Question

solve for x. round to the nearest tenth, if necessary.
answer
x -

Explanation:

Step1: Use trigonometric function

In right - triangle \( \triangle MNO\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 29^{\circ}\), the opposite side to angle \(M\) is \(ON=x\), and the hypotenuse is \(OM\). Given \(ON = x\), \(OM\) is not given directly, but we know that \(\sin29^{\circ}=\frac{x}{OM}\). Wait, no, actually, using \(\tan\) is wrong. Correct: \(\sin M=\frac{ON}{OM}\) is wrong. Wait, in right - triangle \( \triangle MNO\) with right - angle at \(N\), \(\sin M=\frac{ON}{OM}\) (no, \(\sin M=\frac{ON}{OM}\) is wrong. Wait, \(\sin M=\frac{ON}{OM}\) (no, \(\sin\alpha=\frac{\text{opposite}}{\text{hypotenuse}}\). For angle \(M\) (\(29^{\circ}\)), the opposite side is \(ON\) and the hypotenuse is \(OM\). But we can also use \(\tan\) wrong. Wait, correct: \(\sin29^{\circ}=\frac{x}{OM}\) (no). Wait, in right - triangle \( \triangle MNO\), \(\sin M=\frac{ON}{OM}\) (no, \(\sin M=\frac{ON}{OM}\) (no). Wait, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For angle \(M = 29^{\circ}\), opposite side \(ON=x\), hypotenuse \(OM\). But we can use \(\tan\) wrong. Wait, no, \(\tan M=\frac{ON}{MN}\) (no, \(MN\) is adjacent. Wait, no, in right - triangle \( \triangle MNO\) with \(\angle N = 90^{\circ}\), \(\sin M=\frac{ON}{OM}\), \(\cos M=\frac{MN}{OM}\), \(\tan M=\frac{ON}{MN}\). But we know \(ON = x\), \(OM\) is not given. Wait, no, we can use \(\tan\). Wait, \(\tan M=\frac{ON}{MN}\) (no, \(MN\) is adjacent. Wait, no, \(\tan M=\frac{\text{opposite}}{\text{adjacent}}\). Opposite to \(M\) is \(ON\), adjacent is \(MN\). But we know \(ON=x\), \(OM = 6.8\) (wait, no, \(ON\) is \(x\), \(OM\) is hypotenuse. Wait, no, the side given is \(OM = 6.8\) (no, \(ON\) is \(x\), \(OM\) is hypotenuse. Wait, no, \(ON\) is opposite to \(M\), \(OM\) is hypotenuse. \(\sin M=\frac{ON}{OM}\). Given \(M = 29^{\circ}\), \(OM = 6.8\). So \(x=OM\times\sin29^{\circ}\).

Step2: Calculate the value

We know that \(\sin29^{\circ}\approx0.485\), and \(OM = 6.8\). Then \(x = 6.8\times\sin29^{\circ}\). Substitute \(\sin29^{\circ}\approx0.485\) into the formula: \(x=6.8\times0.485 = 3.3\)

Answer:

\(3.3\)