12 Trigonometry This unit facilitates you in, Sides of right angled triangle. identifying the opposite, adjacent and Six ratios of sides of right angled triangle. hypotenuse of right angled triangle with Six trigonometric ratios Values of trigonometric ratio of standard angles. Trigonometric identity Problems on heights and distances. 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°, 30°, 45°, 60° and 90°. finding the complementary angle for given trigonometric ratio. deriving the identity sin2 + cos2 = 1. identifying the angle of elevation and the angle of depression in the given situations. solving the problems related to heights and Hipparchus The creator of trigonometry is said to have been the Greek Mathematician Hipparchus of the 2nd century BC. The word Trigonometry which means triangle measurment is credited to Bastholomam Pitiscus (1561-1613) distances. There is perhaps nothing which so occupies the middle position of mathematics as trigonometry. -J.F. Herbart 288 UNIT-12 You might have wondered several times how great heights and distances are found by using mathematical calculations. Below 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 Babylon. Ancient Greek astronomers were able to use trigonometry to calculate the distance from the earth to the moon. Trigonometry 289 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. All 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. A Hy Consider a right angled triangle. Observe the po sides and their names. te nu se You have also learnt the relationship between Legs or arms 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. of right angle B C Example 1 : In ABC, if B = 900, AC = 5 cm, and AB = 3cm what is the length of BC? 3 cm A B AC2 = AB2 + BC2 5 (By Pythagoras theorem) BC2 = AC2 – AB2 cm = 52 – 32 = 25 – 9 = 16 BC = C 16 = 4 cm Example 2 : In PQR what is the measure of P ? P P + Q + R = 180°(Angle sum property) P = 180° – (Q + R) = 180° – (75° + 35°) = 180° – 110° = 70° P = 700 Now observe the following examples 75° Q 35° R 290 UNIT-12 Example 4: In XYZ find X Example 3: In DEF, Find DE D Z 4c E X F 9 cm 7c m 80° m Y 10 cm Can 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. A Hy Basics of Trigonometry po te n Now, Let us learn about the basics of trigonometry. us e The basics of trigonometry are related to "right angled triangle" only. B C Consider ABC, 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 BAC or ACB, 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. A Adjacent B Hy po ten Opposite us e C Opposite A Hy po ten us B Adjacent e C The side which is opposite to is called opposite side and the other one is adjacent side. If A = , then BC is the opposite side and AB is the adjacent side. If C= , then AB is the opposite side and BC is the adjacent side The measure of angle '' can be expressed in "degrees" or "radians". Here are some examples. Trigonometry 291 Degrees Radians 30° 6 45° 4 60° 3 2 90° 2 180° 360° We have learnt about AAA criteria of similar triangles in unit 10. Using this we can evolve the basics of trigonometry. Study the following examples. ABC and DEF are equiangular triangles.Therefore their corresponding sides are in proportion. 5c m D 2. 5c 1c m 2cm A B E C 4cm m F 2cm AB BC CA = DE EF FD Now consider any two ratios at a time ABC ~ DEF, AB BC = DE EF These ratios can also be written as AB DE = BC EF if a b c a than d c b d In the given figures AB 2cm = BC 4cm 1 DE and 2 EF 1 2 AB DE = BC EF 1 2 This means that in ABC, BC is two times AB and in DEF, EF is two times DE. We can infer that, what ever relationship exists between the lengths of the sides AB and BC of a ABC, the same relationship holds good for the corresponding sides DE and EF of DEF, provided the two triangles are similar. 292 UNIT-12 This conclusion can be extended to the other pairs of corresponding sides also. Hence, AB DE = BC EF 1 , 2 BC CA EF 2 = , FD 2.5 CA FD = AB DE 2.5 1 Let us extend this idea to right angled triangles. There are two special cases of right angled triangles. Case (i): 45°, 45°, 90° right triangle or Isoceles right angled triangle. Consider two equiangular isosceles right angled triangles. 