Answer :
### Question:
4. Four telephone posts P, Q, R, and S stand on a level ground such that Q is 28 meters on a bearing of 060° from P. R is 20 meters to the south of Q and S is 16 meters on a bearing of 140° from P.
(a) Using a scale of 1 cm represents 4 m, show the relative positions of the posts. (4 marks)
(b) Find the distance and bearing of R from S. (3 marks)
(c) If the height of post P is 15.6 m, on a separate scale drawing, draw a diagram and determine the angle of depression of post R from the top of post P. (Same scale as above) (3 marks)
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### Solution:
#### (a) Representing the Relative Positions of the Posts
1. Establish the scale: 1 cm represents 4 meters.
2. Plot Point P: Start by marking point P on your graph paper.
3. Point Q Calculation:
- Q is 28 meters from P at a bearing of 060°.
- Convert to scale: [tex]\(28 \, \text{m} \div 4 \, (\text{m/cm}) = 7 \, \text{cm}\)[/tex].
- Draw a line from P at an angle of 060° from the North (use a protractor) and mark a point at 7 cm to represent Q.
4. Point R Calculation:
- R is directly south of Q and 20 meters away.
- Convert to scale: [tex]\(20 \, \text{m} \div 4 \, (\text{m/cm}) = 5 \, \text{cm}\)[/tex].
- From point Q, draw a vertical line (heading directly south) and mark a point at 5 cm to represent R.
5. Point S Calculation:
- S is 16 meters from P at a bearing of 140°.
- Convert to scale: [tex]\(16 \, \text{m} \div 4 \, (\text{m/cm}) = 4 \, \text{cm}\)[/tex].
- Draw a line from P at an angle of 140° from the North and mark a point at 4 cm to represent S.
#### (b) Finding the Distance and Bearing of R from S
1. Determine Coordinates:
- Coordinates of Q: [tex]\((14, 24.25)\)[/tex] meters.
- Coordinates of R: [tex]\((14, 4.25)\)[/tex] meters.
- Coordinates of S: [tex]\((-12.26, 10.28)\)[/tex] meters.
2. Distance of R from S:
- Apply the distance formula:
[tex]\[ RS_{\text{distance}} = \sqrt{(R_x - S_x)^2 + (R_y - S_y)^2} \][/tex]
[tex]\[ RS_{\text{distance}} = \sqrt{(14 - (-12.26))^2 + (4.25 - 10.28)^2} \][/tex]
[tex]\[ RS_{\text{distance}} = \sqrt{(26.26)^2 + (6.03)^2} \approx 26.94 \, \text{meters} \][/tex]
3. Bearing of R from S:
- Use the formula to find the bearing:
[tex]\[ \text{Bearing} = \left(\tan^{-1} \left( \frac{\Delta y}{\Delta x} \right) + 360 \right) \mod 360 \][/tex]
[tex]\[\Delta x = 14 - (-12.26) = 26.26 \][/tex]
[tex]\[\Delta y = 4.25 - 10.28 = -6.03 \][/tex]
[tex]\[ \text{Bearing} = \left(\tan^{-1} \left( \frac{-6.03}{26.26} \right) + 360 \right) \mod 360 \approx 167.05^\circ \][/tex]
#### (c) Finding the Angle of Depression from P to R
1. Determine Horizontal Distance from P to R:
- Combine the distances PQ and QR:
[tex]\[ \text{Horizontal distance} = \sqrt{PQ^2 + QR^2} \][/tex]
[tex]\[ = \sqrt{28^2 + 20^2} \approx 34.41 \, \text{meters} \][/tex]
2. Calculate Angle of Depression:
- Using the height of P (15.6 m):
[tex]\[ \text{Angle of depression} = \tan^{-1} \left( \frac{\text{Height of P}}{\text{Horizontal distance}} \right) \][/tex]
[tex]\[ = \tan^{-1} \left( \frac{15.6}{34.41} \right) \approx 24.39^\circ \][/tex]
Overall, these steps lead us to the solutions for positions, distances, bearings, and angles as required.
