Some Applications of Trigonometry Fastracl« Revision > Line oF Sight: IF an observer is viewing an object, the straight line joining the eye of the observer to that object is called line of sight. Object » Angle of Elevation: The angle of elevation of an object viewed is the angle Formed by the line of sight with the horizontal when < it is above the horizontal elevation level, ie, the case when we Ao raise our head to look at the observer object: > Angle of Depression: The angle of depression of an object viewed is the angle Formed by the line of sight with the horizontal when it is below the horizontal level, ze, the case when we lower our head to look at the object. Gh, & Angle of Horizontallevel Horizontal level ‘Angleot@ dapiassion Obseever ‘Object » The height or length of an object or the distance between two distant objects can be determined through trigonometric ratios. nowledge BOOSTER 1. The angles of elevotian and depression ore always cute angtes. 2. ifthe angle ofelevation of the toner (or Sun) decreases, the shadow of the tower (or Sun) increases. 3. f the observer moves towards (or moves away) the perpendicular line, the angle of elevation increases (or decreases). 4. Ifthe height of tower is doubled and the distance between the observer and the foot of tower is also “doubled, then the angle of elevation remains some. gy Practice Exercise -@ Multiple choice Questions w Q1. Ifa pole 6 m high casts a shadow 2/3 m long on the ground, then Sun's elevation is: INCERT EXENIPLAR; CBSE 2023; CBSE SOP 2023-24) ase bas 30" From a point P on a level ground, the angle of elevation of the top of tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower aWgm bI56m cam 200m The angle of elevation of the top of the tower is 60° and the horizontal distance from the observer's eye to the foot of the tower is 100 m, then the height of the tower will be: Qe. Qs. 100 2. 50/3m “oe < 100Y3 m 4. 60¥3m Q4. The ratio of the length of a rod and its shadow is 23, then the angle of elevation of the Sun is: 24s" b. a0 cer 6 0 QS. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is: 2am b&m CBM. 2m A boy standing on top of a tower of height 20 m observes that the angle of depression of a car on the road is 60°, The distance between the foot of ‘the tower and the car must be: [Usey3 = 1.73] 2.1045m b.1S4m 255m d.1250m Q6. Q7. Ifthe angle of depression of an object from a 50m high tower is 30°, then the distance of the object from the tower is: 2. 253m b. 3" © 50V3m 450m 8. The angle of depression of a car parked on the road from the top of 120 m high tower is 30°. The distance of the car from the tower (in metres) is a. 203m. b.20m < 40V3m d. None of theseQ9. quo. Qu. giz Qs. A vertical straight tree of 15 m high, is broken by the wind in such a way that its top just touches the {ground and makes an angle of 60° with the ground. ‘At what height from the ground did the tree break? [Use 3 = 1.73] a6sm b 36m 79m ‘An observer 1.6 m tall is 20 m away from a tower. The angle of elevation from his eye to the top of the tower is 45°. The height of the tower is: a. 216m b.2m c72m d. None of these ‘An observer from the top of a 100 m high lighthouse from the sea level observed that the angles of depression of two ships are 30° and 45°. if one ship Is exactly behind the other on the same side of the lighthouse, then the distance between two ships is: a. 1003 +) b. 50(¥3 +1) m © s0(/3-1) m 4. 100 (3-1) m When the Sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m.What Is the height of the tower? a35m bm ¢ 35¥3m d. 2/3. m From the top of a cliff 30 m high, the angle of elevation of the top of a tower from cliff top is found to be equal to the angle of depression of the foot of the tower. The height of the tower is: az0m b60m c20m 450m 59m Assertion & Reason type Questions y Directions (Q. Nos. 14-17): In the following questions, @ statement of Assertion (A) is followed by a statement of a Reason (R). Choose the correct option: Qu. Qus. ‘a. Both Assertion (A) and Reason (R) are true and Reason (R) Is the correct explanation of Assertion (A) b, Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) C. Assertion (A) is true but Reason (A) is false 4d. Assertion (A) is false but Reason (R) Is true Assertion (A): If the length of shadow of a vertical pole is equal to its height, then the angle of elevation of the Sun is 45°. Reason (R): Trigonometric ratio, tangent is defined as tang ~ Perpendicular Base Assertion (A): The angle of elevation of the top of a tower is 60°. Ifthe height of the tower and its base is tripled then angle of elevation of its top will also be tripled. Reason (R): Inan equilateral triangle of side 3/3 cm, the length of the altitude is 4.5 cm. Que. Qu. Que. Qs. Q20. ga. Qzz. Q23. Q24. Q2s. Q26. a7. Assertion (A): Suppose a bird was sitting on a tree. ‘person was sitting ona ground and saw the bird, which makes an angle such that tan 0 The 5 distance from bird to the person is 13 units. Reason (R): In a right-angled triangle, (Hypotenuse)? = (Side)? + (Base)? Assertion (A): The angle of elevation of the top of the tower is 30° and the horizontal distance from the observer's eye to the foot of the tower is 50 m, then the height of the tower will be 2s m. Reason (R): While using the concept of angle of elevation/depression, triangle should be a right angled triangle. ® Fill