A straight highway leads to the top of the base of a 50 meter tall tower. Frm the top of the tower, if yhe angles of depression of two cars on the road are 30° and 60° , then the distence between the tow cars is ................meter.

100squreroot3/3

To solve this question, we can use trigonometry and the concept of angles of depression.

Let's break down the given information:
- The tower is 50 meters tall.
- There are two cars on the road.
- The angle of depression to the first car is 30° (let's call this angle A).
- The angle of depression to the second car is 60° (let's call this angle B).

Now, let's visualize the scenario.
- Draw a horizontal line to represent the road.
- Draw a perpendicular line from the top of the tower to the road, representing the height of the tower.
- Label the distance between the first car and the base of the tower as x (let's call this distance AB).
- Label the distance between the second car and the base of the tower as y (let's call this distance BC).

We can use trigonometric ratios to find the distances x and y.

For angle A (30°):
- The tangent of angle A is equal to the opposite side (50 meters) divided by the adjacent side (x meters). So, tan(30°) = 50/x.
- Rearranging this equation, we have 1/sqrt(3) = 50/x.
- Solving for x, we get x = 50/sqrt(3).

For angle B (60°):
- The tangent of angle B is equal to the opposite side (50 meters) divided by the adjacent side (y meters). So, tan(60°) = 50/y.
- Rearranging this equation, we have sqrt(3) = 50/y.
- Solving for y, we get y = 50/sqrt(3).

Now, we need to find the distance between the two cars. Let's call this distance AC.
- We can use the Pythagorean theorem, which states that the square of the hypotenuse (AC) is equal to the sum of the squares of the other two sides (AB and BC). So, AC^2 = AB^2 + BC^2.
- Substituting the values we found for AB and BC, we get AC^2 = (50/sqrt(3))^2 + (50/sqrt(3))^2.
- Simplifying this equation, we have AC^2 = 2500/3 + 2500/3 = 5000/3.
- Taking the square root of both sides, we find AC = sqrt(5000/3).

Therefore, the distance between the two cars is approximately sqrt(5000/3) meters.