Question

question 11 (multiple choice worth 1 points) (05.02 mc) look at the figure: right triangle with angle x° at one acute angle, right angle at another, sides labeled with g, 6, h (inferred from tan x° = 6/9 and sin x° = 6/h) if \\(\tan x^\circ = \frac{6}{9}\\) and \\(\sin x^\circ = \frac{6}{h}\\), what is the value of \\(\cos x^\circ\\)? \\(\cos x^\circ = \frac{h}{9}\\) \\(\cos x^\circ = \frac{9}{h}\\) \\(\cos x^\circ = 9h\\) \\(\cos x^\circ = 69\\)

Explanation:

Step1: Recall Trigonometric Ratios

In a right - triangle, we know the following trigonometric ratios:

• \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)
• \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)
• \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)

Given that \(\tan x^{\circ}=\frac{6}{g}\), from the definition of tangent, the side opposite to angle \(x\) is \(6\) and the side adjacent to angle \(x\) is \(g\).

Given that \(\sin x^{\circ}=\frac{6}{h}\), from the definition of sine, the hypotenuse of the right - triangle is \(h\) (since \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\) and the opposite side is \(6\)).

Step2: Calculate \(\cos x^{\circ}\)

Using the definition of cosine, \(\cos x^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}\). We know that the adjacent side to angle \(x\) is \(g\) and the hypotenuse is \(h\)? Wait, no, wait. Wait, \(\tan x=\frac{\text{opposite}}{\text{adjacent}}=\frac{6}{g}\), so opposite \( = 6\), adjacent \(=g\). \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{6}{h}\), so hypotenuse \( = h\). Then, by the definition of cosine, \(\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{g}{h}\)? Wait, no, wait. Wait, let's re - check.

Wait, \(\tan x=\frac{\text{opposite}}{\text{adjacent}}=\frac{6}{g}\), so opposite \(a = 6\), adjacent \(b = g\). \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{6}{h}\), so hypotenuse \(c=h\). Then, by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), \(6^{2}+g^{2}=h^{2}\), but we can also use the trigonometric identity \(\tan x=\frac{\sin x}{\cos x}\).

We know that \(\tan x=\frac{6}{g}\) and \(\sin x=\frac{6}{h}\). Since \(\tan x=\frac{\sin x}{\cos x}\), we can solve for \(\cos x\):

\(\cos x=\frac{\sin x}{\tan x}\)

Substitute \(\sin x = \frac{6}{h}\) and \(\tan x=\frac{6}{g}\) into the formula:

\(\cos x=\frac{\frac{6}{h}}{\frac{6}{g}}=\frac{6}{h}\times\frac{g}{6}=\frac{g}{h}\)

Answer:

\(\cos x^{\circ}=\frac{g}{h}\) (the option corresponding to \(\cos x^{\circ}=\frac{g}{h}\))