If sin alpha = frac{8}{17}, 0

Question

(C) Given $\sin \alpha = \frac{8}{17}$. Since $0 < \alpha < 90^{\circ}$,$\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{

Vedclass · Accepted Answer

$\frac{220}{221}$

Vedclass · Answer

(C) Given $\sin \alpha = \frac{8}{17}$. Since $0 Given $	an \beta = \frac{5}{12}$. Since $0 Thus,$\sin \beta = \frac{5}{13}$ and $\cos \beta = \frac{12}{13}$.Using the formula $\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:$\cos (\alpha - \beta) = (\frac{15}{17})(\frac{12}{13}) + (\frac{8}{17})(\frac{5}{13})$$= \frac{180}{221} + \frac{40}{221} = \frac{220}{221}$.

If $\sin \alpha = \frac{8}{17}, 0 < \alpha < 90^{\circ}$ and $\tan \beta = \frac{5}{12}, 0 < \beta < 90^{\circ},$ then $\cos (\alpha - \beta)$ is

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