Trigonometry. Hipparchus

Transcription

1 Trigonometry This unit facilitates you in, Sides of right angled triangle. Six ratios of sides of right angled triangle. Six trigonometric ratios Values of trigonometric ratio of standard angles. Trigonometric identity Problems on heights and distances. identifying the opposite, adjacent and hypotenuse of right angled triangle with respect to. defining the six ratios related to the sides of a right angled triangle. identifying the six trigonometric ratios. finding the values of trigonometric ratios of a given right angled triangle. finding the values of trigonometric ratios of some standard angles 0, 0, 45, 60 and 90. finding the complementary angle for given trigonometric ratio. deriving the identity sin + cos =. identifying the angle of elevation and the angle of depression in the given situations. Hipparchus The creator of trigonometry is said to have been the Greek Mathematician Hipparchus of the nd century. solving the problems related to heights and distances. There is perhaps nothing which so occupies the middle position of mathematics as trigonometry. -J.F. Herbart The word Trigonometry which means triangle measurment is credited to astholomam Pitiscus (56-6)

2 88 UNIT- You might have wondered several times how great heights and distances are found by using mathematical calculations. elow are given some situations from our surroundings, where such calculations are required. Height of Mountain Height of the temple Width of river Height of Pyramid Height of Qutubminar Distance between the tower and the building In all such similar cases heights and distances are calculated by using mathematical techniques, which are studied under a branch of mathematics called "Trigonometry". The word trigonometry is derived from the Greek words Tri, gon and metron. Hence, trigonometry means measuring three sides or measure of three sides of triangle. It is the study of relationship between the sides and angles of a triangle. So, trigonometry... is all about triangles. The earliest known work on trigonometry was recorded in Egypt and abylon. ncient Greek astronomers were able to use trigonometry to calculate the distance from the earth to the moon.

3 Trigonometry 89 Even today, most of the technologically advanced methods based on trigonometrical concepts are extremely useful in physical sciences, different branches of engineering like electrical engineering. etc. surveying, architecture, navigation, astronomy and also social sciences. ll the trigonometrical concepts are based on right angled triangle. It is concerned with the length of the sides and size of the angles of right angled triangles. Some of the situations where knowledge of "triangles" play an important role are shown below: Since trigonometry has a lot to do with angles and sides of a right angled triangle, let us first recall the fundamentals. onsider a right angled triangle. Observe the sides and their names. You have also learnt the relationship between the sides of a right angled triangle, given by Pythagoras theorem. Now, let us take some examples where sides or angles of a triangle are calculated. Legs or arms of right angle Example : In, if = 90 0, = 5 cm, and = cm what is the length of? = + (y Pythagoras theorem) cm 5 cm = = 5 = 5 9 = 6 = 6 = 4 cm Example : In PQR what is the measure of P? P + Q + R = 80 (ngle sum property) P P = 80 (Q + R) = 80 ( ) = 80 0 = 70 P = 70 0 Now observe the following examples Q 75 5 R

4 90 UNIT- Example : In DEF, Find DE D Example 4: In XYZ find X Z 4 cm 80 7 cm E 9 cm F X Y 0 cm an we find the length of the side and measure of the angle using Pythagoras theorem or angle sum property. Discuss with your classmates and teachers in the class. It is evident that we cannot solve the above problems using Pythagoras theorem or angle sum property of triangle. How are such problems solved? It is possible to find the sides and angles using trigonometry. Trigonometry is solving triangles and by solving it means finding sides and angles of the triangles. asics of Trigonometry Now, Let us learn about the basics of trigonometry. The basics of trigonometry are related to "right angled triangle" only. onsider, in which one angle is 90, the side opposite to it is hypotenuse. Suppose we mark the angles other than the right angle. We can mark either or, which are always acute angles. The marked angle is denoted as '' (Greek letter - read as theta) Then, what are the sides related to the angle called? Observe the following figures and study how the sides are named. side. djacent Opposite Opposite djacent The side which is opposite to is called opposite side and the other one is adjacent If =, then is the opposite side and is the adjacent side. If =, then is the opposite side and is the adjacent side The measure of angle '' can be expressed in "degrees" or "radians". Here are some examples.

5 Trigonometry 9 Degrees 0 45 Radians We have learnt about criteria of similar triangles in unit 0. Using this we can evolve the basics of trigonometry. Study the following examples. and DEF are equiangular triangles.therefore their corresponding sides are in proportion. cm 5cm D cm.5cm 4cm E cm F ~ DEF, DE = EF FD Now consider any two ratios at a time DE = EF These ratios can also be written as = DE EF In the given figures = cm 4cm and DE EF if a c a b than b d c d = DE EF This means that in, is two times and in DEF, EF is two times DE. We can infer that, what ever relationship exists between the lengths of the sides and of a, the same relationship holds good for the corresponding sides DE and EF of DEF, provided the two triangles are similar.

