Working with Radial Surveys 1

Years

Year 12

Bearings

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) $m^{2}$ $m^2$

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) $m^{2}$ $m^2$

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 $)$ $)$

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) $m^{2}$ $m^2$

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

$73m$

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

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

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

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

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

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

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

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

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

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

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