Find $\tan \beta$ from the given figure. Find $\tan \beta$ from the given figure. A. 10/39 C. 13/38 B. 12/37 D. 11/38 Log in or register to post comments Solution Click here to… Jhun Vert Wed, 06/12/2024 - 22:38 Solution Click here to expand or collapse this section $\tan \alpha = 2/8$ $\tan (\alpha + \beta) = 5/8$ $\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \, \tan \beta}$ $\dfrac{5}{8} = \dfrac{\frac{2}{8} + \tan \beta}{1 - \frac{2}{8}\tan \beta}$ $\frac{5}{8} - \frac{5}{32}\tan \beta = \frac{2}{8} + \tan \beta$ $\frac{3}{8} = \frac{37}{32} \tan \beta$ $\tan \beta = \frac{12}{37}$ Log in or register to post comments
Solution Click here to… Jhun Vert Wed, 06/12/2024 - 22:38 Solution Click here to expand or collapse this section $\tan \alpha = 2/8$ $\tan (\alpha + \beta) = 5/8$ $\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \, \tan \beta}$ $\dfrac{5}{8} = \dfrac{\frac{2}{8} + \tan \beta}{1 - \frac{2}{8}\tan \beta}$ $\frac{5}{8} - \frac{5}{32}\tan \beta = \frac{2}{8} + \tan \beta$ $\frac{3}{8} = \frac{37}{32} \tan \beta$ $\tan \beta = \frac{12}{37}$ Log in or register to post comments
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