6 9 UNIT- This conclusion can be extended to the other pairs of corresponding sides also. Hence, = DE EF, EF FD =.5, FD.5 = DE Let us extend this idea to right angled triangles. There are two special cases of right angled triangles. ase (i): 45, 45, 90 right triangle or Isoceles right angled triangle. onsider two equiangular isosceles right angled triangles. ~ DEF = DE the sides containing EF 45 the right angle are equal 45 = EF FD = side is times of the hypotenuse F = FD DE hypotenuse is times of the sides. ase (ii): 0, 60, 90 or right angled triangles. onsider and DEF with the given measurements. 0º D E 4 60º 60º ~ DEF 0º F 60º 0º ~ DEF D E 60º 6 0º F = DE EF = DE EF is always times of. is always times of. = EF FD = EF FD (hypotenuse) is always (hypotenuse) is always times of. times divided by FD FD DE DE (hypotenuse) is always (hypotenuse) is always times divided by times of. From the above examples, we can conclude that

7 Trigonometry 9 "In a right angled triangle, for the given acute angles the ratio between any two sides is always constant" So from here onwards, we will consider only one right angled triangle instead of two similar right angled triangles to write the ratios. Trigonometric ratios: Let us consider one right angled triangle and write the possible ratios of its sides with respect to its acute angle.,, Opposite djacent,, triangle has sides. To write the ratio (fraction) numerator can be written in ways and denominator in ways. oth can be written in = 6 ways. So, using the sides of the triangle, we can have six different ratios. In trigonometry, each of these six ratios is given a name. Study the following table, to learn the six basic trigonometric ratios. Ratio Trigonometric name short form of the ratio Opposite Sine of angle sin djacent osine of angle cos Opposite djacent Tangent of angle tan Opposite osecant of angle cosec djacent Secant of angle sec djacent Opposite otangent of angle cot The implication of these six trigonometric ratios is that, "In a right angled triangle, if the acute angle is given, the ratios between any two sides can be found. onversely, if the ratio between any two sides is given, the angle can be found. ll these are possible due to " criteria for similarity".

8 94 UNIT- Know this! : The first use of the idea of 'sine' can be found in (500 D) the work of 'ryabhatiyam' of ryabhata in 500 D. ut, ryabhata used the word rdha-jya for the half-chord, which was shortened to Jya or Jiva in due course. When the ryabhatiyam was translated into rabic, the word Jiva was retained. It was further translated into "Sinus", which means curve in Latin. The word 'Sinus' also used as sine was first abbreviated and used as 'sin' by an English professor of astronomy Edmund Gunter (58-66). The origin of the terms 'osine' and 'tangent' was much later. The cosine function arose from the need to compute the sine of the complementary angle. ryabhata called it Kotijya. The name cosinus originated with Edmund Gunter. In 674, another English mathematician. Sir Jonas Moore first used the abbreviated notation 'cos'. Note:. sin is the abbreviation of sine of angle, which means ratio of the opposite side and hypotenuse with respect to the acute angle. It should not be treated as the product of sin and. So is the case for other trigonometric ratios. sin is something - it has a value. sin is nothing - it has no meaning.. T- ratios (trigonometric ratios) depend only on value of acute angle and not on the size of the triangle. Discuss the above statement with respect to the triangles shown here Let us take some more examples and write the trigonometric ratios. ILLUSTRTIVE EXMPLES Example :onsider a right angled triangle, in which 90 trigonometric ratios w.r.t. and. Write all Sol. Here is an acute angle. Observe the position of the sides with respect to angle. - is the opposite side of ngle. - is the adjacent side with respect to angle. - is the hypotenuse of the right angled triangle. djacent side Opposite side Side

9 Trigonometry 95 The trigonometric ratios of the angle in the right angled triangle can be defined as follows. sin = Side opposite to angle cos = Side adjacent to angle tan = Side opposite to angle Side adjacent to angle cosec = Side opposite to angle sec = Side adjacent to angle cot = Side adjacent to angle Side opposite to angle Now let us define the trigonometric ratios for the acute angle in the right angled triangle, 90 Opposite side djacent Side side sin = Observe that the position of the sides changes when we consider angle '' in place of angle. cos = tan = cosec = sec = cot = Example : Write 6 trigonometric ratios for the following right angled triangle. Sol. K L M In KLM, KLM = 90 0 and LMK= KM is the hypotenuse. KL is the opposite side. LM is the adjacent side. So, sin = Opp KL hyp KM cosec = Hyp Opp KM KL cos = adj hyp LM KM sec = Hyp adj KM LM

