Height & Distance (Class 10)

Cheatsheet Content

1. Introduction to Trigonometry Basics Right-Angled Triangle: A triangle with one angle equal to $90^\circ$. Trigonometric Ratios: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$ $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$ $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$ $\csc \theta = \frac{1}{\sin \theta}$ $\sec \theta = \frac{1}{\cos \theta}$ $\cot \theta = \frac{1}{\tan \theta}$ Mnemonic: SOH CAH TOA (Sine Opposite Hypotenuse, Cosine Adjacent Hypotenuse, Tangent Opposite Adjacent) 2. Important Angle Values $\theta$ $0^\circ$ $30^\circ$ $45^\circ$ $60^\circ$ $90^\circ$ $\sin \theta$ $0$ $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ $1$ $\cos \theta$ $1$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ $0$ $\tan \theta$ $0$ $\frac{1}{\sqrt{3}}$ $1$ $\sqrt{3}$ Undefined 3. Angles of Elevation and Depression Line of Sight: The line drawn from the eye of an observer to the object being viewed. Horizontal Line: A line parallel to the ground, passing through the observer's eye. Angle of Elevation: When an observer looks UP at an object, the angle formed between the line of sight and the horizontal line is called the angle of elevation. Observer's Eye Object $\theta$ Horizontal Line Line of Sight Angle of Depression: When an observer looks DOWN at an object, the angle formed between the line of sight and the horizontal line is called the angle of depression. Observer's Eye Object $\phi$ Horizontal Line Line of Sight Key Relationship: Angle of elevation from point A to point B is equal to the angle of depression from point B to point A (alternate interior angles). 4. Steps to Solve Height & Distance Problems Draw a Neat Diagram: Represent the problem statement visually. Label knowns and unknowns. Identify Right-Angled Triangles: Most problems can be broken down into one or more right triangles. Mark Angles and Sides: Clearly label angles of elevation/depression and the sides (height, distance) involved. Choose the Correct Trigonometric Ratio: Use $\tan \theta$ when dealing with Opposite and Adjacent sides (most common for height/distance). Use $\sin \theta$ for Opposite and Hypotenuse. Use $\cos \theta$ for Adjacent and Hypotenuse. Formulate Equation(s): Set up the trigonometric equation based on the chosen ratio. Solve for Unknowns: Use known angle values and algebraic manipulation to find the required height or distance. 5. Common Scenarios & Tips Single Triangle Problems: Direct application of trigonometric ratios. Multiple Triangle Problems: Often involve two right triangles sharing a common side or a common base. Solve one triangle to find a common side, then use it in the second triangle. Example: Finding the height of a tower observed from two different points. Observer's Height: If the observer's height is given, remember to add or subtract it from the calculated height of the object. Often, it's assumed to be negligible unless specified. Approximation: Use $\sqrt{3} \approx 1.732$ and $\frac{1}{\sqrt{3}} \approx 0.577$ when calculations require it. Units: Always include appropriate units (meters, km, etc.) in the final answer. 6. Example Problem Structure Problem: A tower stands vertically on the ground. From a point on the ground, $15$ m away from the foot of the tower, the angle of elevation of the top of the tower is $60^\circ$. Find the height of the tower. Diagram: Draw a right triangle. Let AB be the tower (height $h$), C be the point on the ground. $\angle ACB = 60^\circ$, $BC = 15$ m. $\angle ABC = 90^\circ$. Identify knowns/unknowns: Adjacent side $BC = 15$ m Angle $\theta = 60^\circ$ Opposite side $AB = h$ (unknown height) Choose Ratio: We have Opposite and Adjacent, so use $\tan \theta$. Formulate: $\tan 60^\circ = \frac{AB}{BC}$ Solve: $\sqrt{3} = \frac{h}{15}$ $h = 15 \sqrt{3}$ m $h \approx 15 \times 1.732 \approx 25.98$ m Answer: The height of the tower is $15\sqrt{3}$ meters (or approx. $25.98$ meters).