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Question Number 192497 by cortano12 last updated on 19/May/23

$$\tan 66°+\tan 12°=\sqrt{3}+\tan \mathrm{x}$$$$\mathrm{x}=?$$

Answered by Tomal last updated on 19/May/23

$$\tan 66°+\tan 12°=\sqrt{3}+\tan \mathrm{x}$$$$\Rightarrow 2.246+0.213=\sqrt{3}+\tan \mathrm{x}$$$$\Rightarrow \tan \mathrm{x}=2.246+0.213-\sqrt{3}$$$$\Rightarrow \tan \mathrm{x}=0.727$$$$\Rightarrow \tan \mathrm{x}=\tan {36}^{\mathrm{o}}$$$$\therefore \mathrm{x}={36}^{\mathrm{o}}$$$$$$$$$$

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