Question

Hokiris ladder has two legs that are each 8 feet long. When the ladder is opened safely and locked for use, the legs are 4 feet apart on the ground. What is the angle that is formed at the top of the ladder where the legs meet

Answer 1

To solve this problem you must apply the proccedure shown below:

1- The ladder forms an isosceles triangle, because it has two equal sides.

2- Let's call the angle $\alpha$. To calculate it , you can divide the triangle into two equal right triangles and the angle $\alpha$ will be divided into two equal parts too. Therefore, you can calculate the half of the angle and then multiply it by two, as following:

$Sin^{-1}(\alpha 1)=opposite/hypotenuse$

$Sin^{-1}(\alpha1)=8/2\\ \alpha1=14.47 degrees$

$\alpha =2\alpha 1\\ 2=2(14.47degrees)\\ \alpha =28.95degrees$

The answer is: $28.95degrees$