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…

1. Analyze the Question

We are given the movement of a bird starting from point P (tree) to Q (building), and then to R (another tree).

• P to Q: Distance = 200 km, Bearing = N60°E.
• Q to R: Bearing = S30°E.
• R is directly East of P: This means the line PR is a horizontal line (a line of latitude).
• Goal: Find the distance QR.

2. Explain the Concept

This is a problem involving bearings and trigonometry.

• Bearings: Always measured clockwise from the North.
• Geometry: Since R is directly East of P, the line PR is horizontal. We can form a triangle PQR.
• Angles: • At point P, the bearing N60°E means the angle between the North line and PQ is 60°. Since the East line is 90° from North, the angle inside the triangle at P (angle QPR) is $90° - 60° = 30°$. • At point Q, the bearing S30°E means the angle between the South line and QR is 30°. Since the North-South line is vertical, the angle between the vertical line at Q and the line QR is 30°. • Using alternate angles, if we draw a vertical line through Q, the angle between PQ and the vertical is 60° (alternate to the angle at P). Therefore, the interior angle at Q (angle PQR) is $60° + 30° = 90°$.

3. Show the Solution

We now have a right-angled triangle PQR where:

• $\angle PQR = 90°$
• $\angle QPR = 30°$
• Hypotenuse $PR$ is the side opposite the 90° angle.
• Side $PQ = 200$ km.
• We need to find side $QR$.

In a right-angled triangle, we use the trigonometric ratio:

For angle $P$ (30°):

Substitute the known values:

Solve for $QR$:

4. Identify the Answer

Comparing our result to the given options, we find that our calculation matches option D.

5. Tip

When solving bearing problems, always draw a neat sketch. Mark your North lines at every point where the direction changes. Remember that "Directly East" or "Directly North" implies a 90-degree relationship with the cardinal axes, which often helps you identify right-angled triangles quickly!