A ladder leans against a wall. Its top touches the wall 24 ft above the ground and its base is 7 ft from the wall. Find the length of the ladder and the angle it makes with the ground.

Step-by-step solution

1. 1

   Model the situation as a right triangle. The wall is vertical and the ground is horizontal, so they meet at $90^\circ$. The ladder is the hypotenuse, the height up the wall ($24$ ft) is the side opposite the ground angle, and the distance from the wall ($7$ ft) is the side adjacent to it. Drawing this once removes all the ambiguity about which side is which.

2. 2

   Apply the Pythagorean theorem for the length. With legs $7$ and $24$ and hypotenuse $c$,

   $c^2=7^2+24^2=49+576=625$
   $c=\sqrt{625}=25\text{ ft}$

   The length comes out a whole number because $(7,24,25)$ is a Pythagorean triple — a common sign that a textbook problem has been set up deliberately.

3. 3

   Choose the trig ratio that uses only the given data. Let $\theta$ be the angle at the ground. The two given sides are the opposite ($24$) and the adjacent ($7$), and the ratio built from exactly those two is the tangent:

   $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{24}{7}$

   Using tangent means an error in step 2 cannot propagate into the angle.

4. 4

   Invert the tangent to get the angle.

   $\theta=\arctan\!\left(\frac{24}{7}\right)=\arctan(3.4286)\approx 73.74^\circ$

   Make sure the calculator is in degree mode; in radian mode the same keystrokes return $1.287$.

5. 5

   Check the angle with a different ratio. Using the computed hypotenuse,

   $\sin\theta=\frac{24}{25}=0.96,\qquad \cos\theta=\frac{7}{25}=0.28$

   Both $\arcsin(0.96)$ and $\arccos(0.28)$ return $73.74^\circ$, and $0.96^2+0.28^2=0.9216+0.0784=1$ confirms the triangle is consistent. A steep angle is also what you expect physically: the ladder rises $24$ ft while moving only $7$ ft out.