45° ABC ~ DEF A AB BC DE = EF 2 2 1 the sides containing 1 45° 1 the right angle are equal 2 2 45° 45° B C 1 BC CA 1 1 EF = = side is times of the hypotenuse. 2 2 FD AC AB = FD DE 2 hypotenuse is 1 2 2 times of the sides. Case (ii): 30°, 60°, 90° or right angled triangles. Consider ABC and DEF with the given measurements. D D A A 30º 30º 2 3 4 2 3 60º 1 60º 3 2 6 30º 30º 60º 60º B 1 E C 2 B F ABC ~ DEF AB DE = BC EF 3 C E ABC ~ DEF AB DE = EF BC 3 AB is always 3 times of BC. BC is always 1 3 3 times of AB. BC EF 1 = CA FD 2 CA (hypotenuse) is always BC EF 3 = CA FD 2 CA (hypotenuse) is always 2 2 times of BC. times BC divided by CA FD 2 AB DE 3 CA (hypotenuse) is always CA FD 2 AB DE 1 CA (hypotenuse) is always 2 2 times AB divided by 3 times of AB. From the above examples, we can conclude that 3 3 3 F F Trigonometry 293 "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: Opposite Let us consider one right angled triangle and write the possible ratios of its sides with respect to its acute angle. A AB AB BC H , , yp AC BC AC ot en us AC BC AC e , , AB AB BC B Adjacent C A triangle has 3 sides. To write the ratio (fraction) numerator can be written in 3 ways and denominator in 2 ways. Both can be written in 3 × 2 = 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 of the ratio short form AB AC Opposite Hypotenuse Sine of angle sin BC AC Adjacent Hypotenuse Cosine of angle cos AB BC Opposite Adjacent Tangent of angle tan AC AB Hypotenuse Opposite Cosecant of angle cosec AC BC Hypotenuse Adjacent Secant of angle sec BC AB Adjacent Opposite Cotangent 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. Conversely, if the ratio between any two sides is given, the angle can be found. All these are possible due to "AAA criteria for similarity". 294 UNIT-12 Know this ! : The first use of the idea of 'sine' can be found in (500 AD) the work of 'Aryabhatiyam' of Aryabhata in 500 AD. But, Aryabhata used the word Ardha-jya for the half-chord, which was shortened to Jya or Jiva in due course. When the Aryabhatiyam was translated into Arabic, 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 (15811626). The origin of the terms 'Cosine' and 'tangent' was much later. The cosine function arose from the need to compute the sine of the complementary angle. Aryabhata called it Kotijya. The name cosinus originated with Edmund Gunter. In 1674, another English mathematician. Sir Jonas Moore first used the abbreviated notation 'cos'. Note: 1. 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. 2. 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. A A A 30° 30° 30° C B B C B C Let us take some more examples and write the trigonometric ratios. ILLUSTRATIVE EXAMPLES Example 1 :Consider a right angled triangle ABC, in which trigonometric ratios w.r.t. A and ABC 90 . Write all C - BC is the opposite side of Angle A. - AB is the adjacent side with respect to angle A. - AC is the hypotenuse of the right angled triangle ABC. Adjacent side Sol. Here CAB is an acute angle. Observe the position of the sides with respect to A angle A. H yp ot en us e B Opposite Side side C Trigonometry 295 The trigonometric ratios of the angle A in the right angled triangle ABC can be defined as follows. sin A = Side opposite to angle A Hypotenuse BC AC cos A = Side adjacent to angle A Hypotenuse tan A = Side opposite to angle A Side adjacent to angle A BC AB cosec A = sec A = Hypotenuse Side adjacent to angle A AC AB cot A = Hypotenuse Side opposite to angle A Side adjacent to angle A Side opposite to angle A AB AC AC BC AB BC Now let us define the trigonometric ratios for the acute angle C in the right angled triangle, ABC 90 Opposite side A H yp B Adjacent side Side sin C = Observe that the position of the sides changes when we consider angle 'C' in place of angle A. ot en us e C AB AC cosec C = AC AB cos C = BC AC tan C = AB BC sec C = AC BC cot C = BC AB Example 2: Write 6 trigonometric ratios for the following right angled triangle. Sol. K In KLM, KLM = 900 and LMK= KM is the hypotenuse. M L KL is the opposite side. LM is the adjacent side. So, sin = Opp hyp KL KM cosec = Hyp Opp KM KL cos = adj hyp LM KM sec = Hyp adj KM LM 296 UNIT-12 tan = opp adj KL LM cot = adj opp Example 3 : Write trigonometric ratios for the LM KL PQR. Sol. In the figure; Hyptoenuse = 5 units, Opposite = 3 units, Adjacent = 4 units. Opp Hyp 3 5 cosec = Adj cos = Hyp 4 5 Hyp sec = Adj sin = Opp tan = Adj 3 4 Hyp Opp 5 4 Adj cot = Opp Example 4: Write the T ratios for the following P 5 3 4 3 5 3 Q R 4 XYZ X 15 Sol. Given XY = 15, YZ = 8, XZ = ? XZ2 = XY2 + YZ2 (By Phythagoras theorem) = 152 + 82 = 225 + 64 = 289 XZ = 289 = 17. sin = 15 17 cosec = 17 15 Example 5: sin A = Y cos = 8 17 tan = 15 8 sec = 17 8 cot = 8 15 Z 8 4 , A being an acute angle. 