A pole 25m long is placed against a vertical wall such that its lower end is 7m from the foot of the wall on the same horizontal ground. If the upper end of the pole is push down by 2m, calculate correct to 2 significant figure: 1) how much further away from the wall the lower end will move; 2) the angle of the pole now makes with the horizontal.
A pole 25cm long is placed against a vertical wall such that its lower end is 7cm from the foot of the wall on the same horizontal ground. If the upper end of the pole is pushed downward by 2cm,calculate to two significant figure: 1)How much further away from the wall the lower end will move:2)The angle the pole now makes with the horizontal.
I don't have any idea so help me out
25^2=base^2+vertical^2
on the fist position, calculate the vertical distance up the wall.
Then subtract 2m, then recalculate the new base.
Hsmdidbevbz
To solve this problem, we can use similar triangles. Let's break it down step by step:
1) How much further away from the wall will the lower end move?
We have a right triangle formed by the pole, the distance from the foot of the wall to the lower end of the pole, and the distance from the foot of the wall to the point where the lower end of the pole will move.
Using the Pythagorean theorem, we can find the original distance of the lower end from the wall (let's call it x):
x² + 7² = 25²
x² + 49 = 625
x² = 576
x = √576
x ≈ 24
Now, when the upper end of the pole is pushed down by 2m, the pole will also move downward, causing the lower end to move away from the wall. To find how much further away it will move, we need to find the new distance of the lower end from the wall (let's call it y):
Using the same concept of similar triangles:
7/25 = y/(25+2)
7(25+2) = 25y
7(27) = 25y
189 = 25y
y = 189/25
y ≈ 7.56
Therefore, the lower end of the pole will move approximately 7.56m further away from the wall.
2) What is the angle the pole now makes with the horizontal?
To find the angle, we can use trigonometry. We know the height of the pole and the base length.
The height of the pole after it is pushed down is 25-2 = 23m.
The tangent of an angle is equal to the opposite side divided by the adjacent side. In this case, the opposite side is the height of the pole (23m), and the adjacent side is the base length (7.56m).
tan(θ) = opposite/adjacent
tan(θ) = 23/7.56
θ = tan^(-1)(23/7.56)
Using a calculator, we find:
θ ≈ 71.11 degrees
Therefore, the angle the pole now makes with the horizontal is approximately 71.11 degrees.