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Question Number 188720 by Rupesh123 last updated on 06/Mar/23

Commented by mr W last updated on 06/Mar/23

$$ithink{it}^{\prime }simpossible.$$

Commented by ajfour last updated on 06/Mar/23

Answered by a.lgnaoui last updated on 07/Mar/23

$$△MOA\tan \alpha =\frac{a}{h}$$$$△MOB\tan \left(2\alpha \right)=\frac{a+6}{h}=\frac{2\tan \alpha }{1-\tan \alpha {}^2}$$$$=\frac{\frac{2a}{h}}{1-{\left(\frac{a}{h}\right)}^2}=\frac{2ah}{h^2-a^2}$$$$\Rightarrow {\mathit{a}}^3+6{\mathit{a}}^2+\mathit{a}{h}^2-6h^2=0\Rightarrow (\mathit{a}=5)$$$$\Rightarrow \tan \alpha =\frac{5}{5\sqrt{11}}=\frac{\sqrt{11}}{11}.$$$$\bullet \tan 3\alpha =\tan (2\alpha +\alpha )=\frac{11+c}{5\sqrt{11}}=\frac{4\sqrt{11}}{11}$$$$\Rightarrow \mathit{c}=9$$$$△MOC$$$$\measuredangle \left(\frac{MOC}{2}\right)=\measuredangle MOH=(\alpha +\frac{\alpha }{2})$$$$MH\oplus ON\Rightarrow \measuredangle MOH=\frac{\pi }{2}-(\alpha +\frac{\alpha }{2})$$$$\theta =\frac{\pi }{2}-[\frac{\pi }{2}-(\alpha +\frac{\alpha }{2})]=\frac{3\alpha }{2}$$$$$$

Commented by a.lgnaoui last updated on 07/Mar/23

Commented by a.lgnaoui last updated on 07/Mar/23

$$\left(suite\right)$$$$△AOC$$$$\tan 3\alpha =\frac{5+6+9}{5\sqrt{11}}=\frac{4}{\sqrt{11}}$$$$\Rightarrow 3\alpha =Arctg\left(\frac{4}{\sqrt{11}}\right)=85,5631..$$$$soit:\mathit{\theta }=42,84$$$$$$

Commented by mr W last updated on 08/Mar/23

$$non-sense!$$$$thisisimpossible:$$

Commented by mr W last updated on 08/Mar/23

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