10 96 UNIT- tan = opp adj KL LM cot = adj opp LM KL Example : Write trigonometric ratios for the PQR. Sol. In the figure; Hyptoenuse = 5 units, Opposite = units, djacent = 4 units. sin = Opp Hyp 5 cos = dj 4 Hyp 5 tan = Opp dj 4 cosec = Hyp 5 Opp sec = Hyp 5 dj 4 cot = dj 4 Opp Q P 4 5 R Example 4: Write the T ratios for the following XYZ Sol. Given XY = 5, YZ = 8, XZ =? XZ = XY + YZ (y Phythagoras theorem) = = = 89 X 5 sin = 5 7 XZ = 89 = 7. cos = 8 7 tan = 5 8 Y 8 Z cosec = 7 5 sec = 7 8 cot = 8 5 Example 5: sin = 4, being an acute angle. 5 Find the value of tan + sec + 4 sec. cosec. Sol. Given sin = Opp 4 Hyp 5 In, = + (y Pythagoras theorem) = = 5 4 = 5 6 = 9 = 9 = 4 5 So tan = Opp 4 Hyp 5, sec = dj dj cosec = Hyp 5 Opp 4

11 Trigonometry 97 y substituting the values in, tan + sec + 4 sec. cosec = = = tan + sec + 4 sec. cosec = 6 Reciprocal Relations onsider the PQR in which PQR = 90 0 We know sin P = Opposite QR PR PR Find the reciprocal of sin P, i.e. sin P QR QR PR ut PR QR Hyp Opp cosec P Q P dj Hyp Opp. R sin P or cosec P = cosec P sin P i.e., sin P and cosec P are reciprocal to each other. Similarly, we can express the relationships between the other trigonometric ratios. cos P = djacent PQ PR PR cos P PQ PQ djacent PR cos P or sec P secp cos P tan P = Opposite QR djacent PQ PQ djacent tan P QR QR Opposite PQ tanp or cot P cot P tan P The reciprocal relations are listed below. cot P sec P sin = cosec = cosec sin cos = sec = sec cos tan = cot = cot tan

12 98 UNIT- Remember Trigonometric ratios can be easily recalled by remembering the acronym as follows: SOH H TO S O H Sin Opposite H os djacent T O Tan Opposite djacent The reciprocals of the above three ratios gives cosec, Sec and cot respectivety. Relation between the trigonometric ratios: onsider in which 90 sin Opp cos dj tan cos cot = tan sin sin cos tan cot sin cos cos sin Discuss : These relationships can be obtained in different ways. Discuss in class and try them. Expressing T - ratios in terms of a given ratio: Now you are familiar with the six trigonometric ratios. n important question arises at this juncture. If we know any one of the ratios, can we find the other ratios? onsider the. sin = K K Let us find the values of other T - ratios. If sin =, it means djacent Opposite side K K

13 Trigonometry 99 This means that, the lengths of the sides and are in the ratio :. So, if is equal to K, then will be K, where K is any positive number. To determine the other ratios we need the lengths of the third side. In, = 90 0 = = (K) K = 9K K = 8K = K = K K cos = K K cosec = K K and tan = K K and sec = K K Think! cot = K K The value of sin and cos are always less than. Discuss the reason for this statement. ILLUSTRTIVE EXMPLES Example : If sin = 5. Write the values of all other T - ratios. Sol. sin = Opp 5 Hyp In, 90, 5 = + (y Pythagoras theorem) = = 5 = 69 5 = 44 = 44 = cos = dj Hyp tan = Opp 5 dj cosec = Hyp Opp 5 sec = Hyp dj cot = dj Opp 5

14 00 UNIT- Example : If cos = 4 Find the values of ther T-ratios 5 Sol. In 90, = (y Pythagoras theorem) x = 5 4 = = 49 x = 49 = 7 = 7 x 5 4 Now we can write other trigonometric ratios as follows sin = Opp 7 Hyp 5 cos = dj 4 Hyp 5 tan = Opp 7 dj 4 cosec = Hyp 5 Opp 7 sec = Hyp 5 dj 4 Example In the, D and D = Sol. cot = dj 4 Opp 7 If = 0 cm. D = 6 cm and D = 5cm then find sin + cos '' is in triangle D. We need to know and D. Let us find them. In the D, D = 90 0 D + D = D +6 = 0 D = 0 6 = =44 D = 44 = cm D 0 cm 5 cm 6 cm Now, in D, D = 90 0 D + D = + 5 = = 69 = 69 = cm sin sin D 5 D and cos cos 5 7 Example 4 : Prove that cos. cosec = cot Sol. onsider LHS of the given equation cos. cosec = cos. sin cosec = cos sin = cot = RHS cos sin sin cot cos. cosec = cot

15 Trigonometry 0 Example 5: In right triangle, right -angled at, if tan =, then verify that sin cos =. Sol. In, =90 0 tan = Since tan = (Given) = = Let = = k, where k is a positive number. Now, = + = + = k +k = K sin = k k, cos = k k sin cos = sin cos = Example 6: D is a rhombus whose diagonal makes an angle with the side D, where cos =. If PD = 4 cm then find the side and the diagonals of the rhombus. Sol. We know, the diagonals of rhombus bisect each other at right angles. DP = P and P = P In DP, DP = 90 0 cos = P D P = k and D = K ut, D = PD +P (k) = 4 + (k) D P 9k = 6 + 4k 5k = 6, k = 6 5 k = 4 5 D = k = 4 5 = 5 cm and P = k = 4 5 = 8 5 cm D = PD = 4 = 8cm and = P = 8 5 = 6 5 cm Each side of the rhombus = cm, diagonal D = 8cm and 5 diagonal = 6 5 cm.

