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Question Number 100815 by john santu last updated on 28/Jun/20

Answered by mr W last updated on 28/Jun/20

$$seeQ47657$$$$$$$$\delta =(2+3+\sqrt{7})(-2+3+\sqrt{7})(2-3+\sqrt{7})(2+3-\sqrt{7})$$$$=108$$$$sidelengthofequilateraltriangle:$$$$l=\sqrt{\frac{2^2+3^2+{\left(\sqrt{7}\right)}^2+\sqrt{3\times 108}}{2}}=\sqrt{\frac{20+18}{2}}=\sqrt{19}$$$$\cos \alpha =\frac{19+4-7}{2\times \sqrt{19}\times 2}=\frac{4\sqrt{19}}{19}$$$$\Rightarrow \tan \alpha =\frac{\sqrt{3}}{4}$$$$\cos \beta =\frac{19+9-7}{2\times \sqrt{19}\times 3}=\frac{7\sqrt{19}}{38}$$$$\Rightarrow \tan \beta =\frac{3\sqrt{3}}{7}$$$$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$$$=\frac{\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{7}}{1-\frac{\sqrt{3}}{4}\times \frac{3\sqrt{3}}{7}}=\sqrt{3}$$$$\Rightarrow AnswerA$$

Commented by bramlex last updated on 29/Jun/20

$$whatis\delta sir?$$

Commented by mr W last updated on 29/Jun/20

$$seeformulainQ47657$$

Commented by mr W last updated on 29/Jun/20

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