I'm a $12^{\text{th}}$ Grader, in our trigonometry class, our prof gave this as an assignment. Please help me understand.
From the top of a building, the angle of depression of a point in the same horizontal plane with the base of the building is observed to be $47^{\circ} 18'$. What will be the angle of depression of the same point when viewed from a position halfway up the building?
I'm not that good at picturing out the figures or diagrams. Also, is it possible not to use the measurements of the sides (height of building etc.)?
Let $\triangle ABC$ be such that $A$ is the top of the building, and side $BC$ serves as the horizontal line on the ground. Thus $AB \perp BC$. Let $\angle ACB = \theta$ (angle of depression),$D$ be the midpoint of $AB$, and denote $\angle DCB = \beta$. We are to find $\beta$. Let $h = AB$ be the height of the building,and $a = BC$ be the distance from the point of interest on the ground to the building. We have: $\tan \beta = \dfrac{BD}{BC} = \dfrac{\frac{h}{2}}{a}, \tan \theta = \dfrac{AB}{BC} = \dfrac{h}{a}\implies \tan \beta = \dfrac{\tan \theta}{2}\implies \beta = \tan^{-1}\left(\dfrac{\tan \theta}{2}\right)$.