Edit to “Solution”
en/13.3.16.md
+1 −1
| ### 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. | |||
| @@ -5,7 +5,7 @@Statement | |||
| ### Solution | |||
| − |  | ||
| #### Answer | |||
| $$ | |||
| \operatorname{tg}\alpha^{\prime}=(1-\alpha/f)\operatorname{tg}\alpha. | |||
| $$ | |||
| unchanged lines 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 | ||
|  | ||
| #### Answer | #### Answer | ||
| $$ | $$ | ||
| \operatorname{tg}\alpha^{\prime}=(1-\alpha/f)\operatorname{tg}\alpha. | \operatorname{tg}\alpha^{\prime}=(1-\alpha/f)\operatorname{tg}\alpha. | ||
| $$ | $$ | ||
| unchanged lines 4 | |||