To solve for [tex]\(\cos(A - B)\)[/tex] given that [tex]\(\tan A = \frac{11}{60}\)[/tex] and [tex]\(\cos B = \frac{5}{13}\)[/tex], follow these steps:
1. First, use the given value of [tex]\(\tan A = \frac{11}{60}\)[/tex] to determine [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex]. Recall that:
[tex]\[
\tan A = \frac{\sin A}{\cos A}
\][/tex]
From [tex]\(\tan A = \frac{11}{60}\)[/tex], we can introduce [tex]\(\sin A = 11k\)[/tex] and [tex]\(\cos A = 60k\)[/tex] for some constant [tex]\(k\)[/tex]. Applying the Pythagorean identity [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex]:
[tex]\[
(11k)^2 + (60k)^2 = 1 \implies 121k^2 + 3600k^2 = 1 \implies 3721k^2 = 1 \implies k^2 = \frac{1}{3721} \implies k = \frac{1}{\sqrt{3721}}
\][/tex]
Thus,
[tex]\[
\sin A = \frac{11}{\sqrt{3721}} \quad \text{and} \quad \cos A = \frac{60}{\sqrt{3721}}
\][/tex]
Since [tex]\(\sqrt{3721} = 61\)[/tex], it simplifies to:
[tex]\[
\sin A = \frac{11}{61} \quad \text{and} \quad \cos A = \frac{60}{61}
\][/tex]
2. Next, use the given value [tex]\(\cos B = \frac{5}{13}\)[/tex] to determine [tex]\(\sin B\)[/tex]. Using the Pythagorean identity [tex]\(\sin^2 B + \cos^2 B = 1\)[/tex]:
[tex]\[
\sin^2 B + \left(\frac{5}{13}\right)^2 = 1 \implies \sin^2 B + \frac{25}{169} = 1 \implies \sin^2 B = 1 - \frac{25}{169} = \frac{144}{169} \implies \sin B = \frac{12}{13}
\][/tex]
3. Finally, use the angle difference formula for cosine:
[tex]\[
\cos(A - B) = \cos A \cos B + \sin A \sin B
\][/tex]
Substitute the known values:
[tex]\[
\cos(A - B) = \left( \frac{60}{61} \right) \left( \frac{5}{13} \right) + \left( \frac{11}{61} \right) \left( \frac{12}{13} \right)
\][/tex]
Calculate each term separately:
[tex]\[
\frac{60}{61} \cdot \frac{5}{13} = \frac{300}{793} \quad \text{and} \quad \frac{11}{61} \cdot \frac{12}{13} = \frac{132}{793}
\][/tex]
Therefore,
[tex]\[
\cos(A - B) = \frac{300}{793} + \frac{132}{793} = \frac{432}{793}
\][/tex]
Thus, the value of [tex]\(\cos(A - B)\)[/tex] is [tex]\(\boxed{\frac{432}{793}}\)[/tex].
