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Question Number 216471 by Ismoiljon_008 last updated on 08/Feb/25

Commented by Ismoiljon_008 last updated on 08/Feb/25

$$help,please$$

Commented by mr W last updated on 08/Feb/25

$$asolutionisgiveninQ207018.$$$$anothersolutionseebelow.$$

Commented by Ismoiljon_008 last updated on 08/Feb/25

$$thankyou$$

Answered by mr W last updated on 08/Feb/25

Commented by mr W last updated on 08/Feb/25

$$\alpha +\beta =\frac{\pi }{2}$$$$p=2r\sin \alpha ,q=2r\sin \beta$$$$\frac{p}{\sin \theta }=\frac{4}{\sin \alpha }$$$$\frac{q}{\sin \theta }=\frac{5}{\sin \beta }$$$$\frac{p}{q}=\frac{4\sin \beta }{5\sin \alpha }\Rightarrow \frac{\sin \alpha }{\sin \beta }=\frac{4\sin \beta }{5\sin \alpha }$$$$\Rightarrow 5{\sin }^2\alpha =4{\sin }^2\beta =4{\cos }^2\alpha$$$$\Rightarrow {\tan }^2\alpha =\frac{4}{5}\Rightarrow \tan \alpha =\frac{2}{\sqrt{5}}\Rightarrow \sin \alpha =\frac{2}{3}$$$$\frac{2r\sin \alpha }{\sin \theta }=\frac{4}{\sin \alpha }$$$$\Rightarrow \sin \theta =\frac{r{\sin }^2\alpha }{2}=\frac{r}{2}\times {\left(\frac{2}{3}\right)}^2=\frac{2r}{9}$$$${\left(2r\right)}^2={10}^2+9^2-2\times 10\times 9\cos (\pi -2\theta )$$$$4r^2=181+180\cos \left(2\theta \right)$$$$4r^2=181+180(1-2\times \frac{4r^2}{81})$$$$\Rightarrow r=\frac{57}{14}✓$$

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