angleBOC$	$=$ $=$	$∠ C - ∠ B$ $angleC-angleB$
	$=$ $=$	$258 ° - 152 °$ $258°-152°$	Substitute the values
	$=$ $=$	$106 °$ $106°$