16 0 UNIT- I. Find sin and cos for the following: EXERISE (i) (ii) (iii) II. Find the following :. If sin x = 5, cosec x =. If cos x = 4 5, sec x =. If tan x = 7 4 cot x = 4. If cosec x = 5 5, sin x = 5. If sin = 5 and cos = 4 5 then, tan = 6. If cot = 8 5 and sin = 5 7 then, cos = III. Solve:. Given tan =, find the value of sin and cos 4. Given cot = 0, determine cos and cosec. Given tan = 7, find the other trigonometric ratios of angle If sin =, find cos, tan and cot + cosec. 5. If tan =, find sin, cos and cot. 6. If sec x =, then find sin x, tan x, cot x and cot x + cosec x. 7. If 4 sin - cos = 0, find sin, cos, sec and cosec.

17 Trigonometry 0 8. If sin = 5 and is acute, find the value of 5sin cos tan 9. If cos = 5 0. If cos 5 = 0, find and is acute, find the value of 5 tan cot 5 tan cot sin cos sin cos Trigonometric ratios of standard angles In this chapter so far we have been discussing about trigonometric ratios of an acute angle of a right anlged triangle. The triangle has a right angle and an acute angle less than 90. The measure of an acute angle can be anything less than 90 and greater than 0. ut the standard angle we quite often construct and use are 0, 45 and 60. Now let us find the trigonometric ratios for these angles and also for 0. () Trigonometric ratios of 45 onside r an Iso sceles right angled triangle, wh ere 90. If we take 45, then is also 45. = ( they are sides opposite to equal angles) Suppose = = unit = + (by Pythagoras theorem) = + = + = 45 = units Now consider the trigonometric ratios with respect to angle. i.e., 45 sin 45 = cos 45 = tan 45 = cosec 45 = () Trigonometric ratios of 0 and 60 sec 45 = Even though trigonometric ratios are studied with respect to right triangles only, they can be applied to any other type of triangle by drawing a perpendicular in the triangle. Let us use this idea to find the trigonometric ratios of 0 and 60. onsider an equilateral triangle, Since, all the angles are equal we get 60 Hence, = = Let us draw a perpendicular D from to. D D ( RHS postulate of congruency) cot 45 = 0 60 D

18 04 UNIT- D = D D = D since 60 we get, D = D = 0 0 Now, observe that, D is a right angled triangle with D 90, D 0 and D 60 onsider once again the equilateral triangle. We need to know the lengths of the sides of the triangle to find the trigonometric ratios. Let us consider, = = = units In D, = units, D = unit, 0 60 D Since D 90, y pythagoras theorem, = D + D = D + D = 4 = D = Trigonometric ratios of 0 0 sin 0 = Opp D Hyp cos 0 = dj D Hyp tan 0 = Opp D dj D cosec 0 = Hyp Opp D Trigonometric ratios of 60 0 sec 0 = Hyp dj D cot 0 = dj D Opp D sin 60 = D cos 60 = D tan 60 = D D cosec 60 = D () Trigonometric ratios of 0 and 90 sec 60 = D onsider the right triangle in which 90. Imagine what happens to the trigonometric ratios of the acute angle, if it is made smaller and smaller till it becomes zero. Observe what happens to the length of and. cot 60 = D D

19 Trigonometry 05 s, gets smaller and smaller, the length of the side decreases. s, a point gets closer to point., coincides with. When is very close to 0, length of gets very close to zero. sin = very close to 0. When is 0, length is nearly the same as. cos = very close to. Let us define sin and cos when 0 we define sin 0 = 0 and cos 0 = Now consider the other trigonometric ratios. tan 0 = sin 0 0 0, cot 0 = cos 0 ( division is not defined by zero.) tan0 0 this value is not defined sec 0 = cos 0 cosec 0 = this value is not defined. sin 0 0 Now imagine what happens to the trigonometric values of larger and larger till it becomes 90. when it is made s, gets larger and larger, gets smaller and smaller and the length of side decreases. s the point gets closer to point and finally when, is very close to 90, becomes very close to 0. and becomes almost the same as.

