A bird flies from a tree P on a bearing of N60º E to a building, Q, a distance of 200 km. It then changes course and flies to another tree R on a bearing of...

Answer Details

Bearing is a way of describing direction using angles measured clockwise from north. For example, N60ºE means 60° east of north, and S30ºE means 30° east of south.

Step 1: Drawing and understanding the path
Let’s clarify the problem with a diagram (you can imagine or sketch this):

• P is the starting tree.
• The bird flies from P to Q (the building) on a bearing of N60ºE, meaning the direction is 60º east of north.
• After reaching Q, it turns and flies toward R (another tree), on a bearing of S30ºE (30º east of south). You are told R is directly east of P.

Step 2: Finding coordinates using trigonometry
Let's place P at the origin \((0,0)\). The bird's first flight to Q covers 200 km at an angle of 60º east of north. In trigonometry, the north direction matches the positive y-axis, and east is the positive x-axis.

To find the coordinates of Q:

• East (x-component): \( 200 \times \sin(60^\circ) \)
• North (y-component): \( 200 \times \cos(60^\circ) \)

Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(60^\circ) = \frac{1}{2} \), the position of Q is:

• x-coordinate: \( 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \)
• y-coordinate: \( 200 \times \frac{1}{2} = 100 \)

So, Q is at \((100\sqrt{3}, 100)\).

Step 3: Coordinates of R
Tree R is directly east of P. Since P is at \((0,0)\), R must be at \((x, 0)\) for some x.

The bird flies from Q to R on a bearing of S30ºE. Bearing S30ºE is 30° east of due south, which means the angle is 30° to the east from the negative y-axis. In terms of vector components from Q to R:

• East (x-component): \( d \times \sin(30^\circ) = d \times \frac{1}{2} \)
• South (y-component): \( -d \times \cos(30^\circ) = -d \times \frac{\sqrt{3}}{2} \) (negative, because south is in the negative y direction)

Setting up the coordinates of R: \[ x_R = 100\sqrt{3} + \frac{d}{2} \] \[ y_R = 100 - d \frac{\sqrt{3}}{2} \] But since R is directly east of P, \( y_R = 0 \). So: \[ 100 - d \frac{\sqrt{3}}{2} = 0 \] Solving for \( d \): \[ d \frac{\sqrt{3}}{2} = 100 \] \[ d = \frac{200}{\sqrt{3}} \]

Conclusion
The distance from the building (Q) to tree R is therefore \(\frac{200}{\sqrt{3}}\) km. This value matches the correct answer in the options.

Underlying concept: The question uses bearings, vector components, and trigonometry to find positions and distances in navigation problems. Recognizing which trigonometric functions to use based on the angle and direction is key to solving these types of problems.