# The angle of elevation of the top of a hill from a pt o the horizontal plane passing through the foot of the hill is found to be $45^{o}$. After walking a distance of 40m towards the top up a slope inclined on the angle of elevation of the top of the hill becomes $75^{o}$. Then the height of the hill in m is  Option: 1 28 Option: 2 56 Option: 3 40 Option: 4 80

$\\\tan 75^{\circ}=\frac{h}{h+40-40 \sqrt{3}} \\ \\\frac{2+\sqrt{3}}{1}=\frac{h}{h+40-40 \sqrt{3}} \\ \\\Rightarrow 2 h+80-80 \sqrt{3}+\sqrt{3} h+40 \sqrt{3}-120=h \\ \\\Rightarrow h(\sqrt{3}+1)=40+40 \sqrt{3} \\ \\\Rightarrow h=40$

Height of hill = 40 + 40 = 80 m

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