From the top of a cliff, the angles of depression of two boats... | Mathematics MCQ | Examoogle

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From the top of a cliff, the angles of depression of two boats on the sea on the same side of the cliff are observed to be 45° and 30°, respectively. If the height of the cliff is 200 m, what is the distance between the two boats (in m)?

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

200( $\sqrt{3} + 1)$

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

100( $\sqrt{3} + 1)$

Correct Answer

📐 MA • Math Direct FormulaMathematicsTrigonometry

Option A

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

Quick Summary:

Use the formula $d = h(\cot \alpha - \cot \beta)$ where $\beta < \alpha$ . Since $\cot 30° = \sqrt{3}$ and $\cot 45° = 1$ , the distance is $200(\sqrt{3} - 1)$ .

📐MAMath SolutionDirect Formula

📋 Given

Height of the cliff (h) = 200 m, angle of depression for boat 1 (theta 1) = 45 degrees, angle of depression for boat 2 (theta 2) = 30 degrees.

🔢 Formula Used

d=h×(cot⁡θ1−cot⁡θ2)ord=h×(tan⁡θ2−tan⁡θ1) depending on the angle configurationd = h \times (\cot \theta_1 - \cot \theta_2) \quad \text{or} \quad d = h \times (\tan \theta_2 - \tan \theta_1) \text{ depending on the angle configuration}d=h×(cotθ1​−cotθ2​)ord=h×(tanθ2​−tanθ1​) depending on the angle configuration

⚡ Exam Hall Shortcut / Speed Trick

Use the formula $d = h(\cot \alpha - \cot \beta)$ where $\beta < \alpha$ . Since $\cot 30° = \sqrt{3}$ and $\cot 45° = 1$ , the distance is $200(\sqrt{3} - 1)$ .

⚠️ Common Student Trap / Pitfall

Students often swap the angles or use the tangent ratios incorrectly, leading to $200(1 - \sqrt{3})$ which results in a negative distance.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Define the geometry

Let the cliff be $AB$ with height $h = 200$ m. Let the two boats be at points $C$ and $D$ . The distance between the boats is $x = CD$ .

h=200,∠CAD=30°,∠CAB=45°h = 200, \quad \angle CAD = 30°, \quad \angle CAB = 45°h=200,∠CAD=30°,∠CAB=45°

Calculate distance for boat 1

In $\triangle ABC$ , $\tan 45° = \frac{AB}{BC}$ , so $BC = \frac{200}{\tan 45°} = 200$ m.

BC=200BC = 200BC=200

Calculate distance for boat 2

In $\triangle ABD$ , $\tan 30° = \frac{AB}{BD}$ , so $BD = \frac{200}{\tan 30°} = 200\sqrt{3}$ m.

BD=2003BD = 200\sqrt{3}BD=2003​

Find the distance between boats

The distance between the boats is $BD - BC$ .

x=2003−200=200(3−1)x = 200\sqrt{3} - 200 = 200(\sqrt{3} - 1)x=2003​−200=200(3​−1)

✅

A is correct because the distance between the two boats is calculated as $200(\sqrt{3} - 1)$ m.

Core Concepts Used

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Trigonometric ratios Angles of depression and elevation Right-angled triangle properties

💡 EXAM TIP

This concept is a precursor to 3D geometry problems and shadow-length calculations in competitive exams like JEE or NDA.

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