$15\sqrt{2}\text{m}$
This problem involves finding the length of a ladder leaning against a wall using basic trigonometry.
We can model this situation as a right-angled triangle where:
We are given:
The trigonometric ratio that relates the adjacent side and the hypotenuse is the cosine function:
\(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
\(\cos(45^\circ) = \frac{15 \text{ m}}{\text{Hypotenuse}}\)
\(\cos(45^\circ) = \frac{1}{\sqrt{2}}\)
\(\frac{1}{\sqrt{2}} = \frac{15 \text{ m}}{\text{Hypotenuse}}\)
\(\text{Hypotenuse} = 15 \text{ m} \times \sqrt{2}\)
\(\text{Hypotenuse} = 15\sqrt{2} \text{ m}\)
The length of the ladder is \(15\sqrt{2}\) m.
There is a 10 m tall tower between two parallel roads. The angles of depression of the roads from the top of the tower are 30° and 45° respectively. Approximately, how far are the roads from each other?
Two pillars are placed vertically 8 feet apart. The height difference of the two pillars is 6 feet. The two ends of a rope of length 15 feet are tied to the tips of the two pillars. The portion of the length of the taller pillar that can be brought in contact with the rope without detaching the rope from the pillars is