Explanation:


Let ED be the building and AC be the tower.

Given that ED = 230 m, $\angle$ADC = b, $\angle$AEB = a
Also given that tan a = 5/12 and tan b = 4/5

Let AC = h

Required Distance = Distance between the top of these buildings = AE

From the right $\triangle$ ABE,
tan(a)=$\dfrac{AB}{BE}$
=> $\dfrac{5}{12}=\dfrac{(h-230)}{BE}$ [∵ tan(a)=5/12(given), AB = (AC-BC) = (AC-ED) = (h-230)]
=> BE =$\dfrac{12(h-230)}{5}$ ⋯(eq:1)
From the right $\triangle$ ACD,
tan(b)=$\dfrac{AC}{CD}$

=> $\dfrac{4}{5}=\dfrac{h}{CD}$ [∵ tan(b) = 4/5(given), AC=h]

=> CD =$\dfrac{5}{h4}$ ⋯(eq:2)


From the diagram, BE = CD
⇒$\dfrac{12(h-230)}{5}=\dfrac{5h}{4}$ (from eq:1 & eq:2)
⇒48h−(4×12×230)=25h
⇒23h=(4×12×230)

⇒h=$\dfrac{(4×12×230)}{23}$=480 m ⋯(eq:3)


AB = (AC - BC)
= (480 - 230) [∵ Since AC=h=480(from eq:3) and BC=ED=230 m(given)]
= 250 m

In the triangle ABE, tan(a) = 5/12. Let s figure out the value of sin(a) now.

Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = 5/12
hypotenuse =$\sqrt{5^{2}+12^{2}}$=13

i.e., sin(a) =$\dfrac{opposite}{ sidehypotenuse}=\dfrac{5}{13}$
We have seen that sin(a) = $\dfrac{5}{13}$

⇒$\dfrac{AB}{AE}=\dfrac{5}{13}$
⇒AE = AB×$\dfrac{13}{5}$
=250×$\dfrac{13}{5}$=650 m


i.e., Distance between the top of the buildings = 650 m