Solving for Bearings
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Question 1 of 7
1. Question
Find the bearing of `C` from `A`
- `AC=` (124)`°T`
Hint
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Well Done!
Incorrect
Sin Ratio
$$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio
$$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio
$$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$A true bearing is an angle measured clockwise from the True North around to the required direction.Notice that when starting from North and heading to East, the two lines form a complementary angle.
Since complementary angles add up to `90°`, we find the value of `theta` and add it to `90°` to get the true bearing of `C` from `A`.To solve for `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 theta``tan theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `tan theta` `=` $$\frac{\color{#004ec4}{4}}{\color{#00880A}{6}}$$ `tan theta` `=` `0.6666…` Remember that we are looking for `theta`, not `tan theta`Use your calculator to find the value of `theta`. The common key combinations on your calculator would be:`\text(Shift) +tan+0.6666…` `or` `tan^(-1)+0.6666…` This will give you the value of `33.69°` or `33°41’`, depending on the calculator.Next, round the value to the nearest degree. For rounding values:Degrees with decimals:Decimal value below 50: Round downDecimal value 50 or above: Round upDegrees with minutes:Minute value below 30: Round downMinute value 30 or above: Round upThis will give us `theta=34°`.
Finally, add `34°` to `90°` to find the true bearing of `C` from `A`.`AC` `=` `90+34` `=` `124°T` 
`AC=124°T` -
Question 2 of 7
2. Question
Bianca leaves her home and cycles due north for `12km`, then `7km` due west to the gym. What is the bearing of her home from the gym?
- `HG=` (150)`°T`
Hint
Help VideoCorrect
Nice Job!
Incorrect
Sin Ratio
$$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio
$$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio
$$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$A true bearing is an angle measured clockwise from the True North around to the required direction.Notice that when starting from North and heading to East, the two lines form a complementary angle.
Since complementary angles add up to `90°`, we find the value of `theta` and add it to `90°` to get the true bearing of `H` from `G`.To solve for `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 theta``tan theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `tan theta` `=` $$\frac{\color{#004ec4}{12}}{\color{#00880A}{7}}$$ Remember that we are looking for `theta`, not `tan theta`Use your calculator to find the value of `theta`. The common key combinations on your calculator would be:`\text(Shift) +tan+(12/7)` `or` `tan^(-1)+(12/7)` This will give you the value of `59.74°` or `59°45’`, depending on the calculator.Next, round the value to the nearest degree. For rounding values:Degrees with decimals:Decimal value below 50: Round downDecimal value 50 or above: Round upDegrees with minutes:Minute value below 30: Round downMinute value 30 or above: Round upThis will give us `theta=60°`.
Finally, add `60°` to `90°` to find the true bearing of `H` from `G`.`HG` `=` `90+60` `=` `150°T` 
`HG=150°T` -
Question 3 of 7
3. Question
Town `A` is `128 \text(km)` east and `165 \text(km)` south of town `B`. Find the true bearing of town `A` from town `B`.
Round your answer to the nearest minute- `BA=` (142)`°` (12)`'T`
Hint
Help VideoCorrect
Excellent!
Incorrect
Sin Ratio
$$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio
$$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio
$$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$A true bearing is an angle measured clockwise from the True North around to the required direction.Notice that when starting from North and heading to East, the two lines form a complementary angle.
Since complementary angles add up to `90°`, we find the value of `theta` and add it to `90°` to get the true bearing of `C` from `A`.To solve for `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 theta``tan theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `tan theta` `=` $$\frac{\color{#004ec4}{165}}{\color{#00880A}{128}}$$ `tan theta` `=` `1.289` Remember that we are looking for `theta`, not `tan theta`Use your calculator to find the value of `theta`. The common key combinations on your calculator would be:`\text(Shift) +tan+1.289` `or` `tan^(-1)+1.289` This will give you the value of `52.1972°` or `52°11’50″`, depending on the calculator.Since we are looking for the value with the nearest degree, convert `52.1972°` to Degrees-Minute-Second form by pressing the `DMS` button on a calculator.Next, round the value to the nearest degree. For rounding values:Degrees with minutes:Minute/Second value below 30: Round downMinute/Second value 30 or above: Round upThis will give us `theta=52°12’`.
Finally, add `52°12’` to `90°` to find the true bearing of `A` from `B`.`BA` `=` `90°+52°12’` `=` `142°12’T` 
`BA=142°12’°T` -
Question 4 of 7
4. Question
Town `X` is `87 \text(km)` west and `29 \text(km)` north of town `Y`. Find the compass bearing of town `X` from town `Y`.
- `XY=` (N) (72)`°` (W)
Hint
Help VideoCorrect
Fantastic!
Incorrect
Sin Ratio
$$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio
$$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio
$$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$To get a compass bearing, take the compass quadrant where the direction belongs to, then get the angle from the vertical line around to the required direction.
