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Question Number 203327 by mr W last updated on 16/Jan/24

Answered by ajfour last updated on 16/Jan/24

Commented by ajfour last updated on 16/Jan/24

$$Whatwouldrbeinthiscase?$$

Commented by mr W last updated on 16/Jan/24

$$seebelow$$$$\tan \varphi =\frac{4}{6}=\frac{2}{3}$$$$2\alpha =45°+\varphi$$$$\tan 2\alpha =\frac{1+\frac{2}{3}}{1-\frac{2}{3}}=5=\frac{2\tan \alpha }{1-{\tan }^2\alpha }$$$$5{\tan }^2\alpha +2\tan \alpha -5=0$$$$\tan \alpha =\frac{\sqrt{26}-1}{5}=\frac{r}{6-r}$$$$r=\frac{6(\sqrt{26}-1)}{4+\sqrt{26}}=3(6-\sqrt{26})$$

Commented by ajfour last updated on 16/Jan/24

$$yessir,excellent!igotthesame.$$

Commented by mr W last updated on 16/Jan/24

$$thankssir!$$

Answered by mr W last updated on 16/Jan/24

Commented by mr W last updated on 16/Jan/24

$$\tan \varphi =\frac{8}{10+6}=\frac{1}{2}$$$$2\alpha =45°+\varphi$$$$\tan \left(2\alpha \right)=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3=\frac{2\tan \alpha }{1-{\tan }^2\alpha }$$$$3{\tan }^2\alpha +2\tan \alpha -3=0$$$$\Rightarrow \tan \alpha =\frac{-1+\sqrt{10}}{3}=\frac{x}{6-x}$$$$\Rightarrow (2+\sqrt{10})x=6(\sqrt{10}-1)$$$$\Rightarrow x=\frac{6(\sqrt{10}-1)}{2+\sqrt{10}}=3(4-\sqrt{10})✓$$

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