Question

look at the figure: if \\( \tan x^{\circ}=\frac{j}{11} \\) and \\( \sin x^{\circ}=\frac{j}{k} \\), what is the value of \\( \cos x^{\circ} \\)? \\( \cos x^{\circ}=\frac{11}{k} \\) \\( \cos x^{\circ}=\frac{k}{11} \\) \\( \cos x^{\circ}=11 j \\) \\( \cos x^{\circ}=11 k \\)

Explanation:

Step1: Recall the trigonometric identity

We know that \(\tan x^{\circ}=\frac{\sin x^{\circ}}{\cos x^{\circ}}\). Given \(\tan x^{\circ}=\frac{j}{11}\) and \(\sin x^{\circ}=\frac{j}{k}\).

Step2: Substitute into the identity

Substitute \(\tan x^{\circ}=\frac{\sin x^{\circ}}{\cos x^{\circ}}\) with the given values: \(\frac{j}{11}=\frac{\frac{j}{k}}{\cos x^{\circ}}\).

Step3: Solve for \(\cos x^{\circ}\)

Cross - multiply: \(\frac{j}{11}\cos x^{\circ}=\frac{j}{k}\). Since \(j
eq0\) (if \(j = 0\), \(\tan x^{\circ}=0\) and \(\sin x^{\circ}=0\), and we can still use the right - triangle definitions. But assuming non - degenerate case for algebraic manipulation), we can divide both sides by \(j\) (if \(j = 0\), the original right - triangle ratios would be based on side lengths. In a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). If \(\tan x^{\circ}=\frac{j}{11}\) (opposite = \(j\), adjacent = \(11\)), \(\sin x^{\circ}=\frac{j}{k}\) (opposite = \(j\), hypotenuse=\(k\)), then by the Pythagorean theorem \(k^{2}=j^{2}+11^{2}\)). After dividing by \(j\) (algebraic approach), we get \(\cos x^{\circ}=\frac{11}{k}\).

Answer:

\(\cos x^{\circ}=\frac{11}{k}\)