Respuesta:
To solve this problem, we can use trigonometric ratios. Let's denote:
- \( L \) as the length of the ladder (5m)
- \( \theta_1 \) as the original angle the ladder makes with the ground (70°)
- \( \theta_2 \) as the new angle the ladder makes with the ground after slipping down the wall
- \( d \) as the distance the ladder slips on the ground
First, we need to find the original distance the ladder is from the wall:
\[ \text{Original distance from wall} = L \times \cos(\theta_1) \]
\[ \text{Original distance from wall} = 5 \times \cos(70°) \]
\[ \text{Original distance from wall} ≈ 5 \times 0.342 ≈ 1.71 \, \text{m} \]
Now, after slipping down 2m, the ladder is still at the same height as the original position, so the new distance from the wall is \( 1.71 - 2 = -0.29 \) meters. However, this negative distance doesn't make sense geometrically, so we need to consider the absolute value, which gives \( 0.29 \) meters.
Now, using trigonometry, we can find the new angle \( \theta_2 \):
\[ \tan(\theta_2) = \frac{\text{Opposite}}{\text{Adjacent}} \]
\[ \tan(\theta_2) = \frac{\text{Opposite}}{5 - 2} \]
\[ \tan(\theta_2) = \frac{0.29}{3} \]
\[ \theta_2 ≈ \arctan\left(\frac{0.29}{3}\right) \]
\[ \theta_2 ≈ \arctan(0.0967) ≈ 5.53° \]
So, the new angle \( \theta_2 \) is approximately \( 5.53° \).
To find the distance the ladder slips on the ground from its original position, we can use trigonometry again:
\[ \text{Distance slipped} = \text{Original distance from wall} - \text{New distance from wall} \]
\[ \text{Distance slipped} = 1.71 - 0.29 \]
\[ \text{Distance slipped} ≈ 1.42 \, \text{m} \]
Therefore:
i. The new angle which the ladder makes with the ground is approximately \( 5.53° \).
ii. The distance the ladder slipped back on the ground from its original position is approximately \( 1.42 \) meters.
Explicación paso a paso:
I hope it works for you.
