Question

which equation can be used to find the measure of angle bac? $\tan^{-1}\left(\frac{5}{12}\
ight)=x$ $\tan^{-1}\left(\frac{12}{5}\
ight)=x$ $\cos^{-1}\left(\frac{12}{13}\
ight)=x$ $\cos^{-1}\left(\frac{13}{12}\
ight)=x$

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) for an acute angle \(\theta\).
For angle \(x = \angle BAC\), the opposite side to \(\angle BAC\) is \(BC = 12\), the adjacent side to \(\angle BAC\) is \(AC=5\), and the hypotenuse \(AB = 13\).

Step2: Check tangent formula

\(\tan x=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\tan x=\frac{BC}{AC}=\frac{12}{5}\), so \(x = \tan^{-1}(\frac{12}{5})\).

Step3: Check cosine formula

\(\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here, \(\cos x=\frac{AC}{AB}=\frac{5}{13}\), not \(\frac{12}{13}\) or \(\frac{13}{12}\) (since \(\frac{13}{12}>1\) and the range of the cosine function for real - valued angles is \([- 1,1]\)).

Answer:

\(\tan^{-1}(\frac{12}{5})=x\) (the second option)