Certainly, let's work through the various parts of this problem step by step.
### Part a) Indicate the position of port [tex]\( Q \)[/tex] on the drawing
Steps:
1. Lindsey sails from port [tex]\( P \)[/tex] on a bearing of [tex]\( 050^\circ \)[/tex].
2. She sails for [tex]\( 1 \frac{1}{2} \)[/tex] hours at an average speed of [tex]\( 5 \)[/tex] km/h.
First, we need to calculate the distance [tex]\( PQ \)[/tex]:
[tex]\[ \text{Distance} = \text{Speed} \times \text{Time} = 5 \text{ km/h} \times 1.5 \text{ h} = 7.5 \text{ km} \][/tex]
Next, we use the scale where [tex]\( 1 \text{ cm} \)[/tex] represents [tex]\( 3 \text{ km} \)[/tex]:
[tex]\[ \text{Distance in cm} = \frac{7.5 \text{ km}}{3 \text{ km/cm}} = 2.5 \text{ cm} \][/tex]
So, port [tex]\( Q \)[/tex] should be marked [tex]\( 2.5 \)[/tex] cm from port [tex]\( P \)[/tex] in the direction of [tex]\( 050^\circ \)[/tex].
### Part b) Find the distance, in km, of port [tex]\( Q \)[/tex] from lighthouse [tex]\( L \)[/tex].
We are directly given this information:
[tex]\[ \text{Distance} = 10.5 \text{ km} \][/tex]
### Part c) Find the bearing of port [tex]\( Q \)[/tex] from lighthouse [tex]\( L \)[/tex].
We are also directly given this information:
[tex]\[ \text{Bearing from \( L \)} = 280^\circ \][/tex]
### Summary
a) The position of port [tex]\( Q \)[/tex] should be marked [tex]\( 2.5 \)[/tex] cm from port [tex]\( P \)[/tex] on the drawing, at a bearing of [tex]\( 050^\circ \)[/tex].
b) The distance from port [tex]\( Q \)[/tex] to the lighthouse [tex]\( L \)[/tex] is [tex]\( 10.5 \)[/tex] km.
c) The bearing of port [tex]\( Q \)[/tex] from the lighthouse [tex]\( L \)[/tex] is [tex]\( 280^\circ \)[/tex].
Total marks: 5
