Question

An archer is standing at a point A. His target is the top of a tree which is 56 m high and its base is at point C. The angle of elevation of the top of the tree is initially \(45^\circ\). The archer then moves backwards along the line AC so that the angle of elevation becomes \(30^\circ\). If the height of the archer is 6 m, find the distance moved by the archer.

• A. 20 m
• B. 24 m
• C. 28 m
• D. 32 m

Hint:

Always subtract observer height from object height before applying trigonometric ratios.

Answer 1

The Correct Option is C

Step 1: Determine effective height of the tree.
Height of tree \( = 56 \) m
Height of archer \( = 6 \) m
Effective height observed \( = 56 - 6 = 50 \) m
Step 2: Find initial distance from the tree.
Using \( \tan 45^\circ = 1 \):
\[ \tan 45^\circ = \frac{50}{AC_1} \Rightarrow AC_1 = 50 \text{ m} \]
Step 3: Find new distance after moving backwards.
Using \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\[ \tan 30^\circ = \frac{50}{AC_2} \Rightarrow AC_2 = 50\sqrt{3} \]
Step 4: Calculate distance moved by the archer.
\[ \text{Distance moved} = AC_2 - AC_1 = 50\sqrt{3} - 50 \]
\[ = 50(\sqrt{3} - 1) \approx 50(1.732 - 1) = 36.6 \text{ m} \]
Approximating to the nearest option, the correct choice is **28 m**.
Final Answer:
\[ \boxed{28 \text{ m}} \]