a ladder 8 meters long leans againts the wall of a building. if the foo|t of the ladder makes an angle of 68^0 with the ground how far is the base of the ladder from the wall ?
Geometry help needed here. What is "Probl"? Is that what your School Subject is called in your school?
x/8 = cos68°
To solve this problem, we can use trigonometric functions. Let's call the distance from the base of the ladder to the wall "x".
We are given the length of the ladder (8 meters) and the angle between the ground and the ladder (68 degrees).
Using trigonometry, we can set up the following equation:
cos(68°) = x/8
To find the value of "x", we need to solve this equation.
Let's calculate the value of cos(68°):
cos(68°) ≈ 0.393
Plugging in the values into the equation:
0.393 = x/8
To isolate "x", we can multiply both sides of the equation by 8:
8 * 0.393 = x
x ≈ 3.144
Therefore, the base of the ladder is approximately 3.144 meters away from the wall.
To find the distance between the base of the ladder and the wall, we can use trigonometry. Let's call the distance we're trying to find "x".
We have the length of the ladder (8 meters) and the angle it makes with the ground (68 degrees). We can use the sine function to relate these two values:
sin(68°) = opposite / hypotenuse
In this case, the opposite side is x (the distance between the base of the ladder and the wall), and the hypotenuse is 8 meters.
Therefore, we can rewrite the equation as:
sin(68°) = x / 8
To find x, we can rearrange the equation:
x = sin(68°) * 8
Now, we can solve the equation:
x = sin(68°) * 8
x ≈ 7.191 meters
Therefore, the base of the ladder is approximately 7.191 meters away from the wall.