The angle of elevation of a tower from a point on the ground is $30^{\circ}$. Moving 40 m towards the tower, the angle becomes $60^{\circ}$. What is the height of the tower?

The correct answer is Option A: 20\sqrt{3} m. Use the ratio formula: $h = d \times \frac{\tan(60^\circ) \times \tan(30^\circ)}{\tan(60^\circ) - \tan(30^\circ)}$, which simplifies to $h = d \times \frac{\sqrt{3}}{2}$ for this specific case, leading directly to $40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}$. Step-by-Step Solution & Concepts: Shortcut Trick: Use the ratio formula: $h = d \times \frac{\tan(60^\circ) \times \tan(30^\circ)}{\tan(60^\circ) - \tan(30^\circ)}$, which simplifies to $h = d \times \frac{\sqrt{3}}{2}$ for this specific case, leading directly to $40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}$. Formula Used: $$h = \frac{d}{\cot(\theta_2) - \cot(\theta_1)}$$ Given: Initial angle of elevation = 30 degrees, Final angle of elevation = 60 degrees, Distance moved towards tower = 40 meters. Steps: Set up equations: Let $h$ be the height of the tower and $x$ be the remaining distance from the tower at the 60 degree angle. Then $\tan(60^\circ) = \frac{h}{x}$ and $\tan(30^\circ) = \frac{h}{40+x}$.. Substitute and Solve: Substitute $x = \frac{h}{\sqrt{3}}$ into the second equation: $\frac{1}{\sqrt{3}} = \frac{h}{40 + \frac{h}{\sqrt{3}}}$.. Simplify: Calculate the coefficient of $h$: $40 = h \left(\frac{3-1}{\sqrt{3}}\right) = h \left(\frac{2}{\sqrt{3}}\right)$. Solving for $h$ gives $h = 40 \times \frac{\sqrt{3}}{2}$. Answer: A is correct because substituting the given values into the height formula results in a height of 20\sqrt{3} meters. Exam Tip: This concept is a precursor to coordinate geometry problems involving lines of sight and is frequently tested in physics problems involving vector components.

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The angle of elevation of a tower from a point on the ground is $30°$ . Moving 40 m towards the tower, the angle becomes $60°$ . What is the height of the tower?

20 $\sqrt{3} m$

20 m

40 $\sqrt{3} m$

10 $\sqrt{3} m$

Correct Answer

📐 MA • Math Direct FormulaMathematicsMathematics

Option A

20 $\sqrt{3} m$

Quick Summary:

Use the ratio formula: $h = d \times \frac{\tan(60°) \times \tan(30°)}{\tan(60°) - \tan(30°)}$ , which simplifies to $h = d \times \frac{\sqrt{3}}{2}$ for this specific case, leading directly to $40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}$ .

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📐MAMath SolutionDirect Formula

📋 Given

Initial angle of elevation = 30 degrees, Final angle of elevation = 60 degrees, Distance moved towards tower = 40 meters.

🔢 Formula Used

h=dcot⁡(θ2)−cot⁡(θ1)h = \frac{d}{\cot(\theta_2) - \cot(\theta_1)}h=cot(θ2​)−cot(θ1​)d​

⚡ Exam Hall Shortcut / Speed Trick

⚠️ Common Student Trap / Pitfall

Confusing the values of $\tan(30°)$ and $\tan(60°)$ , or forgetting to set up the two equations based on the base distance.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Set up equations

Let $h$ be the height of the tower and $x$ be the remaining distance from the tower at the 60 degree angle. Then $\tan(60°) = \frac{h}{x}$ and $\tan(30°) = \frac{h}{40+x}$ .

tan⁡(60°)=hx  ⟹  x=h3,tan⁡(30°)=h40+x\tan(60°) = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}, \quad \tan(30°) = \frac{h}{40+x}tan(60°)=xh​⟹x=3​h​,tan(30°)=40+xh​

Substitute and Solve

Substitute $x = \frac{h}{\sqrt{3}}$ into the second equation: $\frac{1}{\sqrt{3}} = \frac{h}{40 + \frac{h}{\sqrt{3}}}$ .

40+h3=h3  ⟹  40=h(3−13)40 + \frac{h}{\sqrt{3}} = h\sqrt{3} \implies 40 = h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right)40+3​h​=h3​⟹40=h(3​−3​1​)

Simplify

Calculate the coefficient of $h$ : $40 = h \left(\frac{3-1}{\sqrt{3}}\right) = h \left(\frac{2}{\sqrt{3}}\right)$ . Solving for $h$ gives $h = 40 \times \frac{\sqrt{3}}{2}$ .

h=203h = 20\sqrt{3}h=203​

✅

A is correct because substituting the given values into the height formula results in a height of 20 $\sqrt{3}$ meters.

Core Concepts Used

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Trigonometric ratios Heights and Distances Angle of elevation

💡 EXAM TIP

This concept is a precursor to coordinate geometry problems involving lines of sight and is frequently tested in physics problems involving vector components.

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