20 06 UNIT- When is very close to 90 is almost equal to sin = very close to. When is very close to 90 length of reduces nearly to zero. cos = very close to 0. we define sin 90 = and cos 90 = 0 Now consider other trigonometric ratios. tan 90 = cot 90 = sin 90 this value is not defined, cos 90 0 cos 90 0 sin 90 = 0 sec 90 = cos 90 0 this value is not defined, cosec 90 = sin 90 Now let us tabulate the trigonometric ratios of 0, 0, 45, 60 and 90. This table can be used for ready reference to solve problems. Values of trigonometric ratios for the standard angles are tabulated sin 0 cos 0 tan 0 ND cosec ND sec ND cot ND 0

21 Trigonometry 07 Remember Here is an activity to easliy recall or prepare the table of values of sin for standard angles. Step : Write the number 0 to Step : Divide the numbers by Step : Take the square root These are the values for sin Step 4: Write the values in reverse order 0 These are the values for cos sin Step 5: We know tan cos ND Step 6: Take the reciprocals of the values of sin cos and tanfor standard angles; we get values of cosecsec and cot respectively ILLUSTRTIVE EXMPLES Example : Find the value of the following : (i) tan 60 + tan 45 We have, tan 60 0 = and tan 45 0 = tan 60 + tan 45º =.() = + = 5

22 08 UNIT- (ii) osec 60 sec 45 + cot 0 We have cosec 60 =, sec 45 = and cot 0 = onsider cosec 60 sec 45 + cot 0 = (iii) sin 4 + cos 4 tan = = sin sin 45 4 cos cos 45 4 = = = tan tan 60 (iv) cos 0.sin.cos.sec 6 4 tan cot from the table cos 0 = sin = sin 90 = sec sec 45 4 cos cos 0 6 cot cot 60 tan tan 60 y substitution,. 4 = 4 8 Example : If cos = and is acute angle, then what is equal to? Sol. Given cos cos We know, cos coscos Examples : If tanand is acute, find the values of sin and cos Sol. Given tan tan We know tan 0 0 = tan and sin sin 90 0 = cos = cos 60 0 =

23 Trigonometry 09 Example 4 : If = 60 0, = 0 0 then prove that cos ( + ) = cos cos - sin sin Sol. LHS = cos (+) = cos ( ) = cos 90 0 = 0 LHS = RHS RHS = cos cos- sin sin = cos 60 0 cos sin 60 0 sin 0 0 = = 4 4 = 0 I. nswer the following questions: EXERISE. () What trigonometric ratios of angles from 0 0 to 90 0 are equal to 0? () Which trigonometric ratios of angles from 0 0 to 90 0 are equal to? () Which trigonometric ratios of angles from 0 0 to 90 0 are equal to 0.5? (4) Which trigonometric ratios of angles from 0 0 to 90 0 are not defined? (5) Which trigonometric ratios of angles from 0 0 to 90 0 are equal? II. Find if (i) cos (ii) tan= (iii) sin (iv) 5 sin (v) tan III. Find the value of the following: (i) sin0 0 cos tan 45 0 (ii) sin60 0 cos0 0 + cos60 0 sin0 0 (iii) cos60 0 cos0 0 - sin60 0 sin0 0 (iv) sin cos tan sin 90 0 (v) 4 sin tan sin45 0. cos cos45 4 sin 60 - cos 45 (vi) 0 0 (vii) sec0 + cosec0 tan 0 + sin 0 (viii) sin0 0 + tan cosec60 0 (ix) 5 cos sec tan 45 0 Sec0 0 +cos cot45 0 sin cos 0 0 (x) 5 sin cos tan 0 0 sin0 0 +cos0 0 + tan45 0 IV. Prove the following equalites (i) sin0 0. cos cos0 0. sin60 0 = sin90 0 (ii) cos = - sin 0 0 = cos60 0 (iii) If 0 0, prove that 4 cos cos0 0 = cos

24 0 UNIT- (iv) If = 80 0 and = 6 prove that ( - cos) (+ cos) (-sin) (+sin) (v) If = 5 0, prove that 4 sin.cos4.sin6 = (vi) If = 60 0 and = 0 0, then prove that tan ( - ) = Trigonometric identities: Recall that in algebra, you have learnt about identities, tan - tan + tan tan For example, (a + b) = a + ab + b is an identity. It is an equation which is true for all the value of the variables a and b. Similarly an equation involving trigonometric ratios of an angle is called a "Trigonometric identity", if it is true for all the values of the angle involved. Now let us derive same trigonometric identities. onsider a right angled triangle, in which 90 y Pythagoras theorem, we get + = Dividing each term of the equation by, we get = = (cos ) + (sin ) = sin + cos =... (i) This equation is true for all values of angle, such that 0 90 This equation (i) is a fundamental trigonometric identity. Now let us obtain two more trigonometric identities from the fundamental identity. sin + cos = by dividing by sin we get, For convenience (sin ) is written as sin and read as sine squared. sin cos sin sin sin + (cot ) = (cosec ) sin cos sin sin sin cot cosec...(ii) Trigonometric identity (ii) is true for all values of angle. Such that 0 < 90, for 0, cot and cosec are not defined.

