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Question 1 of 3
1. Question
Two ships set sail straight away from point `B`. Ship `A` traveled `130 \text(km)` with a bearing of `33°T` and ship `C` traveled `180 \text(km)` with a bearing of `123°T`. How far away are the two ships `(AC)`?
Round your answer to the nearest kilometre- (222) `\text(km)`
Hint
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Excellent!
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Pythagoras’ Theorem Formula
`a^2``=``b^2``+``c^2``a` is the hypotenuse, and `b` and `c` are the two legsFirst, find the value of the interior angle `ABC` by subtracting the two given bearings.`/_ABC` `=` `123°-33°` `=` `90°` Since the triangle has an angle with a value of `90°`, that means it is a right triangle.Use the Pythagoras’ Theorem to solve for the distance `AC`.
`a` `=` `AC` `b` `=` `AB` `=` `130 \text(km)` `c` `=` `BC` `=` `180 \text(km)` `a^2` `=` `b^2``+``c^2` `AC^2` `=` `130^2``+``180^2` Substitute known values `sqrt(AC^2)` `=` `sqrt(16900+32400)` Get the square root of both sides `AC` `=` `sqrt(49300)` `AC` `=` `222 \text(km)` Rounded to the nearest kilometre 
`222 \text(km)` -
Question 2 of 3
2. Question
Find the distance of point `K` to point `M`
- (10) `\text(km)`
Hint
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Exceptional!
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Pythagoras’ Theorem Formula
`a^2``=``b^2``+``c^2``a` is the hypotenuse, and `b` and `c` are the two legsFirst, find the value of the interior angle `KLM` by subtracting the sum of the two given bearings from the angle of a straight line, which is `180°`.`/_KLM` `=` `180°-(35°+55°)` `=` `180°-90°` `=` `90°` Since the triangle has an angle with a value of `90°`, that means it is a right triangle.Use the Pythagoras’ Theorem to solve for the distance `KM`.
`a` `=` `KM` `b` `=` `KL` `=` `6 \text(km)` `c` `=` `LM` `=` `8 \text(km)` `a^2` `=` `b^2``+``c^2` `KM^2` `=` `6^2``+``8^2` Substitute known values `sqrt(KM^2)` `=` `sqrt(36+64)` Get the square root of both sides `KM` `=` `sqrt(100)` `KM` `=` `10 \text(km)` 
`10 \text(km)` -
Question 3 of 3
3. Question
A yacht is located `43°T` from port `X` and `302°T` from port `Z`. Port `Z` is `180 \text(km)` directly east of port `X`. Find the distance of the yacht from port `X`.
Round your answer to two decimal places- (97.17) `\text(km)`
Hint
Help VideoCorrect
Correct!
Incorrect
Sine Rule
`x/(sinX)=y/(sinY)=z/(sinZ)`
Remember
- Uppercase letters represent angles in the triangle
- Lowercase letters represent the side lengths
First, find the value of `/_ X`.Notice that `/_ X` and the bearing of `Y` from `X` `(43°)` forms a complementary angle.
Since complementary angles are equal to `90°`, we can simply subtract `43°` from `90°` to find `/_ X`.`/_X` `=` `90°-43°` `=` `47°` Next, find the value of `/_ Z`.Notice that the true bearing of `Y` from `Z` includes `/_ Z` along with `3` quadrants.
We can simply subtract the value of three quadrants, which is `270°`, from the given true bearing to find angle `Z`.`/_Z` `=` `302°-270°` `=` `32°` Then, find the value of angle `/_ Y` by subtracting the sum of `/_ X` and `/_ Z` from the total interior angle of a triangle, which is `180°`.
`/_Y` `=` `180°-(``47°``+``32°``)` `=` `180°-79°` `=` `101°` Since we know the values of angle `Y`, angle `Z`, and line `XZ`, we can use the sine law to solve for the distance of `Y` from `X`.
`Y` `=` `101°` `y` `=` `180 \text(km)` `Z` `=` `32°` `y/(sinY)` `=` `z/(sinZ)` `(180)/(sin101°)` `=` `z/(sin32°)` Substitute known values `zxxsin101°` `=` `180xxsin32°` Cross-multiply `zxxsin101°``divsin101°` `=` `180xxsin32°``divsin101°` Divide both sides by `sin101°` `z` `=` `(180xxsin32°)/(sin101°)` Using a calculator, `(180xxsin32°)/(sin101°)=97.17 \text(km)`, rounded to two decimal places
`97.17 \text(km)`
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