With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

\(\displaystyle \cos = \frac{adjacent}{hypotenuse}\)

\(\displaystyle \cos\left ( \theta\right ) = \frac{15}{25} = 0.6\)

\(\displaystyle \arccos\left ( 0.6\right ) = \theta\)

\(\displaystyle 53.1^{\circ}= \theta\)

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

\(\displaystyle \cos = \frac{adjacent}{hypotenuse}\)

\(\displaystyle \cos\left ( \theta\right ) = \frac{12}{30} = 0.4\)

\(\displaystyle \arccos\left ( 0.4\right ) = \theta\)

\(\displaystyle 66.4^{\circ}= \theta\)

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. However, if we plug the given values into the formula for cosine, we get:

\(\displaystyle \cos = \frac{adjacent}{hypotenuse}\)

\(\displaystyle \cos\left ( \theta\right ) = \frac{17}{13} = 1.3\)

\(\displaystyle \arccos\left ( 1.3\right ) = \theta = \textup{no solution}\)

This problem does not have a solution. The sides of a right triangle must be shorter than the hypotenuse. A triangle with a side longer than the hypotenuse cannot exist. Similarly, the domain of the arccos function is \(\displaystyle \left [ -1, 1\right ]\). It is not defined at 1.3.

You can draw your scenario using the following right triangle:

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

\(\displaystyle \small \small \Theta = cos^-^1(\frac{27}{50})=57.31636115374206\) or \(\displaystyle \small 57.32\) degrees.

Recall that the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse of the triangle. You can solve for the angle by using an inverse cosine function:

\(\displaystyle \small \Theta = cos^-^1(\frac{47}{140})=70.38402086459453\) or \(\displaystyle 70.38$^{\circ}$\).

Our answer lies in inverse functions. If the buttress is \(\displaystyle 25\) feet long and is \(\displaystyle 24\) feet up the ladder at the desired angle, then:

\(\displaystyle \textup{cos }x^{\circ} = \frac{24}{25}\)

Thus, using inverse functions we can say that \(\displaystyle x = \textup{cos}^{-1}\frac{24}{25} \approx 16.26\)

Thus, our buttress strikes the buliding at approximately a \(\displaystyle 16^{\circ}\) angle.

Our answer lies in inverse functions. If the monument is \(\displaystyle 8\) meters away and the camera is \(\displaystyle 12\) meters from the monument's top at the desired angle, then:

\(\displaystyle \textup{cos} x^{\circ} = \frac{8}{12}\)

Thus, using inverse functions we can say that \(\displaystyle x = \textup{cos}^{-1}\frac{8}{12} \approx 48.189\)

Thus, our buttress strikes the buliding at approximately a \(\displaystyle 48.19^{\circ}\) angle.