Explanation
Let the distance between the nearer kilometre stone and the hill be '$$x$$' km.
So, the distance between the farther kilometre stone and the hill is '$$1+x$$' km since both are on the same side of the hill.
In triangle APB,
$$\tan 45 ^0=\dfrac hx$$
$$\Rightarrow $$$$1=\dfrac { h }{ x } $$
$$\Rightarrow $$$$h=x$$
In triangle AQB,
$$\tan 30 ^0=\dfrac {h}{1+x}$$
$$\Rightarrow $$$$\dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ 1+x } $$
$$\Rightarrow $$$$1+x=\sqrt { 3 } h$$
From equation 1,
$$1+h=\sqrt { 3 } h \Rightarrow $$$$1=\sqrt { 3 } h-h$$
$$\Rightarrow $$$$h=\dfrac { 1 }{ \sqrt { 3 } -1 } $$
$$\Rightarrow $$$$h=1.365km$$
Hence option A is correct.
Let AB be the tower which is $$50\sqrt { 3 } m$$ high and C and D be the positions of the two towers such that
$$\angle ADB=45 ^0$$ and $$\angle ACB=60 ^0$$
Now, in triangle ABC,
$$\tan {60 ^0}=\dfrac {AB}{BC}$$
$$\Rightarrow $$$$\sqrt { 3 } =\dfrac { 50\sqrt { 3 } }{ BC } $$
$$\Rightarrow $$$$BC=50m$$
Also, in triangle ABD,
$$\tan 45 ^0=\dfrac {AB}{BD}$$
$$\Rightarrow \tan 45 ^0=\dfrac {AB}{BC+CD}$$
$$\Rightarrow $$$$1=\dfrac { 50\sqrt { 3 } }{ 50+CD } $$
$$\Rightarrow $$$$50+CD=50\sqrt { 3 } $$
$$\Rightarrow $$$$CD=50\sqrt { 3 } -50$$
$$\Rightarrow $$$$CD=50\left( \sqrt { 3 } -1 \right) m$$.
Hence option C is correct.
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