5 Find the value of 2 tan A + 3 sec A + 4 sec A . cosec A. Sol. Given sin A = In Opp Hyp 2 C 4 5 2 4 2 ABC, AC = AB + BC (By Pythagoras theorem) AB2 = AC2 – BC2 = 52 – 42 = 25 – 16 = 9 AB = So tan A = B 9 =3 Opp Adj 4 , 3 sec A = Hyp Adj 5 3 cosec A = Hyp Opp 5 4 5 A Trigonometry 297 By substituting the values in, 2 tan A + 3 sec A + 4 sec A . cosec A 4 = 2 3 = 8 3 3 5 5 3 4 5 3 25 8 15 = 3 3 5 4 25 48 3 16 2 tan A + 3 sec A + 4 sec A. cosec A = 16 Reciprocal Relations PQR in which PQR = 900 We know sin P = Opposite Hypotenuse QR PR 1 sin P Find the reciprocal of sin P, i.e. PR QR But sin P Hyp Opp P Adj Consider the 1 QR PR Hyp PR QR Q Opp. R cosec P 1 1 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. 1 PQ cos P PR cos P = Adjacent Hypotenuse cos P 1 or sec P sec P tan P = Opposite Adjacent 1 PQ PR PR PQ Hypotenuse Adjacent 1 QR PQ PQ QR Adjacent Opposite 1 cos P QR 1 tan P PQ 1 1 or cot P cot P tan P The reciprocal relations are listed below. tan P sin A = 1 cosec A cos A = 1 sec A tan A = 1 cot A 1 sin A sec A = 1 cos A cot A = 1 tan A cosec A = sec P cot P 298 UNIT-12 Remember Trigonometric ratios can be easily recalled by remembering the acronym as follows: SOH CAH TOA S O H Sin Opposite Hypotenuse C A H Cos Adjacent Hypotenuse T O A Tan Opposite Adjacent The reciprocals of the above three ratios gives cosec , Sec and cot respectivety. Relation between the trigonometric ratios: cot A = BC AB 1 tan A Opp Adj 1 sin A cos A tan A cos A sin A tan A sin A cos A cot A cos A sin A B se nu BC AC AB AC A 90 te po Hy sin A cos A B Adjacent Consider ABC in which Opposite side C 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. An important question arises at this juncture. If we know any one of the ratios, can we find the other ratios? Consider the ABC. sin A = BC AC K 3K C 1 3 3K Let us find the values of other T - ratios. 1 BC If sin A = , it means 3 AC 1 3 A K B Trigonometry 299 This means that, the lengths of the sides BC and AC are in the ratio 1 : 3. So, if BC is equal to K, then AC will be 3K, where K is any positive number. To determine the other ratios we need the lengths of the third side AB. ABC, ABC = 900 In AB2 = AC2 – BC2 = (3K)2 – K2 = 9K2 – K2 = 8K2 = 2 2K AB = cos A = AB AC cot A = 2 2K 3K AC BC cosec A = 2 2 2K 3K K AB BC 2 2K 2 2 3 3 and and 2 2K K 2 tan A = BC AB sec A = K 1 2 2K 2 2 AC AB 3K 3 2 2K 2 2 2 2 Think! The value of sin A and cos A are always less than 1. Discuss the reason for this statement. ILLUSTRATIVE EXAMPLES Example 1: If sin = Sol. sin = Opp Hyp 5 . Write the values of all other T - ratios. 13 A 5 13 5 In ABC, 90 , B 2 2 C 2 AC = AB + BC (By Pythagoras theorem) B BC2 = AC2 – AB2 = 132 – 52 = 169 – 25 = 144 BC = 144 = 12 cos = Adj Hyp cosec = Hyp Opp 12 13 13 5 13 tan = Opp Adj 5 12 sec = Hyp Adj 13 12 cot = Adj Opp 12 5 C 300 UNIT-12 Example 2 : If cos = Sol. In ABC B 24 Find the values of ther T-ratios 25 90 , C A 25 x AB2 = AC2 – BC2 (By Pythagoras theorem) x2 = 252 – 242 = 625 – 576 = 49 x = B 24 C 49 = 7 AB = 7 Now we can write other trigonometric ratios as follows Opp Hyp sin = cosec = 7 25 Hyp Opp cos = Adj Hyp 25 Hyp sec = 7 Adj 24 Opp tan = 25 Adj 25 Adj cot = 24 Opp 7 24 24 7 Example 3 In the ABC, ADBC and BAD = If AC= 20 cm. CD = 16 cm and BD = 5cm then find sin + cos Sol. '' is in triangle ABD. We need to know AB and AD. Let us find them. A In the ADC, ADC = 900 AD2 + DC2 = AC2 2 2 20 cm 2 AD +16 = 20 AD2 = 202 162 = 400 256 =144 B AD = 144 = 12 cm Now, in ADB, ADB = 900 AD2 + BD2 = 122 + 52 = 144 + 25 = 169 AB = 169 = 13 cm sin BD AB sin cos 5 and cos 13 5 13 12 13 AD 12 AB 13 17 13 Example 4 : Prove