Two-Triangle Problems: Observations from the Ground for CBSE Class 10

In many practical surveying and architectural scenarios, an observer cannot determine a height from a single measurement point—either because the base of a monument is inaccessible (e.g., across a river) or because an object is observed from two different reference locations. These situations give rise to two interconnected right-angled triangles sharing a common vertical height or ground segment.

In CBSE Class 10 Mathematics, Chapter 9 (Some Applications of Trigonometry), two-triangle ground observation problems are standard 4-mark and 5-mark board examination questions in Section D.

What You Will Learn

• How to analyze and sketch two right triangles sharing a common side
• Type 1: Two observation points on the same side of a vertical tower
• Type 2: Two observation points on opposite sides of a vertical tower
• Type 3: Two poles of equal height on opposite sides of a road (NCERT classic 80m road problem)
• Algebraic techniques to eliminate intermediate ground variables
• Complete step-by-step solutions and presentation templates

1. The Strategy for Two-Triangle Problems

When two right-angled triangles appear in a single geometric system:

    Type A: Points on Same Side                  Type B: Points on Opposite Sides
                 A (Top of Tower)                             A (Top of Tower)
                 |\                                           /|                 | \                                         / |                h |  \                                      h/  |  \h
                 |   \                                     /   |                    | 60°\ 30°                              / 30° | 60°                 B-----C-------D                        B------O-----C
                    x      d                               x     (w - x)

The Universal Algebraic Protocol:

1. Identify the Shared Dimension: Usually, both right triangles share the same vertical height $h$ (or the same ground distance).
2. Formulate Equation 1: In the first right triangle, express the shared variable $h$ in terms of the ground distance using $an heta_1$.
3. Formulate Equation 2: In the second right triangle, express $h$ in terms of the second ground distance using $an heta_2$.
4. Equate and Eliminate: Set the two expressions for $h$ equal to each other to solve for the ground distance, and then substitute back to find $h$.

2. High-Yield Solved Board Examination Problems

Solved Example 1: Moving Closer to a Tower (Same Side)

Problem: The shadow of a tower standing on level ground is found to be $40 ext{ m}$ longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.

Solution:

1. Analyze the Geometry:
   • Let $AB$ be the vertical tower of height $h ext{ metres}$.
   • Let $BC = x ext{ metres}$ be the shadow length when the sun's altitude is $60^\circ$.
   • When the altitude is $30^\circ$, the shadow is $BD = x + 40 ext{ metres}$.
   • In right triangle $\Delta ABC$, ngle ACB = 60^\circ.
   • In right triangle $\Delta ABD$, ngle ADB = 30^\circ.
2. From Right Triangle $\Delta ABC$: an 60^\circ = rac{AB}{BC} \implies \sqrt{3} = rac{h}{x} \implies x = rac{h}{\sqrt{3}} \quad ext{--- (1)}
3. From Right Triangle $\Delta ABD$: an 30^\circ = rac{AB}{BD} \implies rac{1}{\sqrt{3}} = rac{h}{x + 40} $x + 40 = h\sqrt{3} \quad ext{--- (2)}$
4. Substitute x = rac{h}{\sqrt{3}} from (1) into (2): rac{h}{\sqrt{3}} + 40 = h\sqrt{3} $$$$

ight) = h\left(rac{3 - 1}{\sqrt{3}} ight) = rac{2h}{\sqrt{3}}5. **Solve for $h$:** rac{2h}{\sqrt{3}} = 40 \implies 2h = 40\sqrt{3} \implies h = \mathbf{20\sqrt{3} ext{ m}}$$ 6. Therefore, <u>the height of the tower is $20\sqrt{3} ext{ metres}$ (or $34.64 ext{ m}$)</u>.

Solved Example 2: Two Poles of Equal Heights on an 80m Road (NCERT Classic)

Problem: Two poles of equal heights are standing opposite each other on either side of the road, which is $80 ext{ m}$ wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.

       A (Pole 1)                                   C (Pole 2)
       |                                            |
     h |                                            | h
       | 60°                                    30° |
       B -------------------- P ------------------- D
                 x                     (80 - x)
       <------------------ 80 m -------------------->

Solution:

1. Geometric Setup:
   • Let the two poles of equal height be $AB = CD = h ext{ metres}$.
   • Width of the road: $BD = 80 ext{ m}$.
   • Let the observation point on the road be $P$.
   • Let $BP = x ext{ metres}$. Then $PD = (80 - x) ext{ metres}$.
   • Given angles: ngle APB = 60^\circ and ngle CPD = 30^\circ.
2. From Right Triangle $\Delta ABP$: an 60^\circ = rac{AB}{BP} \implies \sqrt{3} = rac{h}{x} \implies h = x\sqrt{3} \quad ext{--- (1)}
3. From Right Triangle $\Delta CDP$: an 30^\circ = rac{CD}{PD} \implies rac{1}{\sqrt{3}} = rac{h}{80 - x} \implies h = rac{80 - x}{\sqrt{3}} \quad ext{--- (2)}
4. Equate the Two Expressions for $h$: x\sqrt{3} = rac{80 - x}{\sqrt{3}} $x\sqrt{3} imes \sqrt{3} = 80 - x$ $3x = 80 - x$ $4x = 80 \implies x = \mathbf{20 ext{ m}}$
5. Find the Distances of the Point from Both Poles:
   • Distance from Pole 1 ($BP$): $x = \mathbf{20 ext{ m}}$.
   • Distance from Pole 2 ($PD$): $80 - x = 80 - 20 = \mathbf{60 ext{ m}}$.
6. Find the Height of the Poles ($h$): $h = x\sqrt{3} = \mathbf{20\sqrt{3} ext{ m}}$
7. Conclusion: <u>The height of each pole is $20\sqrt{3} ext{ metres}$ ($34.64 ext{ m}$), and the observation point is $20 ext{ m}$ from the first pole and $60 ext{ m}$ from the second pole</u>.

3. Summary and Examination Tips

Exam Tip: Notice that the closer the observation point is to the base of the tower, the larger the angle of elevation! Point $C$ (closer) has angle $60^\circ$, while Point $D$ (further away) has angle $30^\circ$. If your sketch has the larger angle further away, your geometry is inverted!

Common Mistake: Forgetting to label which triangle you are calculating in. In board exams, always write "In right triangle $\Delta ABC$ ..." before writing trigonometric ratios.