Applications of Trigonometry: Advanced Two-Triangle Problems Class 10

In CBSE Class 10 Mathematics, Chapter 9 (Some Applications of Trigonometry) is unique: every problem is a real-world descriptive puzzle. Single-triangle problems—such as a kite flying on a string or a ladder leaning against a wall—are simple, direct applications of $\sin, \cos,$ or $\tan$. However, the high-weightage questions in Section D (carrying 4 to 5 marks) are universally Two-Triangle Problems.

Whether you are observing a cloud and its reflection in a deep mountain lake, measuring two ships sailing away from a sea lighthouse, or calculating the height of a statue standing on a stone pedestal, you must construct two right-angled triangles, link them via a common shared side, and eliminate unknown variables.

In this guide, we break down the four essential two-triangle problem blueprints and provide full-length solutions.

What You Will Learn

• Horizontal lines of sight: The exact definition of Angle of Elevation and Angle of Depression
• The Common Side Principle: The golden strategy for two-triangle systems
• Blueprint 1: Two Ships on the Same / Opposite Sides of a Lighthouse
• Blueprint 2: The Pedestal and Statue Problem (Statue of height $h$ on base)
• Blueprint 3: Observation from a Multi-Storey Building to a Cable Tower
• Blueprint 4: The Famous Cloud and Lake Reflection Problem (CBSE 5-Mark Heavyweight)
• Presentation standards, radical handling ($\sqrt{3} = 1.732$), and common diagram errors

1. Angle of Elevation vs. Angle of Depression

                            Angle of Elevation                         Angle of Depression
                                   Object (Top)               Observer ──────── (Horizontal Line)
                                      *                         \  θ (Angle of Depression)
                                     /                                                               /                                                                / θ (Angle of Elevation)        v
    Observer ─────────────────────+                                 * Object (Ground)
              (Horizontal Line)

The Golden Visual Rule: <u>Both the angle of elevation and the angle of depression are ALWAYS measured from the HORIZONTAL line of sight! The angle of depression from an elevated observer equals the angle of elevation from the object to the observer (by alternate interior angles between parallel horizontals).</u>

2. Blueprint 1: The Common Side Elimination Strategy

In almost every two-triangle trigonometry problem:

1. Two right-angled triangles share a common base (horizontal distance $x$) or a common height ($h$).
2. In Triangle 1, express the common side in terms of $\tan \theta_1$: $x = \frac{h_1}{\tan \theta_1}$
3. In Triangle 2, express the common side in terms of $\tan \theta_2$: $x = \frac{h_2}{\tan \theta_2}$
4. Equate the two expressions for $x$ to solve for the unknown height!

3. High-Yield Solved Board Examination Problems

Problem 1: Two Ships on Opposite Sides of a Lighthouse (NCERT Classic)

Problem: From the top of a $75\text{ m}$ high lighthouse from the sea level, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

                              A (Lighthouse Top)
                              |                               |                           75 m  |                                   |                                     |   45°   \ 30°
                              +----------+----------------+
                              B          D (Ship 1)       C (Ship 2)
                                 <-- x --><----- d ----->

Solution:

1. Let the lighthouse be $AB = 75\text{ m}$.
2. Let the two ships be at points $D$ (closer ship) and $C$ (farther ship).
   • Angle of depression of Ship $D = 45^\circ \implies \angle ADB = 45^\circ$.
   • Angle of depression of Ship $C = 30^\circ \implies \angle ACB = 30^\circ$.
3. Let $BD = x$ and the distance between the ships $DC = d$.
4. In Right-Angled Triangle $\Delta ABD$: $\tan 45^\circ = \frac{AB}{BD} \implies 1 = \frac{75}{x} \implies \mathbf{x = 75\text{ m}}$
5. In Right-Angled Triangle $\Delta ABC$: $\tan 30^\circ = \frac{AB}{BC} = \frac{AB}{x + d}$ $\frac{1}{\sqrt{3}} = \frac{75}{75 + d}$ $75 + d = 75\sqrt{3}$ $d = 75\sqrt{3} - 75 = \mathbf{75(\sqrt{3} - 1)\text{ m}}$
6. Substitute $\sqrt{3} = 1.732$: $d = 75(1.732 - 1) = 75 \times 0.732 = \mathbf{54.9\text{ metres}}$
7. Therefore, <u>the distance between the two ships is $75(\sqrt{3}-1)\text{ m}$ (or $54.9\text{ m}$)</u>.

