Question

question 8 (multiple choice worth 1 points) (05.02 mc) look at the figure. image of a right triangle with angle x° at the top left if \\(\tan x^\circ = \frac{1}{11}\\) and \\(\sin x^\circ = \frac{1}{k}\\), what is the value of \\(\cos x^\circ\\)? \\(\circ\\ \cos x^\circ = 11k\\) \\(\circ\\ \cos x^\circ = 11j\\) \\(\circ\\ \cos x^\circ = \frac{k}{11}\\) \\(\circ\\ \cos x^\circ = \frac{11}{k}\\)

Explanation:

Step1: Recall trigonometric identity

We know that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Given \(\tan x^{\circ}=\frac{1}{11}\) and \(\sin x^{\circ}=\frac{1}{k}\).
Substitute these values into the identity: \(\frac{1}{11}=\frac{\frac{1}{k}}{\cos x^{\circ}}\)

Step2: Solve for \(\cos x^{\circ}\)

From \(\frac{1}{11}=\frac{\frac{1}{k}}{\cos x^{\circ}}\), we can rewrite it as \(\frac{1}{11}=\frac{1}{k\cos x^{\circ}}\) (by the rule of dividing by a fraction: \(\frac{a}{b}\div c=\frac{a}{b\times c}\)).
Cross - multiply: \(k\cos x^{\circ}=11\times1\)
Then, \(\cos x^{\circ}=\frac{11}{k}\) (divide both sides by \(k\))

Answer:

\(\cos x^{\circ}=\frac{11}{k}\) (corresponding to the option \(\boldsymbol{\cos x^{\circ}=\frac{11}{k}}\))