P,Q and R are points on the same horizontal plane .the bearing of Q from P is 150° and the bearing of R from Q is 060°. if |PQ| =5m and |QR|=3m find the bearing of R from P correct your answer to the nearest degree
In geometry, drawing a sketch from given information is half-way to the solution.
Note bearing is measured clockwise from North.
If you draw the sketch in Cartesian coordinates, using P at the origin, you will have
P(0,0)
Bearing 150°≡ θ=90-150=-60° in Cartesian coordinates.
So
Q(5cos(-60), 5 sin(-60))
=Q(2.5, -2.5&radic/3)
Similarly, the bearing of R from Q is 60°≡ θ=90-60=30°.
So
Δx=3cos(30)=1.5√3
Δy=3sin(30)=1.5
Therefore the coordinates of R relative to P
R(2.5+Δx, -2.5&radic/3+Δy)
=R(2.5+1.5√3, -2.5√3+1.5)
Bearing of R from P
=90-atan(y/x)
=90-atan((-2.5√3+1.5)/(2.5+1.5√3))
=90-(-29)
=119°
To find the bearing of point R from point P, we need to add the bearings together.
The bearing of Q from P is 150°, and the bearing of R from Q is 060°.
To find the bearing of R from P, we add 150° and 060°:
150° + 060° = 210°
Therefore, the bearing of R from P is 210°.
To find the bearing of R from P, we need to determine the angle between the line segment PQ and PR.
Step 1: Draw a diagram:
We have three points on a horizontal plane - P, Q, and R.
PQ has a length of 5m, and QR has a length of 3m.
The bearing of Q from P is 150° (clockwise from the north) and the bearing of R from Q is 60°.
Step 2: Find the angle QPR:
Since PQ and QR are two sides of a triangle, we can use the Law of Cosines to find the angle QPR.
Using the Law of Cosines:
cos(QPR) = (|PQ|^2 + |QR|^2 - |PR|^2) / (2 * |PQ| * |QR|)
Substituting the given values:
cos(QPR) = (5^2 + 3^2 - |PR|^2) / (2 * 5 * 3)
cos(QPR) = (25 + 9 - |PR|^2) / 30
cos(QPR) = (34 - |PR|^2) / 30
Step 3: Solve for |PR|:
To solve for |PR|, we need to rearrange the equation:
|PR|^2 = 34 - (30 * cos(QPR))
|PR|^2 = 34 - 30 * cos(QPR)
Step 4: Substitute the values and solve for |PR|:
We have the values of QPR = 150°, so we substitute it into the equation:
|PR|^2 = 34 - 30 * cos(150°)
Using a calculator:
|PR|^2 = 34 - 30 * (-0.866)
|PR|^2 = 34 + 25.98
|PR|^2 = 59.98
Therefore, |PR| ≈ √59.98
≈ 7.746 m (rounded to 3 decimal places)
Step 5: Find the bearing of R from P:
To find the bearing of R from P, we subtract the bearing of Q from P (150°) from the bearing of R from Q (60°).
Bearing of R from P = Bearing of Q from P - Bearing of R from Q
= 150° - 60°
= 90°
Therefore, the bearing of R from P is approximately 90° (rounded to the nearest degree).