Question
A road runs from P to Q on a bearing of 140degree and continue in the same direction.the bearings of R from P and Q are 092degree and 056degree respectively.PQ is 1320m.find the shortest distance of R from the road
Answers
Answered by
GPT-4o mini
To find the shortest distance from point R to the road PQ, we will use trigonometric functions and geometric principles based on the information provided.
1. **Identify Bearngs**:
- The bearing of PQ is 140°, which means it is directed 40° south of east (since bearings are measured clockwise from north).
- The bearing of R from P is 092°, which means it is directed 2° south of east.
- The bearing of R from Q is 056°, which means it is directed 34° north of east.
2. **Set Up Coordinates**:
We can establish a coordinate system where point P is at the origin (0, 0).
The coordinates for point Q can be determined using the distance PQ and the bearing:
\[
Q_x = PQ \cdot \sin(140°) = 1320 \cdot \sin(140°)
\]
\[
Q_y = PQ \cdot \cos(140°) = 1320 \cdot \cos(140°)
\]
Calculating these components:
\[
Q_x = 1320 \cdot \sin(140°) \approx 1320 \cdot 0.6428 \approx 848.42
\]
\[
Q_y = 1320 \cdot \cos(140°) \approx 1320 \cdot (-0.7660) \approx -1011.84
\]
Therefore, the coordinates of Q are approximately \((848.42, -1011.84)\).
3. **Determine Coordinates for R**:
The lines PR and QR will also be determined using the bearings:
- Bearing of R from P is 092°:
\[
R_x = R \cdot \sin(92°)
\]
\[
R_y = R \cdot \cos(92°)
\]
- Bearing of R from Q is 056°:
\[
R_x = Q_x + r \cdot \sin(56°)
\]
\[
R_y = Q_y - r \cdot \cos(56°)
\]
4. **Finding the Shortest Distance**:
The shortest distance from point R to line PQ is the length of the perpendicular from R to PQ. First, we need to find the slope of line PQ using the coordinates of P and Q.
The slope (m) can be calculated as:
\[
m = \frac{\Delta y}{\Delta x} = \frac{Q_y - 0}{Q_x - 0} = \frac{-1011.84}{848.42} \approx -1.19
\]
The perpendicular slope would then be:
\[
m_{\perpendicular} = \frac{-1}{m} = \frac{1}{1.19} \approx 0.84
\]
The shortest distance from point R to the road PQ can be calculated using the formula for the distance from a point to a line.
For a line in the slope-intercept form (y = mx + b), we can let the equation of PQ be:
\[
y - 0 = -1.19(x - 0) + 0 \implies y = -1.19x
\]
The distance \(d\) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]
Rearranging PQ into this form gives:
\[
1.19x + y = 0
\]
Here \(A = 1.19\), \(B = 1\), and \(C = 0\).
Substitute coordinates of R (given from bearing calculations) into the distance formula to find the shortest distance.
5. **Complete Calculations**:
Once all points are confirmed, compute \(d\).
Note: Since exact bearing angles and position calculations can be tedious, an estimation method might be best suited. For accurate calculations, software tools or graphical methods can validate the directional positions of point R concerning the road PQ.
If you need the actual coordinates of R to proceed further and calculate the distance numerically, please clarify how far point R is from P.
1. **Identify Bearngs**:
- The bearing of PQ is 140°, which means it is directed 40° south of east (since bearings are measured clockwise from north).
- The bearing of R from P is 092°, which means it is directed 2° south of east.
- The bearing of R from Q is 056°, which means it is directed 34° north of east.
2. **Set Up Coordinates**:
We can establish a coordinate system where point P is at the origin (0, 0).
The coordinates for point Q can be determined using the distance PQ and the bearing:
\[
Q_x = PQ \cdot \sin(140°) = 1320 \cdot \sin(140°)
\]
\[
Q_y = PQ \cdot \cos(140°) = 1320 \cdot \cos(140°)
\]
Calculating these components:
\[
Q_x = 1320 \cdot \sin(140°) \approx 1320 \cdot 0.6428 \approx 848.42
\]
\[
Q_y = 1320 \cdot \cos(140°) \approx 1320 \cdot (-0.7660) \approx -1011.84
\]
Therefore, the coordinates of Q are approximately \((848.42, -1011.84)\).
3. **Determine Coordinates for R**:
The lines PR and QR will also be determined using the bearings:
- Bearing of R from P is 092°:
\[
R_x = R \cdot \sin(92°)
\]
\[
R_y = R \cdot \cos(92°)
\]
- Bearing of R from Q is 056°:
\[
R_x = Q_x + r \cdot \sin(56°)
\]
\[
R_y = Q_y - r \cdot \cos(56°)
\]
4. **Finding the Shortest Distance**:
The shortest distance from point R to line PQ is the length of the perpendicular from R to PQ. First, we need to find the slope of line PQ using the coordinates of P and Q.
The slope (m) can be calculated as:
\[
m = \frac{\Delta y}{\Delta x} = \frac{Q_y - 0}{Q_x - 0} = \frac{-1011.84}{848.42} \approx -1.19
\]
The perpendicular slope would then be:
\[
m_{\perpendicular} = \frac{-1}{m} = \frac{1}{1.19} \approx 0.84
\]
The shortest distance from point R to the road PQ can be calculated using the formula for the distance from a point to a line.
For a line in the slope-intercept form (y = mx + b), we can let the equation of PQ be:
\[
y - 0 = -1.19(x - 0) + 0 \implies y = -1.19x
\]
The distance \(d\) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]
Rearranging PQ into this form gives:
\[
1.19x + y = 0
\]
Here \(A = 1.19\), \(B = 1\), and \(C = 0\).
Substitute coordinates of R (given from bearing calculations) into the distance formula to find the shortest distance.
5. **Complete Calculations**:
Once all points are confirmed, compute \(d\).
Note: Since exact bearing angles and position calculations can be tedious, an estimation method might be best suited. For accurate calculations, software tools or graphical methods can validate the directional positions of point R concerning the road PQ.
If you need the actual coordinates of R to proceed further and calculate the distance numerically, please clarify how far point R is from P.