The angle of elevation of the top of a small tree from a point on the ground $ 300 ext{ m}$ away from the foo

Question

The angle of elevation of the top of a small tree from a point on the ground $ 300	ext{ m}$ away from the foot of the tree was $ 30^\circ $. When the tree grew taller, its angle of elevation became $ 60^\circ $ from the same point. By how much did the tree grow?

Examoogle · Accepted Answer

The correct answer is Option B: $200\sqrt{3} 	ext{ m}$. Use the property of 30-60-90 triangles. For a fixed base, growth is $d(	an 60^\circ - 	an 30^\circ) = 300(\sqrt{3} - \frac{1}{\sqrt{3}}) = 300(\frac{3-1}{\sqrt{3}}) = \frac{600}{\sqrt{3}} = 200\sqrt{3}$.

Step-by-Step Solution & Concepts:
Shortcut Trick: Use the property of 30-60-90 triangles. For a fixed base, growth is $d(	an 60^\circ - 	an 30^\circ) = 300(\sqrt{3} - \frac{1}{\sqrt{3}}) = 300(\frac{3-1}{\sqrt{3}}) = \frac{600}{\sqrt{3}} = 200\sqrt{3}$.

Formula Used: $$	ext{Growth} = d 	imes (	an 	heta_2 - 	an 	heta_1)$$

Given: Distance from tree base (d) = 300 m, initial angle (	heta_1) = 30^\circ, final angle (	heta_2) = 60^\circ.

Steps: Calculate initial height: Initial height $h_1$ is given by $h_1 = d 	imes 	an 30^\circ$. Using $d = 300$ and $	an 30^\circ = \frac{1}{\sqrt{3}}$, we get $h_1 = 300 / \sqrt{3} = 100\sqrt{3}$.. Calculate final height: Final height $h_2$ is given by $h_2 = d 	imes 	an 60^\circ$. Using $d = 300$ and $	an 60^\circ = \sqrt{3}$, we get $h_2 = 300\sqrt{3}$.. Calculate growth: The growth is the difference in height: $h_2 - h_1 = 300\sqrt{3} - 100\sqrt{3} = 200\sqrt{3}$.

Answer: B is correct because the difference between the final height $300\sqrt{3}$ m and initial height $100\sqrt{3}$ m is $200\sqrt{3}$ m.

Exam Tip: This concept is foundational for 'Heights and Distances'. Mastering 30-60-90 degree triangle ratios helps avoid trigonometric table lookups in time-bound exams.

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