revathi is standing at the top of abuilding of height 25sqrt3 meters and observes a car at an angle of 30 which is approaching the foot of the building with a uniform speed. five seconds later angle of depression is 60 draw fig, what is the distance of the observer from the car when it makes an angle of 60 find the speed of the car in m/sec

Let's break down the problem into clear steps with a diagram, then solve for the distance and speed.

### Step 1: Diagram Explanation
Revathi is standing on a building of height \( 25\sqrt{3} \) meters. She observes a car at two different times:

- Initially, the angle of depression to the car is \( 30^\circ \).
- After 5 seconds, the angle of depression to the car becomes \( 60^\circ \).

We assume that the car is moving at a uniform speed. From this, we can deduce that we are dealing with two right-angled triangles, where the building height is the vertical side, and the horizontal distance to the car is the other side of the triangle.

### Step 2: Formulating the Triangles

We use trigonometric relationships from the right triangles formed by the height of the building and the distances from the foot of the building to the car. 

#### At \( 30^\circ \) angle of depression:
- The height of the building is \( 25\sqrt{3} \) meters.
- Let the horizontal distance from the car to the building initially be \( x_1 \).

From the triangle, using tangent:
\[
\tan(30^\circ) = \frac{\text{Height}}{\text{Horizontal Distance}}
\]
\[
\tan(30^\circ) = \frac{25\sqrt{3}}{x_1}
\]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[
\frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{x_1}
\]
Solving for \( x_1 \):
\[
x_1 = 75 \text{ meters}
\]

#### At \( 60^\circ \) angle of depression:
Let the new horizontal distance after 5 seconds be \( x_2 \). Using the same tangent relationship:
\[
\tan(60^\circ) = \frac{\text{Height}}{\text{Horizontal Distance}}
\]
\[
\tan(60^\circ) = \frac{25\sqrt{3}}{x_2}
\]
Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[
\sqrt{3} = \frac{25\sqrt{3}}{x_2}
\]
Solving for \( x_2 \):
\[
x_2 = 25 \text{ meters}
\]

### Step 3: Finding the Speed of the Car
The car travels from \( x_1 = 75 \) meters to \( x_2 = 25 \) meters in 5 seconds. The distance traveled by the car is:
\[
\Delta x = x_1 - x_2 = 75 - 25 = 50 \text{ meters}
\]

Since the car covers 50 meters in 5 seconds, the speed of the car is:
\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{50}{5} = 10 \text{ m/s}
\]

### Final Answer:
- The distance of the observer from the car when the angle of depression is \( 60^\circ \) is **25 meters**.
- The speed of the car is **10 m/s**.

Would you like to see the diagram, or do you have any other questions?

Here are 5 related questions to deepen your understanding:
1. How does the tangent function help in calculating distances in such problems?
2. What would happen to the distance if the angle of depression were increased further?
3. How would the speed change if the time to change the angle from 30° to 60° was doubled?
4. How would the calculation change if the height of the building was halved?
5. Could this problem be solved using any other trigonometric functions, such as sine or cosine?

**Tip:** Always draw a clear diagram to visualize angles of depression or elevation—it simplifies understanding and calculation.

Revathi is standing at the top of a building of height 25√3 meters and observes a car at an angle of 30° which is approaching the foot of the building with a uniform speed. Five seconds later, the angle of depression is 60°. Draw a figure. What is the distance of the observer from the car when it makes an angle of 60°? Find the speed of the car in m/s.

Solve a trigonometry problem involving angles of depression. Revathi observes a car from the top of a 25√3-meter-high building, with initial and final angles of 30° and 60° over a span of 5 seconds. Learn how to calculate the car's distance and speed using tangent formulas. Suitable for high school students in Grades 10-12.

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