Question

question 12: standard g.srt.6
solve for x using the appropriate trig function.
(right triangle with vertical leg 20, horizontal leg x, angle at the end of horizontal leg is 40°)

a. 21.9
b. 14.7
c. 53.6
d. 23.8

Explanation:

Step1: Identify trigonometric ratio

We have a right triangle with opposite side to \(40^\circ\) as \(x\) and adjacent side as \(20\)? Wait, no, wait. Wait, the right angle, the side labeled 20 is opposite? Wait, no, the angle is \(40^\circ\), the side adjacent to \(40^\circ\) is 20? Wait, no, the right triangle: one leg is 20 (vertical), one leg is \(x\) (horizontal), angle at the bottom is \(40^\circ\). So \(\tan(40^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{20}\)? Wait, no, wait: opposite to \(40^\circ\) is \(x\)? Wait, no, the vertical side is 20, horizontal is \(x\). So angle at the bottom: the side opposite is 20, adjacent is \(x\)? Wait, no, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So \(\theta = 40^\circ\), opposite side is 20? Wait, no, the vertical side is 20, so if the angle is at the bottom, then the opposite side is 20, adjacent is \(x\). Wait, no, \(\tan(40^\circ)=\frac{20}{x}\)? Wait, no, I think I mixed up. Wait, let's re-express: in a right triangle, for angle \(\theta\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). So if the angle is \(40^\circ\), the side opposite to \(40^\circ\) is the vertical side (length 20), and the adjacent side is the horizontal side (length \(x\))? Wait, no, that would mean \(\tan(40^\circ)=\frac{20}{x}\), so \(x=\frac{20}{\tan(40^\circ)}\). Wait, but \(\tan(40^\circ)\approx0.8391\), so \(x\approx\frac{20}{0.8391}\approx23.8\). Wait, that's option d. Wait, let's check again. Wait, maybe I had opposite and adjacent reversed. Wait, the angle is at the bottom, so the side opposite is the vertical leg (20), adjacent is the horizontal leg (\(x\)). So \(\tan(40^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{20}{x}\), so \(x = \frac{20}{\tan(40^\circ)}\). Let's calculate \(\tan(40^\circ)\approx0.8390996312\). Then \(20\div0.8390996312\approx23.8\). So that's option d.

Step1: Recall tangent ratio

In a right triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 40^\circ\), opposite side to \(40^\circ\) is \(20\), adjacent side is \(x\). So \(\tan(40^\circ)=\frac{20}{x}\).

Step2: Solve for \(x\)

Rearrange the formula: \(x=\frac{20}{\tan(40^\circ)}\). Calculate \(\tan(40^\circ)\approx0.8391\). Then \(x=\frac{20}{0.8391}\approx23.8\).

Answer:

d. 23.8