CPM Homework Help : INT2 Problem 11-14

Hokiri’s 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?

Hint:

Create a diagram of the ladder with a perpendicular bisector from the top ( $∠C$ ) to the base ( $\overline{AB}$ ).

More help:

Next, use the right triangle $ADC$ and a calculator to find the angle whose sine is equal to $\frac{2}{8}$ .

Steps:

$\text{sin}\left(\frac{m\angle ACD}{2}\right)=\frac{2}{8}$ or $\frac{m\angle ACD}{2}=\text{arcsin}\left(\frac{2}{8}\right)$

Since $\angle ACD$ is half of the desired angle, double $m\angle ACD$ to get $m∠ACB$ .

Answer:

$m∠ACB=29°$

Triangle, with vertices labeled, bottom left, A, bottom right, B, top, C. A dashed line segment from, C, perpendicular to base, at point labeled, D. Left and right sides, of triangle, A, B, C, each labeled, 8. Segments, A, D, &, D, B, each labeled 2.