that cos . cosec = cot Sol. Consider LHS of the given equation 1 cos . cosec = cos . sin cos = = cot = RHS sin cos . cosec = cot cosec 1 sin cos sin cot 5 cm D 16 cm C Trigonometry 301 Example 5: In right triangle ABC, right -angled at B, if tan A=1, A then verify that 2 sin A cos A =1. Sol. In ABC, ABC=900 tan A = BC =1 AB Since tan A = 1 (Given) BC AB BC = AB Now, AC2 = AB2 +BC2 AC = C B Let AB = BC = k, where k is a positive number. k 2 +k 2 = AB2 +BC2 AC = K 2 sin A = BC AB k 1 k 2 2 1 2 sin A cosA = 2 , cos A = 1 2 2 AB AC k 1 k 2 2 1 2 sin A cos A = 1 Example 6: ABCD is a rhombus whose diagonal AC makes an angle with the side CD, where cos = Sol. 2 . If PD = 4 cm then find the side and the diagonals of the rhombus. 3 We know, the diagonals of rhombus bisect each other at right angles. DP = PB and AP = PC In DPC, DPC = 90 PC cos = CD 0 D 2 3 C PC = 2k and CD = 3K P But, CD2 = PD2 +CP2 (3k)2 = 42+ (2k)2 9k2 = 16 + 4k2 16 5k = 16, k = 5 2 2 CD = 3k = 3 × 4 5 = k= 12 5 4 cm and PC = 2k = 2 × Each side of the rhombus = 16 5 cm. B 5 BD = 2PD = 2×4 = 8cm and AC = 2PC = 2 × diagonal AC = A 12 5 8 5 = 4 5 16 5 = 8 5 cm cm, diagonal BD = 8cm and cm 302 UNIT-12 EXERCISE 12.1 I. Find sin and cos for the following: 25 24 15 25 20 (i) 26 7 10 (ii) 24 (iii) II. Find the following : 1. If sin x = 3 , 5 cosec x = 2. If cos x = 24 , 25 sec x = 3. If tan x = 7 24 cot x = 4. If cosec x = 25 , 15 sin x = 5. If sin A = 3 4 and cos A = then, tan A = 5 5 6. If cot A = 8 15 and sin A = then, cos A = 15 17 III. Solve: 1. Given tan A = 3 , find the value of sin and cos 4 2. Given cot = 20 , determine cos and cosec 21 3. Given tan A = 7 , find the other trigonometric ratios of angle A. 24 4. If 2 sin = 3 , find cos , tan and cot + cosec . 5. If 3 tan =1, find sin , cos and cot . 6. If sec x = 2, then find sin x, tan x, cot x and cot x + cosec x. 7. If 4 sin A - 3 cos A = 0, find sin A, cos A, sec A and cosec A. Trigonometry 303 5 sin A 2cos A tan A 8. If 13 sin A = 5 and A is acute, find the value of 9. If cos = 5 tan 5 and is acute, find the value of 5 tan 13 10. If 13 cos 5 = 0, find sin sin 12 cot 12 cot cos 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°. But the standard angle we quite often construct and use are 30°, 45° and 60°. Now let us find the trigonometric ratios for these angles and also for 0°. (1) Trigonometric ratios of 45° Conside r an Iso sceles right angled triangle ABC, wh ere B 90 . If we take C 45 , then A is also 45°. A AB = BC ( they are sides opposite to equal angles) Suppose AB = BC = 1 unit AC2 = AB2 + BC2 (by Pythagoras theorem) 45° B C AC2 = 12 + 12 = 1 + 1 = 2 AC = 2 units Now consider the trigonometric ratios with respect to angle C. i.e., sin 45° = AB AC cosec 45° = AC AB 1 cos 45° = 2 2 1 2 sec 45° = BC AC AC BC 1 tan 45° = 2 2 1 2 cot 45° = C AB BC BC AB 45 1 1 1 1 1 1 (2) Trigonometric ratios of 30° and 60° A 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 30° and 60°. Consider an equilateral triangle ABC, Since, all the angles are equal we get Hence, 60° A B C 60 AB = BC = CA Let us draw a perpendicular AD from A to BC. ° 30 ABD ACD ( RHS postulate of congruency) B D C 304 UNIT-12 BD = DC BAD = since CAD 60 we get, BAC BAD = CAD = 300 A Now, observe that, ABD is a right angled triangle with ADB 90 , BAD 30 and ABD 30° 60 We need to know the lengths of the sides of the triangle ABC to find the trigonometric ratios. Let us consider, AB = BC = CA = 2 units In Since 2 2 Consider once again the equilateral triangle ABC. 3 60° B 1 D 2 ABD, AB = 2 units, BD = 1 unit, ADB C 1 90 , By pythagoras theorem, AB2 = AD2 + BD2 22 = AD2 + 12 AD2 = 4 – 1 = 3 AD = 3 Trigonometric ratios of 300 sin 30° = Opp Hyp cosec 30° = BD AB Hyp Opp 1 2 AB BD cos 30° = Adj Hyp AD AB 3 2 2 sec 30° = Hyp Adj AB AD 2 3 tan 30° = Opp Adj BD AD cot 30° = Adj Opp AD BD 1 3 3 Trigonometric ratios of 600 sin 60° = AD AB cosec 60° = AB AD 3 2 2 3 cos 60° = BD AB 1 2 sec 60° = AB BD 2 1 tan 60° = 2 cot 60° = AD BD BD AD 3 1 3 1 3 (3) Trigonometric ratios of 0° and 90° C Consider the right triangle ABC in which ABC 90 . Imagine what happens to the trigonometric ratios of the acute angle A, if it is made smaller and smaller till it