Heights and Distances: Single Triangle Problems for CBSE Class 10

The most fundamental real-world application of trigonometry involves calculating an inaccessible height or distance using a single right-angled triangle. In these scenarios, an observer measures one distance along the ground and one angle using an inclinometer (theodolite), or knows the physical length of an inclined structure (such as a leaning ladder or rope) and determines the remaining unknown dimension algebraically.

In CBSE Class 10 Mathematics, Chapter 9 (Some Applications of Trigonometry), single triangle word problems appear frequently in Section B (2 marks) and Section C (3 marks). Mastering these foundational archetypes ensures error-free execution before progressing to complex multi-triangle systems.

What You Will Learn

• Universal 4-step framework for solving single-triangle word problems
• The classic Broken Tree Storm Problem (calculating total original tree height)
• The Circus Rope Climber Problem (hypotenuse-perpendicular relations)
• The Children's Park Slide Design Problem (double comparison of slides)
• Handling radical values (\sqrt{3} pprox 1.732) and denominator rationalization
• Board exam presentation guidelines for full marks

1. The 4-Step Single Triangle Framework

Whenever you encounter a single-triangle word problem:

1. Step 1 (Read and Sketch): Draw a clean right-angled triangle. Label vertices clearly ($A, B, C$) and identify the $90^\circ$ right angle (usually where the vertical pole, wall, or tree meets the horizontal ground).
2. Step 2 (Assign Variables): Identify what is given ($P, B,$ or $H$) and assign a variable (e.g., $h$ for height, $x$ for distance) to what must be found.
3. Step 3 (Select Trigonometric Ratio):
   • If Perpendicular and Base are involved $\implies$ Use $\tan \theta = \frac{P}{B}$.
   • If Perpendicular and Hypotenuse are involved $\implies$ Use $\sin \theta = \frac{P}{H}$.
   • If Base and Hypotenuse are involved $\implies$ Use $\cos \theta = \frac{B}{H}$.
4. Step 4 (Substitute, Solve, and Rationalize): Substitute standard exact values ($30^\circ, 45^\circ, 60^\circ$), solve algebraically, rationalize radicals, and state the final answer with units.

2. High-Yield Solved Board Examination Problems

Solved Example 1: The Broken Tree Storm Problem (NCERT Classic)

Problem: A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground, making an angle of $30^\circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 ext{ m}$. Find the total height of the tree.

                                  A' (Original Top of Tree)
                                  :
                                  : (Broken part fallen)
                                  A
                                 /|
                                / |
                   Broken part /  | Remaining trunk
                     AC = A'A /   | AB = h
                             /    |
                     (30°) C------+ B (Foot of tree)
                              8 m

Solution:

1. Analyze the Geometry:
   • Let the original unbroken tree be $A'B$.
   • The tree broke at point $A$. The broken portion $A'A$ bent over and touched the ground at point $C$.
   • Therefore, the broken part $AC$ is equal in length to the original top part $A'A$: $AC = A'A$
   • Total original height of the tree $= AB + A'A = \mathbf{AB + AC}$.
   • In right triangle $\Delta ABC$: Base $BC = 8 ext{ m}$, ngle B = 90^\circ, and ngle C = 30^\circ.
2. Find the Vertical Trunk Height ($AB$): an 30^\circ = rac{AB}{BC} rac{1}{\sqrt{3}} = rac{AB}{8} \implies AB = rac{8}{\sqrt{3}} ext{ m}
3. Find the Broken Leaning Part ($AC$): \cos 30^\circ = rac{BC}{AC} rac{\sqrt{3}}{2} = rac{8}{AC} \implies AC imes \sqrt{3} = 16 \implies AC = rac{16}{\sqrt{3}} ext{ m}
4. Calculate Total Tree Height: ext{Total Height} = AB + AC = rac{8}{\sqrt{3}} + rac{16}{\sqrt{3}} = rac{24}{\sqrt{3}}
5. Rationalize the Denominator: ext{Total Height} = rac{24 imes \sqrt{3}}{\sqrt{3} imes \sqrt{3}} = rac{24\sqrt{3}}{3} = \mathbf{8\sqrt{3} ext{ m}}
6. Therefore, <u>the total height of the tree is $8\sqrt{3} ext{ metres}$</u>.

Important: <u>In the broken tree problem, the most common student blunder is calculating ONLY the vertical stump $AB$ ($8/\sqrt{3}$) and forgetting to add the broken hypotenuse part $AC$ ($16/\sqrt{3}$)! The tree's total height is the SUM of the vertical stump and the broken leaning portion.</u>

Solved Example 2: The Park Slide Problem (NCERT Classic)

Problem: A contractor plans to install two slides for children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of $1.5 ext{ m}$, and is inclined at an angle of $30^\circ$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3 ext{ m}$, and inclined at an angle of $60^\circ$ to the ground. What should be the length of the slide in each case?

Solution:

Case 1: Slide for Younger Children (Below 5 Years)

1. Let $AB$ be the vertical ladder: $AB = 1.5 ext{ m}$.
2. Let $AC$ be the slide (hypotenuse): length $= L_1$.
3. Angle of inclination: ngle C = 30^\circ.
4. Apply sine ratio: \sin 30^\circ = rac{AB}{AC} \implies rac{1}{2} = rac{1.5}{L_1} $L_1 = 1.5 imes 2 = \mathbf{3 ext{ m}}$

Case 2: Slide for Elder Children

1. Let $PQ$ be the vertical height: $PQ = 3 ext{ m}$.
2. Let $PR$ be the slide (hypotenuse): length $= L_2$.
3. Angle of inclination: ngle R = 60^\circ.
4. Apply sine ratio: \sin 60^\circ = rac{PQ}{PR} \implies rac{\sqrt{3}}{2} = rac{3}{L_2} L_2 imes \sqrt{3} = 6 \implies L_2 = rac{6}{\sqrt{3}} = rac{6\sqrt{3}}{3} = \mathbf{2\sqrt{3} ext{ m}}

Conclusion: <u>The length of the slide for younger children is $3 ext{ metres}$, and for elder children is $2\sqrt{3} ext{ metres}$</u>.

Solved Example 3: Shadow of a Vertical Pole

Problem: A vertical pole of length $6 ext{ m}$ casts a shadow $4 ext{ m}$ long on the ground, and at the same time a tower casts a shadow $28 ext{ m}$ long. Find the height of the tower.

Solution:

1. At the same time of day, the sun's rays strike the earth at the identical angle of elevation $heta$.
2. For the Pole: an heta = rac{ ext{Height of pole}}{ ext{Length of shadow}} = rac{6}{4} = rac{3}{2}
3. For the Tower: Let the height of the tower be $H ext{ metres}$. The shadow is $28 ext{ m}$. an heta = rac{H}{28}
4. Equating both expressions for $an heta$: rac{H}{28} = rac{3}{2} \implies H = rac{3 imes 28}{2} = 3 imes 14 = 42 ext{ m}
5. Therefore, <u>the height of the tower is $42 ext{ metres}$</u>.

3. Summary and Examination Tips

Exam Tip: Always check if the question specifies a numerical approximation for square roots (e.g., Take $\sqrt{3} = 1.732$). If specified, multiply out your radical answer to two decimal places; if not specified, leave the answer in exact simplified radical form (e.g., $8\sqrt{3} ext{ m}$).

Common Mistake: In the broken tree problem, setting $BC$ as the hypotenuse. The ground distance from the base of the tree to where the top touches is the Base, not the hypotenuse! The bent broken tree trunk itself is the Hypotenuse.