25 Trigonometry y dividing () by cos we get, sin cos cos cos cos sin cos cos cos cos (tan ) + = (sec ) tan sec...(iii) Trigonometric identity (iii) is true for all values of angle, such that 0 90 For 90, tan and sec are not defined. Trigonometric identities (i), (ii) and (iii) are called fundamental relations. () sin + cos = () tan + = sec () + cot = cosec These identities can also be rewritten as follows: sin + cos = sin = cos ; or cos = sin tan + = sec tan = sec ; or sec tan = + cot = cosec cot = cosec -; or cosec cot = We can also use these identities to express each trigonometric ratio in terms of other trigonometric ratios. That is, if any one of the ratios is known, we can find the values of other trigonometric ratios. Example : Prove that coscosec = cot Sol: LHS = cos cosec = cos cosec sin sin cos sin = cot = RHS lternatively, we know that in RHS. cot cos sin we have cos Hence sin in LHS, but sin is written as cosec cos cosec = cos sin =cot ILLUSTRTIVE EXMPLES is not there instead we have cosec.

26 UNIT- Example : Show that tan. sin + cos = sec Sol: onsider LHS = tan sin + cos = sin sin cos cos = = sin cos cos cos = = sec = RHS sin cos cos sin tan cos ) (ut sin +cos =) Example : Find the value of (sin + cos ) + (sin cos ) Sol: (sin + cos) + (sin - cos) =sin + cos + sin cos + sin + cos sin cos = [sin + cos ] = () = ( sin + cos =) Example 4: Prove that cos = tan Method - Sol: LHS = cos = sec tan Method - RHS= tan sec cos = RHS = LHS Example 5 : Show that ( sin ) ( + tan ) = Sol: onsider LHS = ( sin ) ( + tan ) Example 6: Prove that Sol: onsiderlhs = = cos sec [ sin = cos, + tan = sec ] = cos cos sin cos cosec sec sin cosec cos sec = = = RHS sin cos sin cos = sin + cos = = RHS Example 7 : If sin + 5 cos = 4. Show that cos = Sol: sin + 5 cos = 4 [ cos ] + 5 cos = 4 cos + 5 cos = 4 cos + 5 cos 4 = 0 cos + 5 cos = 0 cos 5 cos + = 0

27 Trigonometry This is quadratic expression in the form ax + bx + c = 0 cos 4 cos cos + = 0 cos [cos ] [cos ] = 0 [ cos ] [cos ] = 0 cos = 0 or cos = 0 cos = or cos = cos = Discuss: cos= is not considered. Why? I. Show that EXERISE.. (-sin ) sec =. (+tan ) cos =. (+tan )(-sin (+sin = 4. sin cos cos sin cosec 5. sin sin (sec tan ) 6. (+cos )(-cos (+cot = 7. cos sin tan cot = sin + cos 8. tan tan 9. (sin + cos ) = + sin cos 0.. sin cos tan + cos sin cot =. cos sin sin cos sec sin cos cos cos cos tan sin tan. sin cos 4 cot cosec 4. tan - sin = tan sin 5. cos - sin = cos - 6. If x = r sin cos, y = r sin sin, z = r cos, then x +y +z =r. Trigonometric ratios of complementary angles: We are already familiar with complementary angles. Recall that, if the sum of any two angles is 90, they are said to be complementary angles. How do we define trigonometric ratios involving two angles which are complementary to each other? onsider the in which 90. Identify the pair of complementary angles in Since 90, 90 and are complementary angles. In any right angled triangle, the two acute angles other than right angle always form a pair of complementary angles.

28 4 UNIT- Now let us define trigonometric ratios for these complementary angles. Since or 90 [For convenience sake, these complementary angles are written as (90 ) or (90 )]. The trigonometric ratios for angle and (90 ) are written in the table given below observe them, T-ratios for angle T-ratios for angle (90 ) sin = cos = tan = cot = sec = cosec = sin (90 ) = cos (90 ) = tan (90 ) = cot (90 ) = sec (90 ) = cosec (90 ) = ompare the ratios of angle and its complementary angle. We observe that. sin (90 ) = cos = cos (90 ) = sin = tan (90 ) = cot = We can conclude that, sin (90 ) = cos cos (90 ) = sin tan (90 ) = cot cosec (90 ) = sec = sec (90 ) = cosec = cot (90 ) = tan = cosec (90 ) = sec sec (90 ) = cose cot (90 ) = tan 90º ILLUSTRTIVE EXMPLES 0 sin 65 Example : Evaluate 0 cos 5 Sol. We know, cos 5 0 = sin ( ) = sin sin65 0 cos 5 = cos 5 0 cos 5 =

