Distance moved by archer

Question

An archer, from the point $A$, is aiming at a target, placed on the top of a $56	ext{ feet}$ tall tree. The base of the tree is at the point $C$. From the archer’s position, the angle of elevation to the target is $45^\circ$ from his eye level. The archer, facing the tree, moves backwards on the straight line joining the points $A$ and $C$, to a new position at the point $B$. From the point $B$, the angle of elevation from his eye level to the target becomes $30^\circ$. How far did the archer m

Accepted Answer

C. Height of the tree above eye level = $56 - 6 = 50	ext{ feet}$. Let $R$ be the eye level point vertically aligned above $A$, and $S$ above $B$. Let $P$ be the top of the tree, and $Q$ be the point on the tree at eye level height. In $	riangle PQR$, $	an 45^\circ = \frac{PQ}{QR} = 1 \implies QR = 50	ext{ ft}$. In $	riangle PQS$, $	an 30^\circ = \frac{PQ}{QS} = \frac{50}{50 + RS} = \frac{1}{\sqrt{3}} \implies 50 + RS = 50\sqrt{3} \implies RS = 50(\sqrt{3}-1)	ext{ ft}$. Since $AB = RS$, the distance moved is $50(\sqrt{3}-1)	ext{ feet}$.

Answer

$50\sqrt{3}$

Answer

$56\sqrt{3}$

Answer

$50(\sqrt{3}-1)$

Answer

$50\left(1-\frac{1}{\sqrt{3}}\right)$

Answer

$\frac{50}{\sqrt{3}}$

Distance moved by archer

Explanation

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