becomes zero. Observe what happens to the length of AC and AB. A B Trigonometry 305 As, A gets smaller and smaller, the length of the side BC decreases. As, a point C gets closer to point B., AC coincides with AB. • When A is very close to 0°, length of BC gets very close to zero. sin A = • When BC AC very close to 0. A is 0°, length AC is nearly the same as AB. cos A = AB very close to 1. AC Let us define sin A and cos A when A 0 we define sin 0° = 0 and cos 0° = 1 Now consider the other trigonometric ratios. tan 0° = sin 0 cos 0 0 1 0 , cot 0° = 1 tan 0 1 this value is not defined 0 ( division is not defined by zero.) sec 0° = 1 cos 0 1 1 1 cosec 0° = 1 sin 0 1 this value is not defined. 0 Now imagine what happens to the trigonometric values of larger and larger till it becomes 90°. As, A gets larger and larger, C gets smaller and smaller and the length of side AB decreases. As the point A gets closer to point B and finally when, A is very close to 90°, C becomes very close to 0°. and AC becomes almost the same as BC. A when it is made 306 UNIT-12 • When A is very close to 90° AC is almost equal to BC sin A = BC AC very close to 1. • When A is very close to 90° length of AB reduces nearly to zero. cos A = AB very close to 0. AC we define sin 90° = 1 and cos 90° = 0 Now consider other trigonometric ratios. tan 90° = sin 90 cos 90 1 0 cot 90° = cos 90 sin 90 0 = 0 1 sec 90° = 1 cos 90 1 0 cosec 90° = 1 sin 90 this value is not defined, this value is not defined, 1 1 1 Now let us tabulate the trigonometric ratios of 0°, 30°, 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. 0 30 45 60 90 sin 0 1 2 1 3 2 1 cos 1 1 1 2 0 tan 0 3 2 1 cosec ND sec 1 cot ND 3 2 2 3 3 2 2 1 2 2 1 3 2 3 2 1 3 ND 1 ND 0 Trigonometry 307 Remember Here is an activity to easliy recall or prepare the table of values of sin for standard angles. Step 1: Write the number 0 to 4. 0 1 2 3 4 Step 2: Divide the numbers by 4. 0 4 1 4 2 4 3 4 4 4 Step 3: Take the square root 0 1 4 1 2 3 4 1 0 1 2 1 2 3 2 1 These are the values for sin 00 300 450 600 900 Step 4: Write the values in reverse order 00 These are the values for cos Step 5: We know tan sin cos 1 1 3 1 2 2 2 300 450 600 0 900 1 1 3 2 2 2 0 1 1 3 1 1 0 2 2 2 0 1 1 ND 3 3 Step 6: Take the reciprocals of the values of sin cos and tanfor standard angles; we get values of cosecsec and cot respectively ILLUSTRATIVE EXAMPLES Example 1 : Find the value of the following : (i) tan2 60° + 2 tan2 45° We have, tan 600 = 0 3 and tan 45 = 1 tan2 60° + 2 tan2 45º = 3 2 2.(1)2 =3+2=5 308 UNIT-12 (ii) Cosec 60° – sec 45° + cot 30° 2 We have cosec 60° = 3 , sec 45° = 2 and cot 30° = 3 2 Consider cosec 60° – sec 45° + cot 30° = 2 3 = 2 1 2 1 = 2 3 sin 2 .cos 2 tan 3 5 6 3 cos 3 cot 4 tan .sec 2 6 4 1 sin 45 2 2 3 = 1 – 3 = –2 cos 0.sin (iv) 2 2 1 2 6 3 (iii) sin2 4 + cos2 4 – tan2 3 1 2 3 = cos 45 tan 60 3 1 2 3 4 3 from the table cos 0 = 1 sin cos 2 6 = sin 90° = 1 sec 3 2 cot cos 30 3 2 1 1 By substitution, 3 sec 45 4 3 cot 60 2 tan 3 tan 60 3 1 3 2 . 1 3 2 2 1 1 3 4 3 1 3 2 = 3 2 3 4 3 3 8 Example 2 : If 2 cos = 1 and is acute angle, then what is equal to? Sol. Given 2 cos cos 1 2 1 coscos 2 Examples 3 : If 3 tanand is acute, find the values of sin 3 and cos2 1 1 Sol. Given 3 tan tan We know tan 30 0 = tan 3 3 We know, cos 2 and 3 sin 3sin 900 = 1 cos 2 = cos 600 = 1 2 Trigonometry 309 Example 4 : If A = 600, B = 300 then prove that cos (A + B) = cosA cosB - sinA sinB Sol. LHS = cos (A+B) = cos (600 + 300) = cos 900 = 0 RHS = cosA cosB- sinA sin B = cos 600 cos 300- sin 600 sin 300 = 1 2 3 2 3 2 1 2 = 3 4 3 =0 4 LHS = RHS EXERCISE 12.2 I. Answer the following questions: (1) What trigonometric ratios of angles from 00 to 900 are equal to 0? (2) Which trigonometric ratios of angles from 00 to 900 are equal to 1? (3) Which trigonometric ratios of angles from 00 to 900 are equal to 0.5? (4) Which trigonometric ratios of angles from 00 to 900 are not defined? (5) Which trigonometric ratios of angles from 00 to 900 are equal? II. Find if 0 900 (i) 2 cos (iv) 5 sin (ii) 3 tan = 1 (iii) 2 sin 3 (v) 3 tan 3 III. Find the value of the following: (i) sin300 cos600 - tan245 0 (ii) sin600 cos300 + cos600 sin300 (iii) cos600 