Using Bearings to Find Distance 2
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Question 1 of 4
1. Question
A ship has traveled `68 \text(km)` northwest from point `P` with a true bearing of `319°T`. How far west (`x_w`) has it traveled?
Round your answer to one decimal place- `x_w=` (44.6) `\text(km)`
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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.First, find the value of `theta` below.Notice that it is part of the given bearing, which also consists of `3` quadrants from the North line moving clockwise.
To find the value of `theta`, simply subtract the measure of the angles from the three quadrants, `270°`, from the given bearing.`theta` `=` `319°-270°` `=` `49°` To solve for `x_w`, we can use the known values of the hypotenuse and `theta=49°`.
Use `cos` to find the value of `x_w`.`cos49°` `=` $$\frac{\color{#00880A}{x_w}}{\color{#e85e00}{\text{hypotenuse}}}$$ `cos49°` `=` $$\frac{\color{#00880A}{x_w}}{\color{#e85e00}{68}}$$ `cos49°``xx68` `=` `((x_w)/68)``xx68` Multiply both sides by `68` `68cos49°` `=` `x_w` `x_w` `=` `68cos49°` Using your calculator, `68cos49°=44.6`.Therefore, the speedboat is `44.6 \text(km)` to the West.
`x_w=44.6 \text(km)` -
Question 2 of 4
2. Question
A marathon runner runs `25.3 \text(km)` at a true bearing of `129°T`. Find how far east (`x_e`) the runner has traveled from the starting point (`S`).
Round your answer to one decimal place- `x_e=` (19.7) `\text(km)`
Hint
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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.First, find the value of `theta` below.Notice that it is part of the given bearing, which also consists of a right angle from the North line moving to the East line.
To find the value of `theta`, simply subtract the value of a right angle, `90°`, from the given bearing.`theta` `=` `129°-90°` `=` `39°` To solve for `x_e`, we can use the known values of the hypotenuse and `theta=39°`.
Use `cos` to find the value of `x_e`.`cos39°` `=` $$\frac{\color{#00880A}{x_e}}{\color{#e85e00}{\text{hypotenuse}}}$$ `cos39°` `=` $$\frac{\color{#00880A}{x_e}}{\color{#e85e00}{25.3}}$$ `cos39°``xx25.3` `=` `((x_e)/25.3)``xx25.3` Multiply both sides by `25.3` `25.3cos39°` `=` `x_e` `x_e` `=` `25.3cos39°` Using your calculator, `25.3cos39°=19.7`.Therefore, the runner runs `19.7 \text(km)` to the East.
`x_e=19.7 \text(km)` -
Question 3 of 4
3. Question
A ship sails on a bearing of `200° T` towards `P`. If `P` is `80` nautical miles west from `O`, find how far the ship has sailed `(x)`.
Round your answer to one decimal place- (233.9) `\text(nautical miles)`
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>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.First, find the value of angle `QOP`.Notice that adding angle `QOP` to the given bearing `(200°)` will cover `3` quadrants, which is equal to `270°`
To find the value of angle `QOP`, simply subtract the value of the bearing from `270°`.`/_QOP` `=` `270°-200°` `=` `70°` To solve for `x`, we can use the known values of the line adjacent to angle `QOP`.
Use `cos` to find the value of `x`.`cos70°` `=` $$\frac{\color{#00880A}{\text{adjacent}}}{\color{#e85e00}{x}}$$ `cos70°` `=` $$\frac{\color{#00880A}{80}}{\color{#e85e00}{x}}$$ `cos70°``xxx` `=` `((80)/x)``xxx` Multiply both sides by `x` `xcos70°` `=` `80` `xcos70°``divcos70°` `=` `80``divcos70°` Divide both sides by `cos70°` `x` `=` `(80)/(cos70°)` Using your calculator, `(80)/(cos70°)=233.9`.Therefore, the boat traveled `233.9 \text(nautical miles)` in total.
`233.9 \text(nautical miles)` -
Question 4 of 4
4. Question
Bianca leaves her home and cycles due North for `12` km, then `7` km due West to go to the gym. How far is the gym from her home `(x)`?
Round your answer to one decimal place- (13.9) `\text(km)`
Hint
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Pythagoras’ Theorem Formula
`a^2``=``b^2``+``c^2``a` is the hypotenuse, and `b` and `c` are the two legsUse the Pythagoras’ Theorem to solve for the distance `(``x``)`.
`a` `=` `x` `b` `=` `12 \text(km)` `c` `=` `7 \text(km)` `a^2` `=` `b^2``+``c^2` `x^2` `=` `12^2``+``7^2` Substitute known values `sqrt(x^2)` `=` `sqrt(144+49)` Get the square root of both sides `x` `=` `sqrt(193)` `x` `=` `13.9 \text(km)` Rounded to one decimal place 
`13.9 \text(km)`