Important Formulas - Height and Distance

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Important Formulas - Height and Distance1. Trigonometric Basics

$\sin \theta =\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{y}{r}$

$\cos \theta =\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{x}{r}$

$\tan \theta =\frac{\text{opposite side}}{\text{adjacent side}}=\frac{y}{x}$

$\csc \theta =\frac{\text{hypotenuse}}{\text{opposite side}}=\frac{r}{y}$

$\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{r}{x}$

$\cot \theta =\frac{\text{adjacent side}}{\text{opposite side}}=\frac{x}{y}$

From Pythagorean theorem, $x^2+y^2=r^2$ for the right angled triangle mentioned above.

2. Basic Trigonometric Values

θ in degrees	θ in radians	sin θ	cos θ	tan θ
0°	0	0	1	0
30°	$\frac{\pi }{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
45°	$\frac{\pi }{4}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	1
60°	$\frac{\pi }{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
90°	$\frac{\pi }{2}$	1	0	Not defined

3. Trigonometric Formulas

Degrees to Radians and vice versa

$360°=2\pi \text{radian}$

Trigonometry Quotient Formulas

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

$\cot \theta =\frac{\cos \theta }{\sin \theta }$

Trigonometry - Reciprocal Formulas

$\csc \theta =\frac{1}{\sin \theta }$

$\sec \theta =\frac{1}{\cos \theta }$

$\cot \theta =\frac{1}{\tan \theta }$

Trigonometry - Pythagorean Formulas

${\sin }^2\theta +{\cos }^2\theta =1$

${\sec }^2\theta -{\tan }^2\theta =1$

${\csc }^2\theta -{\cot }^2\theta =1$

4. Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer's eye (the line of sight).

i.e., angle of elevation =  AOP

5. Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer's line of sight

i.e., angle of depression =  AOP

6. Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then

$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., BAD = CAD in the above diagram)

7. Important Values to memorise$\sqrt{2}=1.414$

$\sqrt{3}=1.732$

$\sqrt{5}=2.236$