## Trigonometry Basics

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### Ajita S

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This PPT includes basics of Trigonometry.

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Topic — Introduction to Trigonometry Class -X
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Introduction to Trigonometry Branch of Mathematics that deals with the triangles, mostly with right triangles, used in finding relationship between sides & angles. Tri • three Tri gono metry gonia- Greek: angle métrein• Greek: to measure
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Definition Branch of mathematics that studies relationships involving lengths and angles of triangles, (opposite*
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Origin & Purpose of Trigonometry Name - Hipparchus, known as father of trigonometry Place - Turkey Year -140BC Need - To find the motion of the planets, and the solar and lunar eclipses
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Ove rvi Trigonometric Functions There are three basic trigonometric functions which are . 1. 2. 3. Sine Function (Sin) (j opotenuse) Cosine Function (cos) Tangent Function (Tan) (of-pos ite) b cent) c
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Terminology Hypotenuse The hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle. Hypotenuse right thangle
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The opposite side is always across from the given angle. Side BC is opposite LA.
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Angle of elevation The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (the line of sight). Line Objec t ig ht Ho rizo Ota' O r-ve r fig - "-Angle of elevaton
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Basics Understanding of Trigonometry plane line of sight angle of elevation observer horizontal
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Sine Function Sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse. SOH - Sine is Opposite over Hypotenuse Sin Opposite Hypotenuse
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Cosine Function Cosine function (cos)/ defined as the ratio of the adjacent leg to the hypotenuse. CAH - Cosine is Adjacent over Hypotenuse Cos Adiacent Hypotenuse Adjacent
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Tangent Function (Tan) Tangent function (tan)/ defined as the ratio of the opposite leg to the adjacent leg. Tangent - Opposite over Adjacent Tan Adjacent Opposite Adjacent
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Example: Write expressions for the sme, cosme, and tangent of LA The length of the leg opposite ZA is a. The length of the leg adjacent to ZA is b, and the length of the hypotenuse is c. The sine of the angle is given by the ratio "opposite over hypotenuse." So, sin LA = The cosme is given by the ratio "adjacent over hypotenuse. cos ZA = The tangent is given by the ratio "opposite over adjacent." tan LA =
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Reciprocal Functions The cosecant or is the reciprocal of sin(A); i.e. the ratio of the length of the hypotenuse to the length of the opposite side 1 cscA sin A hypotenuse opposite a
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The secant sec(A) is the reciprocal of cos(A); i.e. the ratio of the length of the hypotenuse to the length of the adjacent side 1 sec A cosA hypotenuse adj acent b
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The cotangent cot(A) is the reciprocal of tan(A); i.e. the ratio of the length of the adjacent side to the length of the opposite side 1 cot A tan A adj acent opposite b a
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Example: IVI'ite expressions for the secant, cosecant, and cotangent of LA. The length of the leg opposite LA is a. The length of the leg adjacent to LA is b, and the length ofthe hypotenuse is c. The secant of the angle is given by the ratio "hypotenuse over adjacent". So, sec LA = The cosecant is given by the ratio "hypotenuse over opposite". csc LA = The cotangent is given by the ratio "adjacent over opposite" cot LA =
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Finding the Functions Trigonometric Ratios of 450 value of the other angle In A ABC, right-angled at a, if one angle is 43, then 45', i.e., (see Fig. 13). so, ac = AB Now, Suppose ac = Ad = a- Then by Pythagoras Theorem, AC2 = Ad2 + ac2 = a2 + a2 = 2a2 And, therefore, Trigonometric c Is also Fig. 13 side opposite to angle 450 sin 450 hypotenuse side adjacent to angle 450 cos 450 hypotenuse side opposite to angle 450 tan 450 side adjacent to angle 450 BC AB BC AB a a a a 1 1
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Trigonometric Ratios of 300 and 600 Let us now calculate the trigonometric ratios of 300 and 60' Consider an equilateral triangle Aac- Since each angle in an equilateral triangle is 60' therefore, z a = z C = 600 Draw the perpendicular AD trom A to the side ac (see Fig. 14). Now Therefore, and A ABD Z BAD A ACD DC Z CAD 300 600 Fig. 14 (CPCT)
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Now observe that: LABO Isa right triangle, right-angled at O with L 340= 30' and L ASO = 60' (see Fig. 14). As you know, tor finding the trigonometric ratios, we need to know the lengths ot the sides ot the triangle So, let us suppose that AB = 2a Then, and Therefore, Now, we have: 1 —BC — a 2 (2a)2 2 (a) AD2 sin 300 — tan 300 BD2 AD 1 3 C12 BD AB BD 20 a , cos 300 2 AB 2
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Similarly AD sin 600 cosec 600 , cos 600 sec 600 — 2 and cot 600 tan 600 AB 300 600 Fig. 14
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Trigonometric Ratios of 00 & 900 Case I — When angle A is 00 BC sin A = AC When LA is very close to CIO, ac gets very close to 0 and so the value ot is very close to O. cos A AC Also, when A is very close to O', AC is nearly the same as Ad and so the value of is very close to 1 - This helps us to see how we can define the values ot sin A and cos A when A = CIO. We define sin —0 and cos 00—1. using these, we have sin 00 tan O' cos 00 0, cot 00 vchicll is not defined. tan 00
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sec 00 cos 00 — 1 and cosec 00 which is again not defined, sin 00 Case Il — When angle A is 900 When L C is very close to 0', LA is very close to 90', side AC Is nearly tha same as side BC, and so sin A is very close to 1. Also when L A is very close to 90', -L C is very close to 0', and the side Ad is nearly zero, so cos A Is very closa to 0. So, we define : sin 900 = 1 and cos 90'-0.
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Angles in 300 Degrees cos tan Not csc defined 2V3 sec Not cot defined 450 600 2V3 900 Not defined Not defined
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Example : In A PQR/ right-angled at Q PQ = 3 cm and PR = 6 cm. Determine L QPR and L PRQ. Solution : PQ PR Z PRQ Z QPR 6 cm 3 cm sin R 300
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Example :-Evaluate the following + 3 cos 60 • -3 tan 45' a) 2 sin 30' Solution : a) 2 sin 30' + 3 cos 60' -3 tan 45'
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Trigonometrical Ratios of Complementary Angles
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Complementaty Angles: Two angles are said to be complementaty if their sum is 900, Thus 8 and (900 - 8) are complementaty angles, - 8): coss 8) : sin 8 (vi) csc 8 -sec 8 (i) sin (goo 8) = cot 8 tan (90' (v) sec 8) : csc8 (ii) cos (900 (iv) cot (90' - 8) tan 8
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Example tan6b• 1. 'Without using trigonometric tables, evaluate cot2s- Solution: tan65• cot25" tan65" at.(00 •-650) tan65• , [Since cot (900 vtan65• tan 8]
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Trigonometric Identities sin B + eos2 B I + cot sec esc
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Example Prove the identities: 1 tan e + cot æc csc a sin a tana+cot cosa cos a. sin cos a sin a sin2a+cosæa cos a•sinc — sec a. csc a
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Applications in Real Life In ancient time it was used for astronomy in finding distance of stars Finding radius of earth Finding height of hills, buildings, trees Navigation — Airplane, Ships etc. Defense

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