From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings if tan a = 512 and tan b = 45

Answer: Option D

Let ED be the building and AC be the tower.
Given that ED = 230 m, ∠ADC = b, ∠AEB = a
Also given that tan a = $$\frac{5}{12}$$ and tan b = $$\frac{4}{5}$$
Let AC = h
Required Distance = Distance between the top of these buildings = AE
$$\begin{array}{rl}& \text{From}\text{the}\text{right}\Delta ABE,\\ & \tan \left(a\right)=\frac{AB}{BE}\end{array}$$
$$\Rightarrow \frac{5}{12}=\frac{\left(h-230\right)}{BE}$$     [∵ tan(a) = $$\frac{5}{12}$$ (given), AB = (AC - BC) = (AC - ED) = (h - 230)]
$$\begin{array}{rl}& \Rightarrow BE=\frac{12\left(h-230\right)}{5}.....\left(1\right)\\ & \text{From}\text{the}\text{right}\Delta ACD,\\ & \tan \left(b\right)=\frac{AC}{CD}\end{array}$$
$$\Rightarrow \frac{4}{5}=\frac{h}{CD}$$    [∵ tan(b) = $$\frac{4}{5}$$ (given), AC = h]
$$\begin{array}{rl}& \Rightarrow CD=\frac{5h}{4}.....\left(2\right)\\ & \text{From}\text{the}\text{diagram,}BE=CD\end{array}$$
$$\Rightarrow \frac{12\left(h-230\right)}{5}=\frac{5h}{4}$$     (from equation 1 & 2)
$$\begin{array}{rl}& \Rightarrow 48h-\left(4\times 12\times 230\right)=25h\\ & \Rightarrow 23h=\left(4\times 12\times 230\right)\\ & \Rightarrow h=\frac{\left(4\times 12\times 230\right)}{23}\\ & =480\text{m}......\left(3\right)\\ & AB=\left(AC-BC\right)\end{array}$$
$$=\left(480-230\right)$$    [∵ Since AC = h = 480 (from 3) and BC = ED = 230 m(given)]
$$=250\text{m}$$
In the triangle ABE, tan(a) =$$\frac{5}{12}$$ . 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) = $$\frac{5}{12}$$
$$\begin{array}{rl}& \text{hypotenuse}=\sqrt{5^2+{12}^2}=13\\ & \text{i}\text{.e}\text{.,}\\ & \sin \left(a\right)=\frac{\text{opposite}\text{side}}{\text{hypotenuse}}=\frac{5}{13}\end{array}$$
We have seen that $$\sin \left(a\right)=\frac{5}{13}$$
$$\begin{array}{rl}& \Rightarrow \frac{AB}{AE}=\frac{5}{13}\\ & \Rightarrow AE=AB\times \frac{13}{5}\\ & =250\times \frac{13}{5}\\ & =650\text{m}\end{array}$$

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