cos300 - sin600 sin300 (iv) 2 sin2300 - 3 cos2300 + tan600 + 3 sin2900 (v) 4 sin2600 + 3 tan2 300 - 8 sin450 . cos450 cos450 sec30 0 + cosec300 (viii) sin300 + tan450 - cosec600 (vi) (vii) 4 sin2 60° - cos 2 45° tan2 30° + sin2 0° (ix) 5 cos2 600 + 4 sec2 300 - tan2 450 Sec300 +cos600 + cot450 (x) 5 sin2300 + cos2450 - 4 tan2 300 2 sin300 +cos300 + tan450 IV. Prove the following equalites (i) sin300 . cos600 + cos300. sin600 = sin900 (ii) 2 cos2300-1 = 1-2 sin2300 = cos600 (iii) If 300 , prove that 4 cos2 3 cos00 = cos 3 sin2 300 + cos2 300 310 UNIT-12 (iv) If = 1800 and A = 6 prove that ( 1- cosA) (1+ cosA) 1 (1-sinA) (1+sinA) 3 0 (v) If B = 15 , prove that 4 sin 2B.cos4B.sin6B = 1 (vi) If A= 600 and B = 300, then prove that tan (A - B) = tan A - tan B 1 + tan A tan B Trigonometric identities: Recall that in algebra, you have learnt about identities, For example, (a + b)2 = a2 + 2ab + b2 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. Consider a right angled triangle ABC, in which B A 90 By Pythagoras theorem, we get AB2 + BC2 = AC2 Dividing each term of the equation by AC2, we get AB2 AC 2 BC 2 AC2 = AC 2 AC2 2 AB 2 AC BC AC 2 = AC AC 2 B C 2 (cos A) + (sin A) = 1 sin2 A + cos2 A = 1 ............ (i) This equation is true for all values of angle A, such that 0 A 90 This equation (i) is a fundamental trigonometric identity. For convenience (sin A)2 is written as sin2A and read as sine squared A. Now let us obtain two more trigonometric identities from the fundamental identity. sin2 A + cos2 A = 1 by dividing by sin2 A we get, sin2 A sin2 A cos 2 A sin2 A 1 sin2 A 1 + (cot A)2 = (cosec A)2 1 cot2 A sin A sin A 2 cos A sin A 2 1 sin A 2 cosec2 A ...............(ii) Trigonometric identity (ii) is true for all values of angle A. Such that 0° < A 90°, for A 0 , cot A and cosec A are not defined. Trigonometry 311 By dividing (1) by cos2 A we get, sin2 A cos 2 A cos 2 A cos 2 A 1 cos 2 A sin2 A cos 2 A cos 2 A cos 2 A 1 cos 2 A (tan A)2 + 1 = (sec A)2 tan2 A 1 sec 2 A ......(iii) Trigonometric identity (iii) is true for all values of angle A, such that 0 For A A 90 90 , tan A and sec A are not defined. Trigonometric identities (i), (ii) and (iii) are called fundamental relations. (1) sin2 A + cos2A = 1 (2) tan2 A + 1 = sec2 A (3) 1 + cot2 A = cosec2 A These identities can also be rewritten as follows: sin2 A + cos2A = 1 sin2 A = 1 – cos2 A; tan2 A + 1 = sec2 A tan2 A = sec2 A – 1; 1 + cot2 A = cosec2 A cot2 A = cosec2 A -1; or cos2 A = 1 – sin2A or sec2 A – tan2 A = 1 or cosec2 A – cot2 A = 1 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. ILLUSTRATIVE EXAMPLES Example 1: Prove that coscosec = cot Sol: LHS = cos cosec = cos 1 sin cosec 1 sin cos = cot = RHS sin Alternatively, we know that in RHS. cot cos sin we have cos in LHS, but Hence cos 1 sin 1 sin is not there instead we have cosec is written as cosec cosec = cos 1 =cot sin . 312 UNIT-12 Example 2 : Show that tan A. sin A + cos A = sec A Sol: Consider LHS = tan A sin A + cos A sin A sin A cos A cos A sin2 A sin2 A cos 2 A cos A = = cos A cos A tan A = = 1 cos A sin A ) cos A (But sin2A+cos2 A=1) = sec A = RHS Example 3 : Find the value of (sin + cos )2 + (sin – cos )2 Sol: (sin + cos)2 + (sin - cos)2 =sin2 + cos2 + 2 sin cos + sin2 + cos2 – 2 sin cos = 2[sin2 + cos2] = 2(1) = 2 ( sin2 + cos2 =1) Example 4: Prove that cos2 = 1 1 tan 2 Method - 1 Sol: LHS = cos2 = 1 sec 2 1 1 tan2 = RHS cos 2 = LHS Method - 2 RHS= 1 1 tan2 1 sec 2 Example 5 : Show that (1 – sin2A) (1 + tan2A) = 1 Sol: Consider LHS = (1 – sin2A) (1 + tan2 A) = cos2 A sec2 A [ 1 – sin2 A = cos2 A, 1 + tan2 A = sec2 A] 1 = 1 = RHS cos 2 A sin cos 1 Example 6: Prove that cosec sec sin cos sin cos 1 1 Sol: ConsiderLHS = = = sin2 + cos2 = 1 = RHS cosec sec sin cos = cos2A Example 7 : If 2 sin2 + 5 cos = 4. Show that cos = Sol: 2 sin2 + 5 cos = 4 2[1 – cos2] + 5 cos = 4 2 – 2cos2 + 5 cos = 4 2 – 2 cos2 + 5 cos – 4 = 0 –2 cos2 + 5 cos – 2 = 0 2 cos2 – 5 cos + 2 = 0 1 2 Trigonometry 313 This is quadratic expression in the form ax2 + bx + c = 0 2 cos2 – 4 cos – cos + 2 = 0 2 cos [cos – 2] – 1 [cos – 2] = 0 [2 cos – 1] [cos – 2] = 0 2 cos – 1 = 0 or cos – 2 = 0 cos = 1 or cos = 2 2 cos = Discuss: cos= 2 is not considered. Why? 1 2 EXERCISE 12.3 I. Show that 1. (1-sin2 ) sec2 = 1 2. (1+tan2 ) cos2 = 1 3. (1+tan2 )(1-sin (1+sin = 1 sin 4. 