Question

1. a custodian secures a ladder of l the school. the ladder makes an a will the custodian be able to reach the ground? justify your answer. 11. hannah wants to make a lean - to shelter against a tree. she starts with a plank that is 2.1 m long. if she wants to have a 45° angle between the ground and the lower end of the plank, how far away from the base of the tree should the lower end of the lean - to be? 12. jeff will use right triangles in his design for the elevator to take resort guests down the cliff. he plans to have an angle of 45°, and a diagonal length of 1.2 m. how long will the vertical piece for this part of the elevator be?

Explanation:

Question 11

Step1: Identify the right - triangle relationship

We have a right - triangle where the length of the hypotenuse (the plank) is $c = 2.1$ m and the angle between the ground and the plank is $\theta=45^{\circ}$. We want to find the adjacent side $x$ to the given angle. In a right - triangle, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$.

Step2: Substitute the values into the cosine formula

Since $\cos\theta=\cos45^{\circ}=\frac{\sqrt{2}}{2}$ and $c = 2.1$ m, and $\cos\theta=\frac{x}{c}$, we can solve for $x$. So $x = c\times\cos\theta$.
$x=2.1\times\cos45^{\circ}=2.1\times\frac{\sqrt{2}}{2}\approx2.1\times0.707 = 1.4847\approx1.48$ m

Step1: Identify the right - triangle relationship

We have a right - triangle with a $45^{\circ}$ angle and the hypotenuse length $c = 1.2$ m. We want to find the length of the vertical side (opposite side to the $45^{\circ}$ angle). In a $45 - 45-90$ right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, and for $\theta = 45^{\circ}$, $\sin45^{\circ}=\frac{\sqrt{2}}{2}$.

Step2: Substitute the values into the sine formula

Let the vertical side be $y$. Since $\sin\theta=\frac{y}{c}$, and $c = 1.2$ m, $\sin45^{\circ}=\frac{\sqrt{2}}{2}$, we have $y=c\times\sin\theta$.
$y = 1.2\times\sin45^{\circ}=1.2\times\frac{\sqrt{2}}{2}=1.2\times0.707 = 0.8484\approx0.85$ m

Answer:

The lower end of the lean - to should be approximately $1.48$ m away from the base of the tree.

Question 12