Two-Triangle Problems: Multi-Tier Elevations and Lighthouses for CBSE Class 10

Not all trigonometric observations originate from ground level. In coastal navigation, a lighthouse keeper surveys approaching ships from an elevated cliff; in architectural surveying, an engineer stands on top of a building and looks up at the peak of a transmission tower while looking down at its base; and in town squares, a statue stands mounted on top of an elevated pedestal.

In CBSE Class 10 Mathematics, Chapter 9 (Some Applications of Trigonometry), multi-tier elevation and depression problems represent the highest-weightage questions in board examinations. These problems require students to establish intermediate horizontal reference planes and solve coupled right-angled triangles.

What You Will Learn

• How to draw accurate geometric sketches with elevated observation points
• Problem Type 1: The Lighthouse Problem (observing two ships at different angles of depression)
• Problem Type 2: The Statue on a Pedestal Problem (observing top and base of an elevated object)
• Problem Type 3: The Building-to-Tower Problem (measuring an angle of elevation AND an angle of depression simultaneously)
• Step-by-step algebraic methods for isolating heights and distances
• Common traps and exam presentation tips

1. Problem Type 1: The Lighthouse and Two Ships (NCERT Classic)

Problem Statement:

As observed from the top of a $75 ext{ m}$ high lighthouse from 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.

    Lighthouse Top A ---------------- Horizontal Datum
                    |\                     | \              75 m   |  \                     |   \                     |45° \ \ 30°
    Sea Level       B-----C--\----------- D
                       x      y (Distance between ships)

Solution:

1. Geometric Setup:
   • Let $AB$ be the lighthouse: $AB = 75 ext{ m}$.
   • Let $C$ be the position of the nearer ship (angle of depression $= 45^\circ \implies$ angle of elevation ngle ACB = 45^\circ).
   • Let $D$ be the position of the farther ship (angle of depression $= 30^\circ \implies$ angle of elevation ngle ADB = 30^\circ).
   • Let $BC = x ext{ metres}$, and distance between the two ships $CD = y ext{ metres}$.
2. In Right Triangle $\Delta ABC$: an 45^\circ = rac{AB}{BC} \implies 1 = rac{75}{x} \implies x = \mathbf{75 ext{ m}}
3. In Right Triangle $\Delta ABD$: an 30^\circ = rac{AB}{BD} = rac{AB}{BC + CD} = rac{75}{x + y} rac{1}{\sqrt{3}} = rac{75}{75 + y}
4. Solve for $y$: $75 + y = 75\sqrt{3}$ $y = 75\sqrt{3} - 75 = \mathbf{75(\sqrt{3} - 1) ext{ m}}$
5. If $\sqrt{3} = 1.732$: $y = 75(1.732 - 1) = 75 imes 0.732 = \mathbf{54.9 ext{ m}}$
6. Therefore, <u>the distance between the two ships is $75(\sqrt{3} - 1) ext{ metres}$ (or $54.9 ext{ m}$)</u>.

2. Problem Type 2: The Statue on a Pedestal (NCERT Classic)

Problem Statement:

A statue, $1.6 ext{ m}$ 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.

                           A (Top of Statue)
                           |
                     1.6 m |
                           |
                           B (Top of Pedestal / Base of Statue)
                           |
                         h |
                           |
       Ground Point D -----+ C (Foot of Pedestal)
                     x

Solution:

1. Geometric Setup:
   • Let $BC$ be the pedestal of height $h ext{ metres}$.
   • Let $AB$ be the statue of height $1.6 ext{ m}$.
   • Total height to top of statue: $AC = h + 1.6 ext{ metres}$.
   • Let $D$ be the ground point at distance $x ext{ metres}$ from the foot $C$.
   • Angle to top of pedestal: ngle BDC = 45^\circ.
   • Angle to top of statue: ngle ADC = 60^\circ.
2. In Right Triangle $\Delta BCD$: an 45^\circ = rac{BC}{CD} \implies 1 = rac{h}{x} \implies x = h \quad ext{--- (1)}
3. In Right Triangle $\Delta ACD$: an 60^\circ = rac{AC}{CD} \implies \sqrt{3} = rac{h + 1.6}{x}
4. Substitute $x = h$ from (1): \sqrt{3} = rac{h + 1.6}{h} $h\sqrt{3} = h + 1.6$ $h\sqrt{3} - h = 1.6 \implies h(\sqrt{3} - 1) = 1.6$
5. Solve for $h$ and Rationalize: h = rac{1.6}{\sqrt{3} - 1} = rac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = rac{1.6(\sqrt{3} + 1)}{3 - 1} = rac{1.6(\sqrt{3} + 1)}{2} = \mathbf{0.8(\sqrt{3} + 1) ext{ m}}
6. Therefore, <u>the height of the pedestal is $0.8(\sqrt{3} + 1) ext{ metres}$</u>.

3. Problem Type 3: Building and Cable Tower (NCERT Classic)

Problem Statement:

From the top of a $7 ext{ m}$ high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower.

                                          C (Top of Tower)
                                         /|
                                        / |
                                       /  | H - 7
        Building Top A --------------E----+ (Horizontal Reference)
                    | \ 45°         60°   |
                7 m |  \                  | 7 m
                    |   \                 |
                    B----+----------------D (Foot of Tower)
                            x

Solution:

1. Geometric Setup:
   • Let $AB$ be the building: $AB = 7 ext{ m}$.
   • Let $CD$ be the cable tower: total height $= H ext{ metres}$.
   • Draw horizontal line $AE$ from building top $A$ to meet tower $CD$ at point $E$.
   • Then $ED = AB = 7 ext{ m}$, so the upper tower section is $CE = (H - 7) ext{ metres}$.
   • Distance between building and tower: $BD = AE = x ext{ metres}$.
   • Angle of elevation of tower top: ngle CAE = 60^\circ.
   • Angle of depression of tower foot: ngle EAD = 45^\circ \implies ngle ADB = 45^\circ.
2. In Right Triangle $\Delta ABD$: an 45^\circ = rac{AB}{BD} \implies 1 = rac{7}{x} \implies x = \mathbf{7 ext{ m}} Thus, horizontal distance $AE = x = 7 ext{ m}$.
3. In Right Triangle $\Delta AEC$: an 60^\circ = rac{CE}{AE} \implies \sqrt{3} = rac{H - 7}{7} $H - 7 = 7\sqrt{3} \implies H = 7\sqrt{3} + 7 = \mathbf{7(\sqrt{3} + 1) ext{ m}}$
4. If $\sqrt{3} = 1.732$: $H = 7(1.732 + 1) = 7(2.732) = \mathbf{19.124 ext{ m}}$
5. Therefore, <u>the total height of the cable tower is $7(\sqrt{3} + 1) ext{ metres}$</u>.

4. Summary and Examination Tips

Exam Tip: Notice the power of $\mathbf{45^\circ}$! Whenever $an 45^\circ = 1$ appears, it immediately equates perpendicular and base ($P = B$). Look for the $45^\circ$ triangle first to isolate a variable instantly!

Common Mistake: In the building-to-tower problem, setting total tower height as $7\sqrt{3}$. Remember that $7\sqrt{3}$ is ONLY the upper section $CE$ above the building; you must add the lower $7 ext{ m}$ ($ED$) to get the complete tower height: $7\sqrt{3} + 7 = 7(\sqrt{3} + 1)$!