1 cos 5. 1 sin 1 sin 7. cos A 1 tan A tan )2 (sec 1 cos sin 2 cosec 6. (1+cos A)(1-cos A(1+cot2 A= 1 sin A = sin A + cos A 1 cot A 9. (sin + cos )2 = 1 + 2 sin cos 8. 10. 1 tan2 A 1 tan2 A 1 cos 1 cos 1 2sin2 A 1 cos 1 cos 4 cot cosec 11. sin A cos A tan A + cos A sin A cot A = 1 12. cos A 1 sin A 1 sin A cos A 2 sec A 13. 14. tan2 A - sin2 A = tan2A sin2 A tan A sin A sin2 A tan A 1 cos A 15. cos2 A - sin2 A = 2 cos2A - 1 16. If x = r sin A cos B, y = r sin A sin B, z = r cos A, then x2+y2+z2=r2. 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? Consider the angles in ABC in which ABC Since A and B 90 , A ABC C 90 . Identify the pair of complementary 90 C are complementary angles. In any right angled triangle, the two acute angles other than right angle always form a pair of complementary angles. 314 UNIT-12 Now let us define trigonometric ratios for these complementary A angles. A Since A C 90 A 90 C or C 90 A [For convenience sake, these complementary angles are written as (90° – C) or (90° – A)]. 90º–A B C The trigonometric ratios for angle A and (90° – A) are written in the table given below observe them, T-ratios for angle A T-ratios for angle (90° – A) BC sin A = AC AB cos A = AC BC tan A = AB AB cot A = BC AC sec A = AB AC cosec A = BC AB AC BC cos (90° – A) = AC AB tan (90° – A) = BC BC cot (90° – A) = AB AC sec (90° – A) = BC AC cosec (90° – A) = AB sin (90° – A) = Compare the ratios of angle A and its complementary angle. We observe that. AB AC BC cos (90° – A) = sin A = AC AB tan (90° – A) = cot A = BC We can conclude that, AC AB AC sec (90° – A) = cosec A = BC BC cot (90° – A) = tan A = AB sin (90° – A) = cos A = cosec (90° – A) = sec A = sin (90° – A) = cos A cosec (90° – A) = sec A cos (90° – A) = sin A sec (90° – A) = cose A tan (90° – A) = cot A cot (90° – A) = tan A ILLUSTRATIVE EXAMPLES Example 1 : Evaluate sin 650 cos 250 Sol. We know, cos 250 = sin (900-250) = sin 650 cos 250 sin 650 = =1 cos 250 cos 250 Trigonometry 315 Example 2 : If tan 2A = cot (A 180), where 2A is an acute angle, find the value of A. Sol. Since tan 2 A = cot (900 2A) we get cot (900 2A) = cot (A 180) 900 2A = A 180 3A = 1080 A = 1080 = 360 3 A = 360 Example 3 : Prove that : [cosec (90º – ) – sin (90º – )] [cosec – sin ] [tan + cot ] = 1 Sol. LHS = [cosec (90º – ) – sin (90º – )] [cosec – sin ] [tan + cot ] = [sec – cos ] [cosec – sin ] [tan + cot ] = 1 cos = 1 cos 2 cos = sin2 cos 1 sin cos 1 sin2 sin cos 2 sin Example 4 : Prove that sin cos sin cos sin 1 cos 1 cos sin sin(90º 1 sin = 1 = RHS ) 1 cos cos(90º sin(90º 1 sin ) = cos 1 sin cos 1 sin = cos (1 sin ) cos (1 sin ) cos = 1 sin2 Sol. LHS = = 2cos cos 2 cos 1 cos(90º sin 2 cos ) = 2 sec ) cos sin cos 1 sin2 cos sin 2sec = 2 sec = RHS EXERCISE 12.4 1. Evaluate : (i) tan 65 0 cot 25 0 (ii) 0 (iv) cosec 31 sec 59 0 sin180 cos 720 0 (iii) cos 480 sin 420 0 (v) cot 34 tan 56 (vii) sec 700 sin 200 cos 700 cosec 200 sin 360 (vi) cos 540 sin 540 cos 360 (viii) cos2 130 sin2 770 316 UNIT-12 2. Prove that (i) sin 350 sin 550 - cos 350 cos 550 = 0 (ii) tan 100 tan 150 tan 750 tan 800 = 1 (iii) cos 380 cos 520 - sin 380 sin 520 = 0 3. If sin 5 = cos 4, where 5 and 4 are acute angles, find the value of 4. If sec 4A = cosec (A-200), where 4A is an acute angle, find the value of A. 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. Trigonometry 317 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. But, 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. Angle 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. Angle of our heads are measured from horizontal downwards. Note: * Angle 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. 