The angle of elevation of the summit of a mou | vedclass

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Question

(A) Let $h$ be the height of the summit. From the initial point,$	an 45^{\circ} = \frac{h}{d} \Rightarrow d = h$,where $d$ is the horizontal distance

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}-1}$

Vedclass · Answer

(A) Let $h$ be the height of the summit. From the initial point,$	an 45^{\circ} = \frac{h}{d} \Rightarrow d = h$,where $d$ is the horizontal distance to the mountain base.After climbing $1 \ km$ at $30^{\circ}$,the new position is at a height $x = 1 \cdot \sin 30^{\circ} = \frac{1}{2} \ km$ and a horizontal distance $z = 1 \cdot \cos 30^{\circ} = \frac{\sqrt{3}}{2} \ km$ from the starting point.The new horizontal distance to the mountain is $y = d - z = h - \frac{\sqrt{3}}{2}$.The new height relative to the current position is $h - x = h - \frac{1}{2}$.Given the new angle of elevation is $60^{\circ}$,we have $	an 60^{\circ} = \frac{h - x}{y}$.$\sqrt{3} = \frac{h - 1/2}{h - \sqrt{3}/2}$.$\sqrt{3}(h - \frac{\sqrt{3}}{2}) = h - \frac{1}{2}$.$\sqrt{3}h - \frac{3}{2} = h - \frac{1}{2}$.$h(\sqrt{3} - 1) = \frac{3}{2} - \frac{1}{2} = 1$.$h = \frac{1}{\sqrt{3} - 1}$.

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