29 Trigonometry 5 Example : If tan = cot ( 8 0 ), where is an acute angle, find the value of. Sol. Since tan = cot (90 0 ) we get cot (90 0 ) = cot ( 8 0 ) 90 0 = 8 0 = 08 0 = 0 08 = 60 = 6 0 Example : Prove that : [cosec (90º ) sin (90º )] [cosec sin ] [tan + cot ] = Sol. LHS = [cosec (90º ) sin (90º )] [cosec sin ] [tan + cot ] = [sec cos ] [cosec sin ] [tan + cot ] = = sin cos cos sin cos sin cos sin cos sin cos sin cos sin = sin cos cos sin cos sin = = RHS Example 4 : Prove that sin(90º ) cos sin cos(90º ) = sec Sol. LHS = = sin(90º ) cos sin cos(90º ) cos cos sin sin cos ( sin ) cos ( sin ) = sin = cos cos sin cos cos sin sin cos = cos cos sec = sec = RHS EXERISE.4. Evaluate : (i) tan 65 cot (ii) sin8 cos (iii) cos 48 0 sin 4 0 (iv) cosec 0 sec 59 0 (v) cot 4 0 tan 56 0 (vi) sin 6 sin54 cos 54 cos (vii) sec 70 0 sin 0 0 cos 70 0 cosec 0 0 (viii) cos 0 sin 77 0

30 6 UNIT-. Prove that (i) sin 5 0 sin cos 5 0 cos 55 0 = 0 (ii) tan 0 0 tan 5 0 tan 75 0 tan 80 0 = (iii) cos 8 0 cos sin 8 0 sin 5 0 = 0. If sin 5 = cos 4, where 5 and 4 are acute angles, find the value of 4. If sec 4 = cosec (-0 0 ), where 4 is an acute angle, find the value of. Heights and Distances We are familiar with measuring the height of objects like height of a person, a statue, a plant etc by using a scale or measuring tape. Measurements of this type are called direct measurements. There are many real life situations where heights cannot be found by direct measurements. For example, height of a tall building, height of a tower, tree, distant mountain or width of a river etc. Such measurements are computed by indirect measurements or methods. One of the best methods for the indirect measuremement of length or height involves ratios. Trigonometry has wide applications in solving problems realated to real life situations. The main application of triginometry is in finding heights and distances. Trignometric ratios are also used in costructing maps, determining the position of an island in relation to longitude and latitude.etc. Let us define a few terms which we use very often in finding heights and distances. Line of sight observe the figures given below.

31 Trigonometry 7 Here, the distance involved is quite large and hence the view is assumed to be viewing a point on the object or the object is treated as a point. In the above diagram, The line drawn from the eye of an observer to the point on the object viewed is the line of sight. The line of sight is an horizantal line parallel to the ground level. ut, not always the line of sight will be a horizontal line. Observe the point of view in the following figures. Fig. (i) Fig. (ii) In fig (i), the boy is viewing the flag for saluting it. The flag is above the horizantal line and the boy has raised his head to view the object. In this process the eyes move through an upward angle formed by the horizontal line and the line of sight. This angle is called angle of elevation. In fig (ii), the boy is viewing the boat which is below the horizontal line and the boy has to lower his head (eyes) to view the object. In this process also, the eyes move through an downward angle formed by the horizantal line and the line of sight. This angle is called angle of depression. From the above examples, we can conclude that The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal line. It is the case where we raise our heads to look at the objects. ngle of elevation are measured from horizontal upwards. - The angle of depression of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal line to look at the objects. ngle of our heads are measured from horizontal downwards. Note: * ngle of elevation and angle of depression are always measured with the horizontal. * The angle of elevation of an object as seen by the observer is same as the angle of depression of the observer as seen from the object. * If the height of the observer is not given, the observer is taken as a point.

32 8 UNIT- Know this: Surveyors have used trigonometry for centuries. One such large surverying project of the nineteenth century was the Great Trigonometric Survey of ritish India for which the two largest-ever theodolites were built. During the survey in 85, the highest mountain in the world was discoverd. From a distance of over 60 km, the peak was observed from six diffrerent stations. In 856, this peak was named after Sir George Everest, who had commissioned and first used the giant theodolites (see the figure alongside). The theodolites are now on display in the Museum of the Survey of India in Dehradun. It is a surveying instrument which is used in measuring the angle between an object and the eye of the observer. It is based on the principles of trigonometry and the angle is read on the rotating telescope scale. ILLUSTRTIVE EXMPLES Example : tower stands vertically on the ground. From a point on the ground, which is 50m away from the foot of the tower, the angle of elevation to the top of the tower is 0. Find the height of the tower. Sol : Draw a figure to represent the given information. Let be the height (h) of the tower, and is the distance between the tower and the point of the viewer. In, = 0 0 tan tan 0 = h 50 h 50 h = 50 m Example : Find the angle of elevation if an object is at a height of 50 m on the tower and the distance between the observer and the foot of the tower is 50 m. Sol: Let PQ be the height of the object = 50 m QR is the distance between tower and observer = 50 m In PQR, PQR = 90 0 P tan = PQ 50 QR 50 50m tan = = tan 45 0 ( tan 45 = ) tan = tan 45 0 = 45 the angle of elevation is 45 Q 50m R Example : From the top of a building 50 m high the angle of depression of a car on the ground is observed to be 60. Find the distance of the car from the building.