4. Four telephone posts P, Q, R, and S stand on a level ground such that Q is 28 meters on a bearing of 060° from P. R is 20 meters to the south of Q and S is 16 meters on a bearing of 140° from P.
(a) Using a scale of 1 cm represents 4 m, show the relative positions of the posts. (4 marks)
(b) Find the distance and bearing of R from S. (3 marks)
(c) If the height of post P is 15.6 m, on a separate scale drawing, draw a diagram and determine the angle of depression of post R from the top of post P. (Same scale as above) (3 marks)
---
### Solution:
#### (a) Representing the Relative Positions of the Posts
1. Establish the scale: 1 cm represents 4 meters.
2. Plot Point P: Start by marking point P on your graph paper.
3. Point Q Calculation:
- Q is 28 meters from P at a bearing of 060°.
- Convert to scale: [tex]\(28 \, \text{m} \div 4 \, (\text{m/cm}) = 7 \, \text{cm}\)[/tex].
- Draw a line from P at an angle of 060° from the North (use a protractor) and mark a point at 7 cm to represent Q.
4. Point R Calculation:
- R is directly south of Q and 20 meters away.
- Convert to scale: [tex]\(20 \, \text{m} \div 4 \, (\text{m/cm}) = 5 \, \text{cm}\)[/tex].
- From point Q, draw a vertical line (heading directly south) and mark a point at 5 cm to represent R.
5. Point S Calculation:
- S is 16 meters from P at a bearing of 140°.
- Convert to scale: [tex]\(16 \, \text{m} \div 4 \, (\text{m/cm}) = 4 \, \text{cm}\)[/tex].
- Draw a line from P at an angle of 140° from the North and mark a point at 4 cm to represent S.
#### (b) Finding the Distance and Bearing of R from S
1. Determine Coordinates:
- Coordinates of Q: [tex]\((14, 24.25)\)[/tex] meters.
- Coordinates of R: [tex]\((14, 4.25)\)[/tex] meters.
- Coordinates of S: [tex]\((-12.26, 10.28)\)[/tex] meters.
2. Distance of R from S:
- Apply the distance formula:
[tex]\[ RS_{\text{distance}} = \sqrt{(R_x - S_x)^2 + (R_y - S_y)^2} \][/tex]
[tex]\[ RS_{\text{distance}} = \sqrt{(14 - (-12.26))^2 + (4.25 - 10.28)^2} \][/tex]
[tex]\[ RS_{\text{distance}} = \sqrt{(26.26)^2 + (6.03)^2} \approx 26.94 \, \text{meters} \][/tex]
3. Bearing of R from S:
- Use the formula to find the bearing:
[tex]\[ \text{Bearing} = \left(\tan^{-1} \left( \frac{\Delta y}{\Delta x} \right) + 360 \right) \mod 360 \][/tex]
[tex]\[\Delta x = 14 - (-12.26) = 26.26 \][/tex]
[tex]\[\Delta y = 4.25 - 10.28 = -6.03 \][/tex]
[tex]\[ \text{Bearing} = \left(\tan^{-1} \left( \frac{-6.03}{26.26} \right) + 360 \right) \mod 360 \approx 167.05^\circ \][/tex]
#### (c) Finding the Angle of Depression from P to R
1. Determine Horizontal Distance from P to R:
- Combine the distances PQ and QR:
[tex]\[ \text{Horizontal distance} = \sqrt{PQ^2 + QR^2} \][/tex]
[tex]\[ = \sqrt{28^2 + 20^2} \approx 34.41 \, \text{meters} \][/tex]
2. Calculate Angle of Depression:
- Using the height of P (15.6 m):
[tex]\[ \text{Angle of depression} = \tan^{-1} \left( \frac{\text{Height of P}}{\text{Horizontal distance}} \right) \][/tex]
[tex]\[ = \tan^{-1} \left( \frac{15.6}{34.41} \right) \approx 24.39^\circ \][/tex]
Overall, these steps lead us to the solutions for positions, distances, bearings, and angles as required.
