Question

practice solving for angle measures of right triangles. which equation could be used to find ( mangle e ) in ( \triangle efg )? ( mangle e=cos ^{-1}left(\frac{3}{4.6}
ight) ) ( mangle e=cos ^{-1}left(\frac{4.6}{3}
ight) ) ( mangle e=\tan ^{-1}left(\frac{3}{4.6}
ight) ) ( mangle e=\tan ^{-1}left(\frac{4.6}{3}
ight) )

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 \(\angle E\) in \(\triangle EFG\) with right - angle at \(F\), the side opposite to \(\angle E\) is \(FG = 3\), the side adjacent to \(\angle E\) is \(EF=4.6\).

Step2: Apply the tangent ratio formula

The formula for the tangent of an angle \(\theta\) in a right - triangle is \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). If \(\theta = \angle E\), then \(\tan\angle E=\frac{FG}{EF}\). Using the inverse - tangent function \(m\angle E=\tan^{-1}(\frac{FG}{EF})\). Substituting \(FG = 3\) and \(EF = 4.6\), we get \(m\angle E=\tan^{-1}(\frac{3}{4.6})\).

Answer:

\(m\angle E=\tan^{-1}(\frac{3}{4.6})\) (corresponds to the third option \(m\angle E=\tan^{-1}(\frac{3}{4.6})\))