This image presents a problem involving a ladder that leans against a house wall, forming a right triangle. The ladder's length is 3.70 meters, and it leans at an angle of 75° to the ground. The problem is divided into several parts that ask for different calculations based on this setup.

Problem Summary:

• The length of the ladder $c = 3.70 \, m$.
• The angle between the ladder and the ground $\theta = 75^\circ$.
• The ladder forms a right-angled triangle with the wall (height $a$) and the ground (base $b$).

Solutions:

a) Does the ladder reach the windowsill at a height of 3 meters?

To determine if the ladder reaches a height of 3 meters, we can calculate the vertical height $a$ using trigonometry:

$a = c \cdot \sin(\theta)$ $a = 3.70 \, m \cdot \sin(75^\circ)$ $a \approx 3.57 \, m$

Since the ladder reaches approximately 3.57 meters, it does reach the windowsill, which is at 3 meters.

b) Distance of the foot of the ladder from the house?

We can find the horizontal distance $b$ (from the foot of the ladder to the wall) using:

$b = c \cdot \cos(\theta)$ $b = 3.70 \, m \cdot \cos(75^\circ)$ $b \approx 0.96 \, m$

Thus, the foot of the ladder is approximately 0.96 meters away from the wall.

c) For which heights does the angle fall between 68° and 83°?

We need to determine the heights corresponding to angles $68^\circ \leq \theta \leq 83^\circ$. For each angle $\theta$, the height $a$ can be calculated using the same formula $a = c \cdot \sin(\theta)$.

1. For $\theta = 68^\circ$: $a = 3.70 \, m \cdot \sin(68^\circ) \approx 3.42 \, m$

2. For $\theta = 83^\circ$: $a = 3.70 \, m \cdot \sin(83^\circ) \approx 3.68 \, m$

Hence, the height of the ladder's top moves between 3.42 meters and 3.68 meters when the angle is between 68° and 83°.

d) Calculate the angle when the ladder reaches 3.50 meters height.

We need to find the angle $\theta$ such that the height $a = 3.50 \, m$. Using:

$\sin(\theta) = \frac{a}{c}$ $\sin(\theta) = \frac{3.50 \, m}{3.70 \, m}$ $\theta = \arcsin\left( \frac{3.50}{3.70} \right) \approx \arcsin(0.946) \approx 71.85^\circ$

The angle is approximately $71.85^\circ$.

e) How long must a ladder be to reach a 6.5-meter-high window with an angle less than 80°?

We need to find the length $c$ of the ladder that reaches a height of 6.5 meters with an angle $\theta \leq 80^\circ$.

$a = c \cdot \sin(80^\circ)$ $6.50 \, m = c \cdot \sin(80^\circ)$ $c = \frac{6.50 \, m}{\sin(80^\circ)} \approx \frac{6.50 \, m}{0.985} \approx 6.60 \, m$

So, the ladder must be approximately 6.60 meters long.

Would you like further details on any of the solutions, or do you have any questions?

Related Questions:

1. What other trigonometric identities can be used to solve ladder-related problems?
2. How would the answers change if the ladder's length increased to 4 meters?
3. Can the cosine rule be used in any part of this problem? Why or why not?
4. How does the height reached by the ladder vary with different angles of inclination?
5. What real-world factors, such as friction or material strength, could influence the ladder's stability?

Tip:

Always double-check your angle measurements and unit consistency when working with trigonometric functions to avoid calculation errors.