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CBSE Class 10 Maths Chapter 9: Some Applications of Trigonometry Notes

📖 Chapter Notes ✏️ NCERT Solutions 📥 PDF Notes

1. Line of Sight

The line drawn from the eye of an observer to the point object being viewed by the observer is called the Line of Sight.

2. Angle of Elevation

When you look upwards at an object (like a bird, plane, or top of a tower), the angle formed between the line of sight and the horizontal line is called the Angle of Elevation.

3. Angle of Depression

When you stand in a high position (like a balcony or cliff) and look downwards at an object on the ground, the angle formed between the line of sight and the horizontal line is called the Angle of Depression.

💡 Golden Alternate Interior Angle Rule:
Because the horizontal line in the sky is parallel to the ground line, the Angle of Depression from the top is always exactly equal to the Angle of Elevation from the bottom object! This helps turn every depression problem instantly into a simple triangle.
CBSE Board Exam Secret

95% of Questions Use Tan: Since these problems almost always give or ask for the height of a building (Perpendicular) and its distance on the ground (Base), you will use $\tan \theta = P/B$ to solve them. Memorize these two values, as they cover almost every question:
• $\tan 30^\circ = 1/\sqrt{3}$
• $\tan 60^\circ = \sqrt{3}$

✏️ Complete NCERT Solutions Class 10 Heights & Distances

Exercise 9.1
Q1. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Step 1: Set up the imaginary triangle ($\Delta ABC$)
Let $AB$ be the height of the tower ($h$).
Let $BC$ be the distance on the ground = $15\text{ m}$.
Given Angle $\theta = 60^\circ$.
Step 2: Apply the Tan formula
$\tan 60^\circ = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$
Step 3: Plug in values and calculate
$\sqrt{3} = \frac{h}{15} \implies \mathbf{h = 15\sqrt{3}\text{ m}}$
Final Answer: The height of the tower is $\mathbf{15\sqrt{3}\text{ m}}$ (or $25.98\text{ m}$ if $\sqrt{3} = 1.732$ is requested).