The angle of elevation of the top of a hill f | vedclass

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(B) Let the height of the hill be $H$. The initial point is at a distance $H$ from the foot of the hill since the angle of elevation is $45^{\circ}$.A

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(B) Let the height of the hill be $H$. The initial point is at a distance $H$ from the foot of the hill since the angle of elevation is $45^{\circ}$.After walking $80 \ m$ along a slope of $30^{\circ}$,the new position is at a horizontal distance of $80 \cos 30^{\circ} = 80 	imes \frac{\sqrt{3}}{2} = 40\sqrt{3} \ m$ from the initial point,and at a vertical height of $80 \sin 30^{\circ} = 80 	imes \frac{1}{2} = 40 \ m$ from the horizontal plane.The remaining horizontal distance to the foot of the hill is $H - 40\sqrt{3}$.The new height above the current position is $H - 40$.Given the new angle of elevation is $75^{\circ}$,we have $	an 75^{\circ} = \frac{H - 40}{H - 40\sqrt{3}}$.Since $	an 75^{\circ} = 2 + \sqrt{3}$,we have $2 + \sqrt{3} = \frac{H - 40}{H - 40\sqrt{3}}$.$(2 + \sqrt{3})(H - 40\sqrt{3}) = H - 40$.$2H - 80\sqrt{3} + \sqrt{3}H - 120 = H - 40$.$H(1 + \sqrt{3}) = 80 + 80\sqrt{3} = 80(1 + \sqrt{3})$.$H = 80 \ m$.

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