Trigonometry

Contents

1. INTRODUCTION
2. MEASUREMENT OF ANGLES
3. TRIGONOMETRIC RATIOS
4. FUNDAMENTAL IDENTITIES
5. HEIGHTS AND DISTANCES
   1. ANGLE OF ELEVATION
   2. ANGLE OF DEPRESSION

Trigonometry

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INTRODUCTION

Trigonometry is a branch of mathematics that deals with the measurements of the sides and the angles of triangles and the relationships between them.

MEASUREMENT OF ANGLES

An angle is the union of two rays having a common end-point, called the vertex. The magnitude of the angle is the amount of rotation that separates the two rays. Angles are measured in degrees or radians. The system of measuring angles in degrees is called the sexagesimal system and the system of measuring angles in radians is called the circular system.

DEGREES

One complete rotation around a point is measured as 360 degrees (360).

RADIAN

One radian is the angle subtended at the centre of a circle by an arc of length equal to the radius of the circle. One complete rotation around a circle is measured as 2 radians. One radian is written as 1 rad or 1^c.

360 = 2 rad

TRIGONOMETRIC RATIOS

The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec) and cotangent (cot).

TRIGONOMETRIC RATIOS OF ACUTE ANGLES

Consider a right triangle XYZ where the lengths of sides are as shown. is the angle between the base and the hypotenuse.

For a given value of , the values of trigonometric ratios are constant, irrespective of the lengths of sides.

The angles 0, 30 (/6 rad), 45 (/4 rad), 60(/3 rad), 90(/2 rad), 180 ( rad), 270 (3/2 rad) and 360(2) rad are called standard angles and their trigonometric ratios are as follows. These values should be memorised as they are quite useful in solving problems in geometry.

Example 1:

Find the value of if 2sin2 – 3sin + 1 = 0. is not obtuse. Solution: 2sin2 – 3sin + 1 = 0 This is a quadratic equation. Let sin = x. 2x2 – 3x + 1 = 0 (2x – 1)(x – 1) = 0 = 30 or = 90

Example 2:

Solution: We construct a right triangle as shown. By Pythagoras theorem, AC = 13 units.

Example 3:

then what is the value of ? [FMS 2009] Solution: Let cot = x Hence, option 3.

TRIGONOMETRIC RATIOS OF ANGLES GREATER THAN 90

The definitions of trigonometric ratios of angles that are greater than 90 are out of scope of this chapter. The values of such ratios, however, can be easily calculated using certain properties of the ratios.

As 360 0,

sin (–) = –sin

cos (–) = cos

tan (–) = –tan

Trigonometric ratios of angles greater than 90 can be found using the above figure as shown in the following.

REMEMBER

• All trigonometric ratios of angles in the I quadrant are positive.

• Only sine and cosecant are positive for angles in the II quadrant.

• In the III quadrant, only tangent and cotangent are positive.

• Whereas, only cosine and secant are positive in the IV quadrant.

Example 4:

If , , and are four angles of a cyclic quadrilateral, then the value of cos + cos + cos + cos is: [IIFT 2007] (1) 1(2) 0 (3) 1(4) None of these Solution: Let (, )and (, )be the two pairs of opposite angles in the cyclic quadrilateral. = 0 Hence, option 2.

FUNDAMENTAL IDENTITIES

For all values of ,

sin² + cos² = 1

1 + tan² = sec²

1 + cot² = cosec²

Example 5:

[IIFT 2008] Solution: Similarly, Hence, option 2.

HEIGHTS AND DISTANCES

ANGLE OF ELEVATION

Consider an object that is at a higher level than the point from which it is observed. Then the angle that the line of sight makes with the horizontal is called the angle of elevation.

Example 6:

Amar is standing at a point C, 10 metres away from the foot of a tower AB. The angle of elevation of the top of the tower (point A) is 60. Find the height of the tower. Solution: Assume that the height of the tower is h.

ANGLE OF DEPRESSION

Consider an object that is at a lower level than the point from which it is observed. Then the angle that the line of sight makes with the horizontal is called the angle of depression.

Example 7:

Niraj stands at the point A, which is the top of a tower AB, 20 metres in height. At a certain distance away from the foot of the tower, at point C, stands a cat. The angle of depression of the cat from the top of the tower is 30. Find the distance of the cat from the foot of the tower. Solution: Assume that the distance of the cat from the foot of the tower is d.

Example 8:

Jayesh is standing on top of a railway bridge watching an approaching train. The angle of depression of the topmost point of the start of the train as it comes into sight is 30. It changes to 45by the time it comes to a halt. The height of the train is 10 feet and the bridge is 60 feet tall. Find the horizontal distance between the point that the train comes into sight and the point at which it halts. Neglect Jayesh’s height. Solution: From the figure, JP = 50 feet. JQP = 45 PQ = 50 feet Also, JKP = 30

Example 9:

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45 to 60. After how much more time will this car reach the base of the tower? [CAT 2003 Re-Test] Solution: Let x be the distance from the later position of the car and the tower (i.e. when the angle of elevation was 60). Since the triangle formed (i.e. ∆ABD) is a 30°-60°-90° triangle, we have, height of the tower, h = Now, since the triangle formed by the initial position of the car (i.e. ∆ABC) is an isosceles triangle, AB = BC i.e.BC = DC = x = x ( 1) Time taken to travel distance DC is 10 minutes, thus, Time taken to travel distance x Hence, option 1.

Example 10:

When the sunray’s inclination increases from 30 to 60, the length of the shadow of a tower decreases by 60 m. Find the height of the tower. [SNAP 2008] (1) 50.9 m(2) 51.96 m (3) 48.8 m(4) None of these Solution: When the angle of inclination of the sun’s rays = 60, the shadow of the tower OP is OA. When = 30, the shadow is OB. Let OA = x AB = OB – OA = 2x = 60, x = 30 metres. OP = 1.732 30 = 51.96 metres. Thus, height of tower = 51.96 metres. Hence, option 2.

Example 11:

Find the length of the shorter diagonal of a rhombus having side 10 cm and one pair of opposite angles measuring 30 each. Solution: ADC = 30 and AD = 10 cm AM DC AM = 5 cm ΔAMC is a right triangle. AC2 = AM2 + CM2 AC 5.17 cm

Example 12:

The diagonals of rectangle ABCD intersect at the origin O. AOB = 60. AD = 10 cm. Find the area of the rectangle. Solution: Consider the figure. P is the mid point of AD. AP = 5 cm PAO = 30 In ΔAOB, AO = OB. As AOB = 60, ΔAOB is an equilateral triangle.

Example 13:

A warship and a submarine (completely submerged in water) are moving horizontally in a straight line. The Captain of the warship observers that the submarine makes an angle of depression of 30, and the distance between them from the point of observation is 50 km. After 30 minutes, the angle of depression becomes 60. [IIFT 2009] Question 1: Find the distance between them after 30 min from the initial point of reference. Solution: Refer to the following figure: The vertical distance between the warship and the submarine = 25 km When the angle of depression of the submarine is 60, the distance between the warship and the submarine is Hence, option 1. Question 2: If both are moving in same direction and the submarine is ahead of the warship in both the situations, then the speed of the warship, if the ratio of the speed of warship to that of the submarine is 2: 1, is: Solution: Let the speed of the submarine be x kmph. Then the speed of the warship is 2x kmph. Distance travelled by the warship in 30 minutes = x km Distance travelled by the submarine in 30 minutes = 0.5x km From the figure, Hence, option 4.