Bearings from Opposite Direction

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Question 1 of 5

1. Question
Find the bearing of $O$ $O$ from $A$ $A$
- $O A =$ $OA=$ (310) $° T$ $°T$
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A true bearing is an angle measured clockwise from the True North around to the required direction.

In bearings, the crosshair is drawn on the point where the bearing comes from.

First, draw a crosshair for point $A$ $A$ .

Notice that when starting from the North line and heading clockwise all the way to line $A O$ $AO$ , it almost forms a revolution.

To find the true bearing of $O$ $O$ from $A$ $A$ , we can find the value of the remaining angle and subtract it from the value of a revolution, which is $360 °$ $360°$ .

Notice that points $O$ $O$ and $A$ $A$ create parallel lines going to the North.

Then, parallel lines create alternate angles that are equal, as seen here.

Each alternate angle is part of an angle formed by a straight line, which is equal to $180 °$ $180°$

We can subtract the known angle, $130 °$ $130°$ , from the straight line, which is $180 °$ $180°$ , to get the value of the alternate angle.

$180 ° - 130 °$ $180°-130°$ $=$ $=$ $50 °$ $50°$

Since alternate angles are equal, the missing value in the revolution is also $50 °$ $50°$ .

Finally, we can subtract $50 °$ $50°$ from $360 °$ $360°$ to find the true bearing of $O$ $O$ from $A$ $A$ .

Don’t forget to add $T$ $T$ to the bearing to indicate that it’s true North.

$O A$ $OA$ $=$ $=$ $360 ° - 50 °$ $360°-50°$

$=$ $=$ $310 ° T$ $310°T$

$O A = 310 ° T$ $OA=310°T$
Question 2 of 5

2. Question
If the bearing of $P$ $P$ from $O$ $O$ is $117 ° T$ $117° T$ , what is the bearing of $O$ $O$ from $P$ $P$
- $O P =$ $OP=$ (297) $° T$ $°T$
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A true bearing is an angle measured clockwise from the True North around to the required direction.

In bearings, the crosshair is drawn on the point where the bearing comes from.

First, notice that when starting from the North line and heading clockwise all the way to line $P O$ $PO$ , it passes by a straight line, which has an angle of $180 °$ $180°$ .

To find the true bearing of $O$ $O$ from $P$ $P$ , we can find the value of the remaining angle and add it to $180 °$ $180°$ .

Notice that angle $N O P$ $NOP$ and angle $S P O$ $SPO$ are alternate angles.

Since alternate angles have equal values, this means that angle $S P O$ $SPO$ is also $117 °$ $117°$

Finally, add the value of angle $S P O$ $SPO$ to the value of a straight angle, which is $180 °$ $180°$ , to find the bearing of $O$ $O$ from $P$ $P$ .

Don’t forget to add $T$ $T$ to the bearing to indicate that it’s true North.

$O P$ $OP$ $=$ $=$ $180 ° + 117 °$ $180°+117°$

$O P$ $OP$ $=$ $=$ $297 ° T$ $297°T$

$O P = 297 ° T$ $OP=297°T$
Question 3 of 5

3. Question
The given bearing of $P$ $P$ from $O$ $O$ is $052 ° T$ $052° T$ . Find $O$ $O$ from $P$ $P$
- $O P =$ $OP=$ (232) $° T$ $°T$
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A true bearing is an angle measured clockwise from the True North around to the required direction.

In bearings, the crosshair is drawn on the point where the bearing comes from.

First, notice that when starting from the North line and heading clockwise all the way to line $P O$ $PO$ , it passes by a straight line, which has an angle of $180 °$ $180°$ .

To find the true bearing of $O$ $O$ from $P$ $P$ , we can find the value of the remaining angle and add it to $180 °$ $180°$ .

Notice that angle $N O P$ $NOP$ and angle $S P O$ $SPO$ are alternate angles.

Since alternate angles have equal values, this means that angle $S P O$ $SPO$ is also $52 °$ $52°$

Finally, add the value of angle $S P O$ $SPO$ to the value of a straight angle, which is $180 °$ $180°$ , to find the bearing of $O$ $O$ from $P$ $P$ .

Don’t forget to add $T$ $T$ to the bearing to indicate that it’s true North.

$O P$ $OP$ $=$ $=$ $180 ° + 52 °$ $180°+52°$

$O P$ $OP$ $=$ $=$ $232 ° T$ $232°T$

$O P = 232 ° T$ $OP=232°T$
Question 4 of 5

4. Question
Find the bearing of $O$ $O$ from $A$ $A$
- $O P =$ $OP=$ (62) $° T$ $°T$
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A true bearing is an angle measured clockwise from the True North around to the required direction.

