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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

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\\\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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himanshu.meshram

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