The angle of elevation of the summit of a mountain from a point on the ground is $45^{\circ}$.
After climding up one $\mathrm{km}$ towards the summit at an inclination of $30^{\circ}$ from the ground, the angle of elevation of the summit is found to be $60^{\circ}$. Then the height (in $\mathrm{km}$ ) of the summit from the ground is :
Correct Option: 1
$\sin 30^{\circ}=x \Rightarrow x=\frac{1}{2}$
$\cos 30^{\circ}=z \Rightarrow z=\frac{\sqrt{3}}{2}$
$\tan 45^{\circ}=\frac{\mathrm{h}}{\mathrm{y}+\mathrm{z}} \Rightarrow \mathrm{h}=\mathrm{y}+\mathrm{z}$
$\tan 60^{\circ}=\frac{\mathrm{h}-\mathrm{x}}{\mathrm{y}} \Rightarrow \tan 60^{\circ}=\frac{\mathrm{h}-\mathrm{x}}{\mathrm{h}-\mathrm{z}}$
$\sqrt{3}(\mathrm{~h}-\mathrm{z})=\mathrm{h}-\mathrm{x}$
$(\sqrt{3}-1) h=\sqrt{3} z-x$
$\Rightarrow(\sqrt{3}-1) \mathrm{h}=\frac{3}{2}-\frac{1}{2}$
$\Rightarrow(\sqrt{3}-1) \mathrm{h}=1$
$\mathrm{~h}=\frac{1}{\sqrt{3}-1}$
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