Question

solve for $x$. round to the nearest tenth of a degree, if necessary.
answer attempt 1 out of 2
$x=$

Explanation:

Step1: Use the sine function

In a right - triangle, the sine of an angle \(x\) is defined as \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, the side opposite to angle \(x\) is \(FG\) (but we can also use the formula \(\sin x=\frac{\text{length of the side opposite to }x}{\text{length of the hypotenuse}}\). For angle \(x\) in right - triangle \(FGE\) (right - angled at \(F\)), \(\sin x=\frac{FG}{GE}\) is not correct. Wait, correct formula: \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). In right - triangle \(FGE\) (right - angled at \(F\)), \(\sin x=\frac{FG}{GE}\) is wrong. Correct: \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). Let's use \(\sin x=\frac{FG}{GE}\) no. Wait, in right - triangle \(FGE\) (right - angled at \(F\)), \(\sin x=\frac{FG}{GE}\) is wrong. Wait, \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite to \(x\) is \(FG\) (no, wait, \(FG = 54\), \(GE=78\). \(\sin x=\frac{FG}{GE}\) is wrong. Wait, \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite to \(x\) is \(FG\) (no, in right - triangle \(FGE\) (right - angled at \(F\)), \(\sin x=\frac{FG}{GE}\) is wrong. Wait, \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite to \(x\) is \(FG\) (no, \(x\) is at \(E\), so the side opposite to \(x\) is \(FG = 54\), hypotenuse \(GE = 78\). So \(\sin x=\frac{54}{78}\).

Step2: Calculate \(x\)

\(x=\sin^{- 1}(\frac{54}{78})\). First, simplify \(\frac{54}{78}=\frac{9}{13}\approx0.6923\). Then \(x = \sin^{-1}(0.6923)\). Using a calculator, \(x\approx43.8^{\circ}\).

Answer:

\(43.8\)