Problem 2: The Statue and Pedestal Problem (NCERT Classic)

Problem: A statue, $1.6\text{ m}$ tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

Solution:

1. Let the height of the pedestal be $h\text{ metres}$ ($BC = h$).
2. The statue stands on the pedestal: $AB = 1.6\text{ m}$.
   • Total height from ground to top of statue: $AC = h + 1.6$.
3. Let the observer be at point $D$ on the ground, at a distance $x$ from the base ($CD = x$).
4. In Right-Angled Triangle $\Delta BCD$ (Pedestal): $\tan 45^\circ = \frac{BC}{CD} \implies 1 = \frac{h}{x} \implies \mathbf{x = h}$
5. In Right-Angled Triangle $\Delta ACD$ (Statue $+$ Pedestal): $\tan 60^\circ = \frac{AC}{CD} = \frac{h + 1.6}{x}$ Since $x = h$: $\sqrt{3} = \frac{h + 1.6}{h}$ $h\sqrt{3} = h + 1.6$ $h\sqrt{3} - h = 1.6 \implies h(\sqrt{3} - 1) = 1.6$ $h = \frac{1.6}{\sqrt{3} - 1}$
6. Rationalize the Denominator: $h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{1.6(\sqrt{3} + 1)}{3 - 1} = \frac{1.6(\sqrt{3} + 1)}{2} = \mathbf{0.8(\sqrt{3} + 1)\text{ m}}$
7. Substitute $\sqrt{3} = 1.732$: $h = 0.8(1.732 + 1) = 0.8 \times 2.732 = \mathbf{2.185\text{ metres}}$
8. Therefore, <u>the height of the pedestal is $0.8(\sqrt{3}+1)\text{ m}$ (or $2.19\text{ m}$)</u>.

Problem 3: The Cloud and Lake Reflection Problem (The 5-Mark Heavyweight)

Problem: The angle of elevation of a cloud from a point $h$ metres above a lake is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$. Prove that the height of the cloud above the lake is: $\mathbf{\frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha}}$

Proof:

1. Analyze the Physics of Water Reflection:
   • In a plane mirror (the lake surface), Object distance = Image distance.
   • If the cloud is at a height $H$ above the water surface, its virtual reflection lies at an identical depth $H$ below the water surface!
2. Analyze Geometry from the Observation Point:
   • Let the observer be at point $O$, which is $h$ metres above the water surface.
   • The height of the cloud above the observer's horizontal line of sight is $(H - h)$.
   • The depth of the cloud's reflection below the observer's horizontal line of sight is $(H + h)$.
   • Let the horizontal distance from the observer to the vertical cloud line be $x$.
3. In the Upper Triangle (Angle of Elevation $\alpha$): $\tan \alpha = \frac{H - h}{x} \implies \mathbf{x = \frac{H - h}{\tan \alpha}} \quad \text{--- (1)}$
4. In the Lower Triangle (Angle of Depression $\beta$): $\tan \beta = \frac{H + h}{x} \implies \mathbf{x = \frac{H + h}{\tan \beta}} \quad \text{--- (2)}$
5. Equate the Two Expressions for $x$: $\frac{H - h}{\tan \alpha} = \frac{H + h}{\tan \beta}$ $(H - h)\tan \beta = (H + h)\tan \alpha$ $H\tan \beta - h\tan \beta = H\tan \alpha + h\tan \alpha$
6. Group Terms with $H$ on the Left: $H\tan \beta - H\tan \alpha = h\tan \beta + h\tan \alpha$ $H(\tan \beta - \tan \alpha) = h(\tan \beta + \tan \alpha)$ $\mathbf{H = \frac{h(\tan \beta + \tan \alpha)}{\tan \beta - \tan \alpha}}$ Hence Proved.

4. Summary and Examination Tips

Exam Tip: In the lake reflection problem, the single most critical geometric fact is: the reflection is at a depth of $H$ below the water surface, making its distance from the observer $(H + h)$! Clearly state this in your proof to secure full marks.

Common Mistake: Leaving radicals in the denominator. Never leave an answer as $\frac{1.6}{\sqrt{3}-1}$; always rationalize by multiplying numerator and denominator by $(\sqrt{3}+1)$!