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Quantitative Aptitude :: Height and Distance

Height and Distance Important Formulas

1. Trigonometry:
In a right angled $$\Delta$$ABC, where $$\angle$$BAC = $$\theta$$,

$$\mathbb{i.} \ \sin \theta = \frac{Perpendicular}{Hypotenuse} = \frac{BC}{AB}$$

$$\mathbb{ii.} \ \cos \theta = \frac{Base}{Hypotenuse} = \frac{AC}{AB}$$

$$\mathbb{iii.} \ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{Perpendicular}{Base} = \frac{BC}{AC}$$

$$\mathbb{iv.} \ cosec \ \theta = \frac{1}{\sin \theta} = \frac{Hypotenuse}{Perpendicular} = \frac{AB}{BC}$$

$$\mathbb{v.} \ \sec \theta = \frac{1}{\cos \theta} = \frac{Hypotenuse}{Base} = \frac{AB}{AC}$$

$$\mathbb{vi.} \ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} = \frac{cosec \ \theta}{\sec \theta} = \frac{Base}{Perpendicular} = \frac{AC}{BC}$$

Note: In school we were use to taught a simple formula to learn it, lets refresh that.

So taking the initials below that sin, cos and tan, we can derive their values.
cosec is simply reciprocal to sin;
sec is reciprocal to cos;
cot is reciprocal to tan;

2. Trigonometrical Identities:

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

$$\mathbb{ii.} \ 1 \ + \ \tan^{2} \theta = \sec^{2} \theta$$

$$\mathbb{iii.} \ 1 \ + \ \cot^{2} \theta = cosec^{2} \theta$$

3. Values of T-ratios:

4. Line of Sight:
A line of sight is a line drawn from the observer's eye to the point, where the object is seen by the observer.

In this diagram, $$\theta$$ marks the angle of elevation of the top of the object as seen from a point on the ground.

5. Horizontal Line:

The line of sight which is parallel to ground level is known as horizontal line.

6. Angle of Elevation:

Suppose that from a point O a person sees an object P, which is placed above his eye level. Then, the angle of elevation is the angle between the horizontal and the line from the object to the observer's eye (line of sight).
i.e., Angle of Elevation = $$\angle$$ AOP

Angle of Depression:

Suppose that from a point O a person sees an object P, which is placed below the level of his eye. Then, the angle between the horizontal and the observer's line of sight is the angle of depression.
i.e., Angle of Depression = $$\angle$$ AOP

7. 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
$$\frac{BD}{DC} = \frac{AB}{AC}$$
(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., $$\angle$$BAD = $$\angle$$CAD in the above diagram)