33 Trigonometry 9 P = 60 M 50 Q Sol: Let PQ represent the height of the building, PQ = 50 m, QR be the distance between the building and the car. ngle of depression is 60 0? 60 since, PM QR, so MPR PRQ ( alternate angles) MPR 60 PRQ 60 In PQR, PQR 90, PRQ 60 PQ 50 tan 60 = QR QR R = 50 QR QR = 50 = 50 the car is 50 m away from the building. Example 4: Two windmills of height 50 m and 40 m are on either side of the field. person observes the top of the windmills from a point in between the towers. The angle of elevation was found to be 45º in both the cases. Find the distance between the windmills. Sol: Read the problem carefully, convert the data into meaningful diagram. D Let be a tower of height = 50 m D be another tower of height = 40 m P The observe is at 'P' and angle of elevation P PD 45 Distance between D = P + PD.

34 0 UNIT- In In P, tan = P = 50 P P = 50 m PD, tan = D PD tan 45 = 40 PD tan 45 = 50 P ( tan 45 = ) ( tan 45 = ) = 40 PD PD = 40m D = P + PD = = 90 m The distance between the windmills on either side of field is 90 m EXERISE.5 I. Find the value of 'x' R x m K. X 90 P x 60 Q D II.. L 0 00 x M 4. Y x. tall building casts a shadow of 00 m long when the sun's altitude (elevation) is 0. Find the height of the tower.. From the top of a building 50 m high, the angle of depression of an object on the ground is observed to be 45. Find the distance of the object from the building.. tree is broken over by the wind forms a right angled triangle with the ground. If the broken part makes an angle of 60 with the ground and the top of the tree is now 0 m from its base, how tall was the tree? Z F 75 x E

35 Trigonometry 4. The angle of elevation of the top of a flagpost from a point on a horizontal ground is found to be 0. On walking 6 m towards the post, the elevation increased by 5. Find the height of the flagpost. 5. Two observers who are km apart in the same plane are observing a balloon at a certain height. They are on opposite sides of it and found the angles of elevation to be 60 and 45. How high is the balloon above the ground? 6. The angles of elevation of the top of a cliff as seen from the top and bottom of a building are 45 and 60 respectively. If the height of the building is 4 m, find the height of the cliff. 7. From the top of a building 6 m high, the angular elevation of the top of a hill is 60 and the angular depression of the foot of the hill is 0. Find the height of the hill. 8. Find the angle of depression if an observer 50 cm tall looks at the tip of his shadow which is 50 cm from his foot. 9. From a point 50 m above the ground the angle of elevation of a cloud is 0 and the angle of depression of its reflection in water is 60. Find the height of the cloud above the ground. 0. From a light house the angles of depression of two ships on opposite sides of the light house were observed to be 45 0 and If the hight of the light house is 0m and the line joining the two ships passes through the foot of the light house, find the distance between to two ships. Trigonometry Trigonometric ratios sin cos tan sec cot Reciprocals of trigonometric ratios Relation between trigonometric ratios Trigonometric identities sin +cos = +cot =cosec tan +=sec Finding heights and distances angle of elevation angle of depression Trigonometric ratios of standard angles

36 UNIT- NSWERS Exercise. I. (i) 4 5, 7 5 (ii) 5, 4 5 (iii) 5, II. (i) 5 (ii) 5 4 (iii) 4 7 (iv) 5 (v) 4 (vi) 8 7 III. ], ] 0 9, 9 ] 7, 4, , 5 7, 5 4, 4 7 4],, 5], 0 0, 6],,, 7], 4, 5, ] 65 9] 7 7 0] 7 7 Exercise. II. (i) 45 (ii) 0 (iii) 60 (iv) 0º (v) 0 III. (i) 4 (vii) 5 (ii) (iii) 0 (iv) 5 4 (viii) 4 4 (ix) 67 (x) (v) (vi) Exercise.4 ] (i) (ii) (iii) 0 (iv) 0 (v) 0 (vi) 0 (vii) 0 (viii) 0 ] = 0 4] = Exercise.5 I (i) 60m (ii) 0 units (iii) 00 units (iv) 00 units (v) 4 units II ] 00 m ] 50 m ] 0 m 4] 6 m (5) km 6] 4 4 m 7] 64m 8] 0º 9] 00m 0] 0 m