A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the gr

Q: Wait, why is the height constant? Doesn't the bird fly at an angle?

The problem explicitly states the bird flies 'horizontally straight away.' This is a key piece of information! In physics, 'horizontal' means the vertical component of the motion is zero, so the height remains constant at $20\text{ m}$ throughout the flight.

Q: Can I use sine or cosine instead of tangent?

You could, but it would be much harder! Sine and cosine involve the hypotenuse (the line of sight), which changes as the bird moves. Tangent is the most efficient tool here because it directly relates the constant height (opposite side) to the horizontal distance (adjacent side) we need to find.

Q: Why does the angle of elevation decrease as the bird flies away?

Think of it this way: as the bird gets further away, you have to look 'less' upwards to see it. The line of sight becomes flatter, closer to the ground, which mathematically means the angle of elevation must decrease. It's a beautiful geometric consequence of moving further from an object of fixed height.

Question

A bird is sitting on the top of a vertical pole $20	ext{ m}$ high and its elevation from a point $O$ on the ground is $45^\circ$. It flies off horizontally straight away from the point $O$. After one second, the elevation of the bird from $O$ is reduced to $30^\circ$. Then the speed (in m/s) of the bird is

Options:

(A) $20\sqrt{2}$

(B) $20(\sqrt{3}-1)$

(C) $40(\sqrt{2}-1)$

(D) $40(\sqrt{3}-\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat field. In front of you stands a vertical pole, exactly $20	ext{ m}$ tall. Perched at the very top is a bird, observing the world.

You are standing at a point $O$ on the ground, looking up at this bird. The angle of elevation—the angle your line of sight makes with the horizontal ground—is $45^\circ$.

In the triangle formed by the pole, the ground, and your line of sight, the height of the pole and the distance from the pole's base to you are equal. Since the pole is $20	ext{ m}$ high, the distance from the base of the pole to your observation point $O$ must also be $20	ext{ m}$.

### The Moment of Departure

Suddenly, the bird takes flight. It flies perfectly horizontally, straight away from you, meaning its altitude remains a constant $20	ext{ m}$ above the ground.

As it flies, the distance between the bird and the pole increases, and consequently, the distance between the bird and your observation point $O$ increases. After exactly one second, you look up again, and the angle of elevation has dropped to $30^\circ$.

The triangle has stretched. The height remains $20	ext{ m}$, but the base of this new, larger triangle is longer.

### The Trigonometric Bridge

We now have two right-angled triangles sharing the same height of $20	ext{ m}$. Let the initial position of the bird be $A$ and the new position be $B$.

The base of the first triangle is $OA'$, where $A'$ is the base of the pole. We already know $OA' = 20	ext{ m}$.

For the second triangle $OBB'$, where $B'$ is the point on the ground directly below the bird's new position $B$, we use the tangent function:

$$	an 30^\circ = \frac{BB'}{OB'}$$

Given $	an 30^\circ = \frac{1}{\sqrt{3}}$ and $BB' = 20	ext{ m}$, we substitute these values:

$$\frac{1}{\sqrt{3}} = \frac{20}{OB'}$$

Solving for $OB'$, we find that the new horizontal distance is:

$$OB' = 20\sqrt{3}	ext{ m}$$

### Final Calculation

The distance the bird traveled in that one second is the difference between these two horizontal distances: the distance from $O$ to the new position $B'$ minus the distance from $O$ to the original position $A'$.

$$d = OB' - OA' = 20\sqrt{3} - 20$$

Factoring out the $20$, we get:

$$d = 20(\sqrt{3} - 1)	ext{ m}$$

Since the bird covered this distance in exactly $1	ext{ s}$, its speed is the distance divided by time:

$$	ext{Speed} = \frac{20(\sqrt{3} - 1)}{1} = \mathbf{20(\sqrt{3} - 1)	ext{ m/s}}$$

There is a profound elegance in how trigonometry allows us to track the motion of an object simply by observing the changing angles of our perspective. You have successfully mapped the bird's flight using the properties of triangles and the steady hand of mathematics.

Select Answer:

Visualized Solution

Visualizing the Setup

Drawing the Initial Line of Sight

Applying Trigonometry in $Δ O A A^{'}$

Calculating Initial Distance $O A^{'}$

Visualizing the Horizontal Flight

Drawing the New Line of Sight

Applying Trigonometry in $Δ O B B^{'}$

Calculating Final Distance $O B^{'}$

Finding the Distance Traveled

Calculating the Speed of the Bird

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Moment of Departure

The Trigonometric Bridge

Final Calculation

Similar Questions

An aeroplane flying at a constant speed, parallel to the horizontal ground, $3$ km above it, is observed at an elevation of $6 0^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $3 0^{\circ}$ , then the speed (in km/hr) of the aeroplane, is :

The angle of elevation of a jet plane from a point A on the ground is $6 0^{\circ}$ . After a flight of 20 seconds at speed of 432 km/ hour, the angle of elevation changes to $3 0^{\circ}$ . If the jet plane is flying at a constant height, then its height is:

A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of $3 0^{\circ}$ . After some time, the angle of depression becomes $6 0^{\circ}$ . The distance (in metres) travelled by the car during this time is

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is