A boat leaves port P and steers 5 km due east and then 10 km on a bearing of 060 degrees. Find its distance and bearing from P
did you make your sketch??
If you follow the standard definition of "bearing", I see it as a cosine law problem
solve:
x^2 = 5^2 + 10^2 - 2(5)(10)cos 150°
....
To find the distance and bearing of the boat from port P, we can use the concept of vectors. Let's break down the boat's movement into two components:
1. Eastward component: The boat moves 5 km due east. We can represent this as a vector (5, 0), where the first value represents the distance eastward (x-axis) and the second value represents the distance northward (y-axis).
2. Northeast component: After moving east, the boat changes its direction and travels 10 km on a bearing of 060 degrees. This means it's moving northeast. We can calculate the northeastward (x-component) and northward (y-component) distances using trigonometry.
To find the northeastward component, we use the cosine function:
northeastward component = distance * cos(60 degrees)
= 10 km * cos(60 degrees)
= 10 km * 0.5
= 5 km
To find the northward component, we use the sine function:
northward component = distance * sin(60 degrees)
= 10 km * sin(60 degrees)
= 10 km * (√3 / 2)
= 5√3 km
Now, we can add up the eastward and northeastward components:
Eastward distance + Northeastward distance = 5 km + 5 km = 10 km
Northward distance = Northward component = 5√3 km
To find the total distance from port P, we can use the Pythagorean theorem:
total distance = √[(Eastward distance)^2 + (Northward distance)^2]
= √[(10 km)^2 + (5√3 km)^2]
= √[100 km^2 + 75 km^2]
= √[175 km^2]
= √175 km
≈ 13.23 km (rounded to two decimal places)
To find the bearing from port P, we can use the tangent function:
bearing = arctan(Northward distance / Eastward distance)
= arctan(5√3 km / 10 km)
= arctan(√3 / 2)
≈ 30.96 degrees (rounded to two decimal places)
Therefore, the boat is approximately 13.23 km away from port P at a bearing of 30.96 degrees.