jstalexx said:
A car sets off in a direction of 65km/h in a direction of N20E and at the same time a bike rider goes in the direction of N70E at a speed of 30km/h. They travel for 2 hours.
a) What is the bearing of the car from the bike rider?
Could someone please explain how to do it ?
Ok. The best way to approach this question is to draw a diagram first.
STEP 1: Draw a horizontal and a vertical line.
STEP 2: The first line originates from the origin and extends to 130 km (in 2 hours, car travels 130 km) in the direction of north-east. The angle between the first line and the vertical line is 20 degrees (because it extends in a direction of N20E).
STEP 3: The second line originates from the origin and extends to 60 km (in tw hours, bike travels 60 km) in the direction of north-east. The angle between the first line and the vertical line is 70 degrees (because it extends in a direction of N70E). Therefore, it can be seen that the angle between the second line and the horizontal line is 20 degrees.
STEP 4: Join the lines and you will make up a triangle.
STEP 5: Find the side BC by using the cosine rule: A^2 = B^2 + C^2 - 2BC . cos a
STEP 6: Find the angle CBO by using the cosin rule: cos a = b^2 + c^2 -a^2 / 2bc
Step 7: From CBD, make another vertical and horizontal line to see where the C is located in relation to B. And then work it out.