In bearings, the crosshair is drawn on the point where the bearing comes from.

First, notice that the bearing of $O$ $O$ from $A$ $A$ and angle $S O A$ $SOA$ are alternate angles

Since alternate angles are equal, we can find the bearing of $O$ $O$ from $A$ $A$ by finding the value of angle $S O A$ $SOA$ .

Notice that the bearing of $A$ $A$ from $O$ $O$ consists of a straight angle, which is $180 °$ $180°$ , and the value of angle $S O A$ $SOA$ .

Knowing these values, we can simply subtract $180 °$ $180°$ from the bearing of $A$ $A$ from $O$ $O$ to find the value of angle $S O A$ $SOA$

$∠ S O A$ $/_SOA$ $=$ $=$ $242 ° - 180 °$ $242°-180°$

$=$ $=$ $62 °$ $62°$

Recalling that angle $S O A$ $SOA$ and the bearing of $O$ $O$ from $A$ $A$ are alternate angles, the bearing of $O$ $O$ from $A$ $A$ is also equal to $62 °$ $62°$ .

$O A = 62 ° T$ $OA=62°T$

Question 5 of 5

What is the bearing of the ship

A

$A$ as seen from ship

C

$C$ ?

Round your answer to the nearest degree

(339) $° T$ $°T$

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Sin Ratio

s i n = \frac{opposite}{hypotenuse}

$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

Cos Ratio

c o s = \frac{adjacent}{hypotenuse}

$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$

Tan Ratio

t a n = \frac{opposite}{adjacent}

$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$

A true bearing is an angle measured clockwise from the True North around to the required direction.

The outcome we are looking for is the bearing of

A

$A$ from

C

$C$ .

This is the same as the angle from

C

$C$ ’s True North spanning clockwise to line

A C

$AC$ .

First, find the value of the angle inside the triangle in the side of ship

C

$C$ . Label it as

θ

$theta$ .

To solve for

θ

$theta$ , we can use the known values that are opposite and adjacent to it.

Since we have the opposite and adjacent values, we can solve for

\tan θ

$tan theta$

$\tan θ$ $tan theta$	$=$ $=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$

$\tan θ$ $tan theta$	$=$ $=$	$\frac{\color{#004ec4}{130}}{\color{#00880A}{180}}$

Remember that we are looking for

θ

$theta$ , not

\tan θ

$tan theta$

Use your calculator to find the value of

θ

$theta$ . The common key combinations on your calculator would be:

Shift + \tan + (\frac{130}{180})

$\text(Shift) +tan+(130/180)$

or

$or$

\tan^{- 1} + (\frac{130}{180})

$tan^(-1)+(130/180)$

This will give you the value of

35.8377 °

$35.8377°$ , rounded to the nearest degree, $θ = 36 °$ $theta=36°$

Next, notice that a part of the bearing we are looking for is an angle formed by the North line to the South line.

Since it forms a straight line, this is equal to $180 °$ $180°$ .

Now, all we have missing is the angle below the triangle.

Notice that this angle is an alternate angle to

∠ N B C

$angle N B C$ from ship

A

$A$ .

Since alternate angles are equal, the missing angle measures $123 °$ $123°$ .

Finally, we can add the three angles to find the true bearing of ship

A

$A$ from ship

C

$C$ .

Don’t forget to add

T

$T$ to the bearing to indicate that it’s true North.

$C A$ $CA$	$=$ $=$	$36 ° + 180 ° + 123 °$ $36°+180°+123°$
	$=$ $=$	$339 ° T$ $339°T$

339 ° T

$339°T$

Bearings from Opposite Direction

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Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

Hint

Keep Going!

3. Question

Hint

Excellent!

4. Question

Hint

Well Done!

5. Question

Hint

Correct!

Sin Ratio

Cos Ratio

Tan Ratio

Quizzes

$O A$ $OA$	$=$ $=$	$360 ° - 50 °$ $360°-50°$
	$=$ $=$	$310 ° T$ $310°T$

$O P$ $OP$	$=$ $=$	$180 ° + 117 °$ $180°+117°$
$O P$ $OP$	$=$ $=$	$297 ° T$ $297°T$

$O P$ $OP$	$=$ $=$	$180 ° + 52 °$ $180°+52°$
$O P$ $OP$	$=$ $=$	$232 ° T$ $232°T$

$∠ S O A$ $/_SOA$	$=$ $=$	$242 ° - 180 °$ $242°-180°$
	$=$ $=$	$62 °$ $62°$