The horizontal distance between two buildings is $70 \; m$. The angle of depression of the top of the first building when seen from the top of the second building is $45^\circ$. If the height of the second building is $120 \; m$, find the height of the first building.
$AA'$, $BB'$ are two buildings
$AB = $ distance between the two buildings $= 70 \; m$
$BB' =$ building of height $120 \; m$
Let height of building $AA' = h$ meter
Draw $A'P \perp BB'$.
Then $A'P = AB = 70 \;m$ and $AA' = BP$.
In $\triangle PB'A'$, $\;$ $\dfrac{PB'}{A'P} = \tan 45^\circ = 1$
$\implies$ $PB' = A'P = 70 \; m$
Now, $\;$ $BB' = BP + PB'$
i.e. $\;$ $BB' = AA' + PB'$
$\therefore \;$ $AA' = h = BB' - PB' = 120 - 70 = 50 \; m$
$\therefore \;$ Height of the building $= 50 \; m$