If a 16' ladder is placed correctly on a level surface, how high up will the ladder reach?

Question

So i have just began learning about sin cos and tan, and i came across this problem and for some reason I'm having trouble figuring it out. *** When using a straight ladder, it is recommended that the base of the ladder by place approximately 1/4 length of the entire ladder away from the wall.

a) If a ladder is placed correctly on a level surface, what is the angle formed between the ground and the ladder?

b) If a 16' ladder is placed correctly on a level surface, how high up will the ladder reach?

So i understand that cos = adjacent / hypotenuse And that the distance of the ladder is 4' away from the wall but I'm lost from there.

Answer 1

A. Here, you should use cosine as you stated. Set up an equation in the format you described: cos(theta) = 4/16. Then, you use something called arccos, which is just cos^(-1). This crosses out the cos on the left side, and you end up with theta = arccos(1/4). Plug this into your calculator to get your answer: 75.52248781.

B. Finding the how high the ladder goes is easy. You just use the Pythagorean Theorem. One leg is 4 feet and the hypotenuse is 16 feet, so the other leg needs to be 4*sqrt(15).

Answer 2

We can think of the ground and the wall as the legs of a right triangle and the ladder as the hypotenuse. Then, if we let $\theta$ be the angle between the ground and the ladder, $\cos(\theta) = \frac{adj}{hyp} = \frac{4}{16} = \frac{1}{4}$. We then find the inverse cosine of $\frac{1}{4}$. (what angle, when cosined, gives $\frac{1}{4}$?) Using a calculator gives the value $1.318...$ radians or about $75.52$ degrees.

Part b) needs no trigonometry. We know that one leg has length $4$ and the hypotenuse has length $16$, so we can use the Pythagorean Theorem to find the length of the other leg (the effective height) : $4^2 + b^2 = 16^2$ or $b = \sqrt{240}\approx 15.49$