What can I do next? How to find alpha?
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$\begingroup$ It's better to write in mathjax $\endgroup$Andre Lin– Andre Lin2025-11-12 18:44:35 +00:00Commented Nov 12 at 18:44
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1$\begingroup$ but I'll try... $\endgroup$Andre Lin– Andre Lin2025-11-12 18:45:41 +00:00Commented Nov 12 at 18:45
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$\begingroup$ $\alpha$ seems not a simple number under degree measure. $\endgroup$JC Q– JC Q2025-11-12 19:04:06 +00:00Commented Nov 12 at 19:04
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1$\begingroup$ If my algebra is right, you should be able to solve for $\tan\alpha$ if you use the other right triangles in the picture to get equations for ratios of more sides. For example, what are $\tan 10^\circ$ and $\tan 60^\circ$ in terms of this picture? $\endgroup$Eli Seamans– Eli Seamans2025-11-12 19:04:12 +00:00Commented Nov 12 at 19:04
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$\begingroup$ @JCQ Use coordinate system may work $\endgroup$Andre Lin– Andre Lin2025-11-12 19:06:17 +00:00Commented Nov 12 at 19:06
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1 Answer
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Here is an answer for completion. Looking at right triangles we have $$\tan\alpha=\frac{DF}{FE},$$ $$\tan 10^\circ=\frac{DF}{AF},$$ $$\tan 50^\circ=\frac{BF}{FE},$$ and $$\tan 60^\circ=\frac{BF}{AF}.$$ Then noticing that $$\frac{DF}{FE}=\frac{\frac{DF}{AF}\cdot \frac{BF}{FE}}{\frac{BF}{AF}},$$ in sight of the above equations we have $$\tan\alpha=\frac{(\tan 10^\circ)(\tan 50^\circ)}{\tan 60^\circ}.$$
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$\begingroup$ You're right the approximation is 7 $\endgroup$Andre Lin– Andre Lin2025-11-12 19:21:00 +00:00Commented Nov 12 at 19:21