Правка раздела «Solution»
en/13.3.16.md
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| ### Statement | |||
| $13.3.16^*.$ The tip of a cone with an angle of $2\alpha$ is viewed through a lens with a focal length $f$ located at a distance a from the tip of the cone $(a < f)$. How is the angle of the cone visible through the lens? The axis of the lens passes through the axis of symmetry of the cone. | |||
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| ### Solution | |||
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| #### Answer | |||
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| \operatorname{tg}\alpha^{\prime}=(1-\alpha/f)\operatorname{tg}\alpha. | |||
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| ещё строк без изменений 4 | |||
| ### Statement | ### Statement | ||
| $13.3.16^*.$ The tip of a cone with an angle of $2\alpha$ is viewed through a lens with a focal length $f$ located at a distance a from the tip of the cone $(a < f)$. How is the angle of the cone visible through the lens? The axis of the lens passes through the axis of symmetry of the cone. | $13.3.16^*.$ The tip of a cone with an angle of $2\alpha$ is viewed through a lens with a focal length $f$ located at a distance a from the tip of the cone $(a < f)$. How is the angle of the cone visible through the lens? The axis of the lens passes through the axis of symmetry of the cone. | ||
| @@ -5,7 +5,7 @@Statement | |||
| ### Solution | ### Solution | ||
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| #### Answer | #### Answer | ||
| $$ | $$ | ||
| \operatorname{tg}\alpha^{\prime}=(1-\alpha/f)\operatorname{tg}\alpha. | \operatorname{tg}\alpha^{\prime}=(1-\alpha/f)\operatorname{tg}\alpha. | ||
| $$ | $$ | ||
| ещё строк без изменений 4 | |||