Question

a support beam needs to be placed at a 28° angle of elevation so that the top meets a vertical beam 1.6 meters above the horizontal floor. the vertical beam meets the floor at a 90° angle.
law of sines: (\frac{sin(a)}{a} = \frac{sin(b)}{b} = \frac{sin(c)}{c})
approximately how far from the vertical beam should the lower end of the support beam be placed along the horizontal floor?
3.4 meters
4.4 meters
3.0 meters
3.9 meters

Explanation:

Step1: Identify the triangle type

We have a right triangle (since one angle is \(90^\circ\)) with the vertical side (opposite the \(28^\circ\) angle) being \(1.6\) meters, and we need to find the adjacent side (horizontal distance from vertical beam), let's call it \(x\). We can use the tangent function, where \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\).

Step2: Apply the tangent formula

Given \(\theta = 28^\circ\), opposite side \(= 1.6\) m, adjacent side \(= x\). So \(\tan(28^\circ)=\frac{1.6}{x}\). We know that \(\tan(28^\circ)\approx0.5317\).

Step3: Solve for \(x\)

Rearranging the formula: \(x=\frac{1.6}{\tan(28^\circ)}\). Substituting \(\tan(28^\circ)\approx0.5317\), we get \(x=\frac{1.6}{0.5317}\approx3.0\) meters? Wait, no, wait. Wait, maybe I mixed up opposite and adjacent. Wait, the angle of elevation is \(28^\circ\), so the vertical side is opposite? Wait, no: the support beam is the hypotenuse, the vertical beam is the opposite side to the \(28^\circ\) angle, and the horizontal distance is the adjacent side. Wait, no, let's re-examine. The vertical beam is 1.6 m (height), the horizontal distance is \(x\) (adjacent), and the support beam is the hypotenuse. The angle of elevation is between the horizontal and the support beam, so \(\tan(28^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{1.6}{x}\). Wait, but if we calculate \(\frac{1.6}{\tan(28^\circ)}\), let's compute \(\tan(28^\circ)\approx0.5317\), so \(1.6\div0.5317\approx3.0\)? But wait, maybe I made a mistake. Wait, no, wait, maybe it's a right triangle, so we can also use the law of sines. Let's check the angles. The triangle has angles: \(28^\circ\), \(90^\circ\), so the third angle is \(62^\circ\). Let's denote the vertical side as \(a = 1.6\) m (opposite \(28^\circ\)), the horizontal side as \(b\) (opposite \(62^\circ\)), and the hypotenuse as \(c\). By law of sines: \(\frac{a}{\sin(28^\circ)}=\frac{b}{\sin(62^\circ)}\). So \(b=\frac{1.6\times\sin(62^\circ)}{\sin(28^\circ)}\). \(\sin(62^\circ)\approx0.8829\), \(\sin(28^\circ)\approx0.4695\). So \(b=\frac{1.6\times0.8829}{0.4695}\approx\frac{1.4126}{0.4695}\approx3.0\) meters? Wait, but the options include 3.0, 3.4, 3.9, 4.4. Wait, maybe my initial assumption is wrong. Wait, no, wait, maybe the vertical beam is 1.6 m, and the angle of elevation is 28 degrees, so the horizontal distance is adjacent, vertical is opposite. So \(\tan(28^\circ)=\frac{opposite}{adjacent}=\frac{1.6}{x}\), so \(x=\frac{1.6}{\tan(28^\circ)}\). Let's calculate \(\tan(28^\circ)\approx0.5317\), so \(1.6\div0.5317\approx3.0\) meters. Wait, but let's check with calculator: \(\tan(28) \approx 0.5317094317\), so \(1.6 / 0.5317094317 \approx 3.0\) meters. So the answer should be 3.0 meters? Wait, but maybe I messed up the angle. Wait, the vertical beam meets the floor at 90 degrees, so the triangle is right-angled. The support beam makes 28 degrees with the horizontal, so the angle between support beam and vertical beam is \(90 - 28 = 62\) degrees. So using tangent: \(\tan(28^\circ) = \frac{1.6}{x}\), so \(x = \frac{1.6}{\tan(28^\circ)} \approx \frac{1.6}{0.5317} \approx 3.0\) meters. So the correct option is 3.0 meters. Wait, but let's check again. Wait, maybe I made a mistake in the tangent. Wait, no, tangent of angle of elevation is opposite over adjacent, where opposite is the height (1.6 m), adjacent is the horizontal distance (x). So yes, \(\tan(28) = 1.6 / x\), so \(x = 1.6 / \tan(28)\). Calculating that: 1.6 divided by 0.5317 is approximately 3.0. So the answer is 3.0 meters.

Answer:

3.0 meters