Compass Bearing Quadrants
Northwest Quadrant`=N` __ `W`Northeast Quadrant`=N` __ `E`Southwest Quadrant`=S` __ `W`Southeast Quadrant`=S` __ `E`The blank is filled with the distance of the bearing from the North-South (vertical) lineSince we are looking for the compass bearing, we need to find the value of the angle spanning from the North line. Mark it as `theta`.To solve for `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 theta``tan theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `tan theta` `=` $$\frac{\color{#004ec4}{87}}{\color{#00880A}{29}}$$ `tan theta` `=` `3` Remember that we are looking for `theta`, not `tan theta`Use your calculator to find the value of `theta`. The common key combinations on your calculator would be:`\text(Shift) +tan+3` `or` `tan^(-1)+3` This will give you the value of `71.56505°` or `71°34’`, depending on the calculator.Next, round the value to the nearest degree. For rounding values:Degrees with minutes:Minute/Second value below 30: Round downMinute/Second value 30 or above: Round upThis will give us `theta=72°`.
Finally, notice that town `X` is located on the Northwest Quadrant.
Using the chart, we can write the compass bearing of `X` from `Y` by using `N` __ `W`. Then get the angle that spans from the vertical line (`N`) to `X`, which is `72°`Therefore, the compass bearing of `X` from `Y` is `N 72° W``XY=N 72°W` -
Question 5 of 7
5. Question
Town `A` is `420 \text(km)` west and `760 \text(km)` north of town `B`. Find the true bearing of town `A` from town `B`.
Round your answer to the nearest degree- `AB=` (331)`°T`
Hint
Help VideoCorrect
Keep Going!
Incorrect
Sin Ratio
$$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio
$$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio
$$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$A true bearing is an angle measured clockwise from the True North around to the required direction.Notice that when starting from North and moving clockwise to line `A`, it passes through `3` quadrants which are `90°` each.
Now, to solve for `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 theta``tan theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `tan theta` `=` $$\frac{\color{#004ec4}{760}}{\color{#00880A}{420}}$$ `tan theta` `=` `1.80952` Remember that we are looking for `theta`, not `tan theta`Use your calculator to find the value of `theta`. The common key combinations on your calculator would be:`\text(Shift) +tan+1.80952` `or` `tan^(-1)+1.80952` This will give you the value of `61.07°` or `61°04’`, depending on the calculator.Next, round the value to the nearest degree. For rounding values:Degrees with decimals:Decimal value below 50: Round downDecimal value 50 or above: Round upDegrees with minutes:Minute value below 30: Round downMinute value 30 or above: Round upThis will give us `theta=61°`.
Finally, add the angles that span from the North line clockwise to line `A` to find the true bearing of `A` from `B`.`AB` `=` `90+90+90+61` `AB` `=` `270+61` `=` `331°T` 
`AB=331°T` -
Question 6 of 7
6. Question
Find the bearing of `B` from `C`
- `BC=` (310)`°T`
Hint
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Exceptional!
Incorrect
A true bearing is an angle measured clockwise from the True North around to the required direction.Notice that when starting from the North line on the right and heading clockwise to line `AC`, we go around three quadrants, which are `90°` each.
This means we can simply add those three `90°` angles to `40°`Don’t forget to add `T` to the bearing to indicate that it’s true North.`BC` `=` `90+90+90+40` `=` `270+40` `=` `310°T` 
`BC=310°T` -
Question 7 of 7
7. Question
Find the true bearing of `M` from `K`
Round your answer to the nearest minute- `M` from `K=` (198)`°` (8)`'T`
Hint
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Correct!
Incorrect
Sin Ratio
$$sin=\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$Cos Ratio
$$cos=\frac{\color{#00880a}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$Tan Ratio
$$tan=\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880a}{\text{adjacent}}}$$A true bearing is an angle measured clockwise from the True North around to the required direction.Co-Interior Angles are when two angles have a sum of `180°`.First, find the value of angle `MKL`. Label it as `theta`.Determine that angle `KLM` is a right angle.`180°-35°-55°` `=` `90°` 
To solve for `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 theta``tan theta` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$ `tan theta` `=` $$\frac{\color{#004ec4}{8}}{\color{#00880A}{6}}$$ Remember that we are looking for `theta`, not `tan theta`Use your calculator to find the value of `theta`. The common key combinations on your calculator would be:`\text(Shift) +tan+(8/6)` `or` `tan^(-1)+(8/6)` This will give you the value of `53.13°` or `53°7’48″`, depending on the calculator.Since we are looking for the value with the nearest degree, convert `53.13°` to Degrees-Minute-Second form by pressing the `DMS` button on a calculator.Next, round the value to the nearest degree. For rounding values:Degrees with minutes:Minute/Second value below 30: Round downMinute/Second value 30 or above: Round upThis will give us `theta=53°8’`.
Next, draw a cross hair for point `K` since this is where we are getting the bearing from.
Notice that the remaining missing angle and angle `NLK` are co-interior angles.Since co-interior angles sum up to `180°`, we can simply subtract `35°` from `180°` to get the value of the missing angle.`180°-35°` `=` `145°` 
Finally, add `145°` to the value of `theta`, which is `53°8’`, to find the true bearing of `M` from `K`.`MK` `=` `53°8’+145°` `=` `198°8’T` 
`M` from `K=198°8’T`
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