# The angle of elevation of the summit of a mountain from a point on the ground is $45^{o}$. After climbing up one km towards the summit at an inclination of $30^{o}$ from the ground, the angle of elevation of the summit is found to be $60^{o}$. Then the height (in km) of the summit from the ground is : Option: 1 Option: 2   Option: 3 Option: 4

$\\\sin 30^{\circ}=x \Rightarrow x=\frac{1}{2} \\ \\\cos 30^{\circ}=z \Rightarrow z=\frac{\sqrt{3}}{2}$

$\\\tan 45^{\circ}=\frac{\mathrm{h}}{\mathrm{y}+\mathrm{z}} \Rightarrow \mathrm{h}=\mathrm{y}+\mathrm{z} \\ \\\tan 60^{\circ}=\frac{\mathrm{h}-\mathrm{x}}{\mathrm{y}} \Rightarrow \tan 60^{\circ}=\frac{\mathrm{h}-\mathrm{x}}{\mathrm{h}-\mathrm{z}}$

$\\\sqrt{3}(h-z)=h-x \\\\ (\sqrt{3}-1) h=\sqrt{3} z-x \\\\ \Rightarrow(\sqrt{3}-1) h=\frac{3}{2}-\frac{1}{2} \\\\ \Rightarrow(\sqrt{3}-1) h=1 \\\\ h=\frac{1}{\sqrt{3}-1}$

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