Suppose that the earth is a sphere of radius | vedclass

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(D) Let $R$ be the radius of the earth and $H$ be the height of the observer at point $P$ above the surface.The area of the spherical cap visible from

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$6400$

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(D) Let $R$ be the radius of the earth and $H$ be the height of the observer at point $P$ above the surface.The area of the spherical cap visible from point $P$ is given by $A = 2 \pi R h'$,where $h'$ is the height of the cap.The total surface area of the sphere is $4 \pi R^2$.Given that one-fourth of the earth's surface is visible,we have:$2 \pi R h' = \frac{1}{4} (4 \pi R^2) = \pi R^2$$h' = \frac{R}{2}$From the geometry of the sphere,the height of the cap $h'$ is related to the angle $	heta$ (the semi-vertical angle of the cone of vision) by $h' = R(1 - \cos 	heta)$.Equating the two expressions for $h'$:$R(1 - \cos 	heta) = \frac{R}{2}$$1 - \cos 	heta = \frac{1}{2}$$\cos 	heta = \frac{1}{2}$$	heta = 60^{\circ}$In the right-angled triangle formed by the center of the earth $O$,the point of tangency $B$,and the observer $P$,we have:$\cos 	heta = \frac{OB}{OP} = \frac{R}{R + H}$$\frac{1}{2} = \frac{R}{R + H}$$R + H = 2R$$H = R$Given $R = 6400 \, km$,we get $H = 6400 \, km$.

Suppose that the earth is a sphere of radius $6400 \, km$. The height from the earth's surface from where exactly a fourth of the earth's surface is visible,is $...... \, km$.

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