318 UNIT-12 Know this: Surveyors have used trigonometry for centuries. One such large surverying project of the nineteenth century was the Great Trigonometric Survey’ of British India for which the two largest-ever theodolites were built. During the survey in 1852, the highest mountain in the world was discoverd. From a distance of over 160 km, the peak was observed from six diffrerent stations. In 1856, 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. ILLUSTRATIVE EXAMPLES Example 1: A 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 30°. Find the height of the tower. Sol : Draw a figure to represent the given information. Let AB be the height (h) of the tower, and BC is the distance between the tower and the point of the viewer. In ABC, ACB = 300 tan 30° = 1 h 3 50 tan h 50 AB BC h = 50 m 3 Example 2: 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 P PQR, PQR = 900 tan = PQ QR 50 50 50m In 1 ( tan 45° = 1) tan = 1 = tan 450 tan = tan 45 0 = 45° Q 50m R the angle of elevation is 45° Example 3: From the top of a building 50 3 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. Trigonometry 319 P M = 60° 50 3 60° Q ? R Sol: Let PQ represent the height of the building, PQ = 50 3 m, QR be the distance between the building and the car. Angle of depression is 600 since, PM ||QR, so In MPR 60 PQR, tan 60° = 3 = PQR PQ QR 50 3 QR PRQ ( alternate angles) MPR PRQ 60 90 , PRQ 60 50 3 QR QR = 50 3 3 = 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. A 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. A C B D P Let AB be a tower of height = 50 m CD be another tower of height = 40 m The observe is at 'P' and angle of elevation BD = BP + PD. APB CPD 45 Distance between 320 UNIT-12 In ABP, tan = AB BP 50 BP BP = 50 m ( tan 45° = 1) 1= In CPD, tan = tan 45° = 1= 50 BP tan 45° = CD PD 40 PD 40 PD ( tan 45° = 1) PD = 40m BD = BP + PD = 50 + 40 = 90 m The distance between the windmills on either side of field is 90 m EXERCISE 12.5 I. Find the value of 'x' A R 90 x 1. 2. 45° B 60m K 3. Q x D X 100 x 75 4. L II. 60° P C 30° 100 M 5. Y 45° x Z F x 75 E 1. A tall building casts a shadow of 300 m long when the sun's altitude (elevation) is 30°. Find the height of the tower. 2. From the top of a building 50 3 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. 3. A 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 20 m from its base, how tall was the tree? Trigonometry 321 4. The angle of elevation of the top of a flagpost from a point on a horizontal ground is found to be 30°. On walking 6 m towards the post, the elevation increased by 15°. Find the height of the flagpost. 5. Two observers who are 1 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 24 m, find the height of the cliff. 7. From the top of a building 16 m high, the angular elevation of the top of a hill is 60° and the angular depression of the foot of the hill is 30°. Find the height of the hill. 8. Find the angle of depression if an observer 150 cm tall looks at the tip of his shadow which is 150 3 cm from his foot. 9. From a point 50 m above the ground the angle of elevation of a cloud is 30° and the angle of depression of its reflection in water is 60°. Find the height of the cloud above the ground. 10. From a light house the angles of depression of two ships on opposite sides of the light house were observed to be 450 and 600. If the hight of the light house is 120m 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 Trigonometric identities sec Reciprocals of trigonometric ratios Relation between trigonometric ratios Trigonometric ratios of standard angles cot Finding heights and distances sin 2 A+cos2 A=1 angle of elevation 1+cot2 A=cosec2 A angle of depression tan2 A+1=sec2 A 322 UNIT-12 ANSWERS Exercise 12.1 I. (i) II. (i) III. 1] 24 7 , 25 25 5 3 1 3 , ,3 10 10 10] 24 7 (iii) 3] 3 , 2 6] 5 12 , 13 13 (iii) 20 29 , 29 21 2] 17 7 3 4 , 5 5 25 24 (ii) 3 4 , 5 5 5] 9] (ii) (iv) 3 5 3 4 (v) 8 17 (vi) 7 24 7 25 25 24 , , , , , 25 25 24 7 24 7 3, 1 , 3 3 7] 4] 1 , , 2 3 3 3 4 5 5 , , , 5 5 4 3 8] 12 65 17 7 Exercise 12.2 II. (i) 45° III. (i) (ii) 30° 3 4 (vii) (ii) 1 15 2 (iii) 60° (iv) 0º (iii) 0 (viii) 3 3 4 3 3 4 (iv) (ix) 5 4 (v) 30° 3 3 (vi) 2 2 1 (v) 0 3 5 67 12 (x) 6 4 3 Exercise 12.4 1] (i) 1 (ii) 1 3] = 10° (iii) 0 (iv) 0 (v) 0 (vi) 0 (vii) 0 (viii) 0 4] A = 22° Exercise 12.5 I (i) 60m (ii) 30 3 units (iii) 2] 50 3 m 3] 20 II 1] 100 3 m 6] 24 24 3 1 m 7] 64m 100 units 3 3 8] 30º 2 m (iv) 100 units 2 4] 6 m 3 1 9] 100m (v) (5) 10] 120 1 4 units 3 1 3 km 1 m 3

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