in the Blanks type Questions y The.. of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. The angle of of an object viewed is the angle formed by the line of sight with the horizontal when it is below the horizontal level. I the length of the shadow of a tower is-/5 times its height, the angle of elevation of the Sun is INCERT EXEMPLAR} From a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°, then the height of the tower is [Use V3 = 1.73] In the given figure, the angles of depressions from the observing positions 0; and 0, respectively of the y True/False type Questions y ‘The height of the tower if Length of the shadow is 10 m and Sun's altitude is 45°, is 10m. Hf length of shadow of tower is 20m and angle of elevation is 60°, then the height of tower is 3 m. 5 Alittle boy is flying a kite. The string of kite makes ‘an angle of 30° with the ground. Ifthe height of the kite is 21m, then the length of the string is 35 m. The angle of elevation of the top of a tower is 30°. I the height of the tower is doubled then angle of elevation ofits top will also be doubled. INCERT EXERCISE] A person walking 20 m towards a chimney in a horizontal line through its base observer that its angle of elevation changes from 30° to 45°, then 20 eal" height of chimney is (Solutt ions 1. (@) Let PQ = 6 m be the height of 6 the pole and RQ=2V3 m be its Polo shadow, 6m Let angle of elevation of the Sun A do bea. Sheae In right-angled aPQR. PQ tend = Fe SRB Seno = 0=60" 2. (Q Let QR = 100 m be the R height of the tower and point P makes an angle of 100 m elevation of the top of the tower Le. ZQPR = 30° a a In right-angled aPOR. 22 RO tan30"= 55 1 _ 100 = NB PQ 2 PQ=100/3m 100x173 m= 173m 3. (| Let BC = h metre be the height of the tower and distance from observer to the foot of the tower be AB= 100m. ¢ Tara Se AC ye , BC enea=S nate A Bo%05 hare = h=100¥3m 4, (0) Let Cbe the postion of the Sun. Let BC and AB be the length af rad and length of the shadow. Length of rod___1._ 8C Length of shadow fy AB 0 (Sun) Given, K B In right-angled aABC, ac tana=50 8 (from eq. (1) 5. (d) Let BC be the height of the wall and AC=m be the length of the ladder leaning against a wall. c wy im Iwan! A | tem Ladder AC makes an angle of elevation of 60° to the wall le. ZCAB = 60° Let AB = 46 m be the foot of the ladder from the wall. In right-angled aABC. AB, gor= 48 cos60"= = 5 146 2 > [292m 6. (b) Let BC= 20 mbe the height of the tower: Let A be the position of car and C be the position of boy. At point C, boy makes an angle of depression of 60° le. ZHCA = 60" He © (Boy) Toner l20m act 8 (can Here, ZBAC = ZHCA= 60" In right-angled AABC. (alternate angles) wearin AB = 6.67 x 173=1154m 7. {¢) Let Abe the position of an object and BC = 50.m be the height of tower, Then ZXCA = ZCAB = 30° (alternate angles) In right-angled AABC ad tango" =F 120 = Es = AB=50V3m_8. (2) Inright-angled AABC. van30° = 120 > Wee = AB = 12003 m 9 (a) Let BC =15 mbe the height of the tree. Let at point 0, tree breaks and touches the top point C of tree to the ground at point A TRICK The broken part CD of tree is equal to the stope line AD, ie, CD=AD. Let BD = xm be the height of broken tree. Then CO = AD =15-x Given, broken part of tree CD makes an angle of 60" with the ground fe. 2DAB = 60". In right-angled sABD. sinor= $2 Bx - 2 =x n 153 - Vx -2« ss x(2+¥3) = 153 eo INE 208 ~ 2B 2-8 3-45 © QP=(3P (-(0+ 6) (o-b)o ab?) =519-45=69m 10. (a) Let AE = 16 m be the 0, height of an observer and | 8D = hm be the height of i the tower. Let EC = AB = 20 m be the i distance from observer to , 4 & the toner. * In right-angled a€co, cera co tangs - 2 and: = (AB = EC = 20 m and AE = BC = 1.6 m) ete co=80-8¢) > 7 h-16 [cCOMMONJERRO)R Sometimes students make an error of taking an angle Jrom point A instead of taking at point E. So, continuous practice is required to make stronger concept. Tl (d) Let CD = 100 m be the height of the lighthouse. Let D be the position of observer. and A and B be the position af two ships. The angles af depression fram point D to the ships A and B are ADA = 30° and ZEDB = 45° Here, ZCAD = Z€DA = 30° and ZC8D = ZEDB= 45" (by alternate angles) Let AB = xm and BC = ym. Then in right-angled ABCD, co a5e= tanag= . = q— 0, - y AWA = yeloom Akme ym And in right-angled SACD. © tanz0°-2 an g0°= 2 = ee acne 8C= x+y) = Em + +) > x+y=100V3 = x+100= 100V3 (> y=100 m) > x=100(V3—1)m 12, (Q Let D be the position of e-~ the Sun and DC» hm be the height of the tower. Given that hs “EDA =30" and 208 = 60" Here, ZCADs ZE0A x c 230" and CBD = ZE08= 60" (alternate angles) Length of the shadow decrease, AB = 70 m In right-angled AAC, a) tan30"= 2 i GE Ae+ac 1 A Bo yore om And in right-angled ABCD, o tangor= 2 an 60" = = ind a BO 3 BC ale From eq, (I), we getB Ln ~ Wa 7003+h = T0Y3+h=3h > 70V3=2h = n= 353m (b) Let C be the position E of top of the cliff, then AC = 30 m. Let 8E = hm be the height ofthe tower. i RX h elevation from paint Cto | 3 the point €. Then t wt Z0LE =0 Also given, angle of depression ZDCB =. Here, ZABC=2DCB=0 —(byalternate angle) In right-angled aBAC. aC tana = AC 5. tande2® 5 ee. a x ‘and nace. 20-90" Then tand=£2 (Let AB = CD = xm) fat eD=BE-BD > tang =F =(h-30)m 30 = 2) tand 4 From eqs. () and (2), we get h-30__30 ‘and tang = P= 60m (2) Assertion (A): Let 8C = hm be the height of the pole and AB = I mbe the length of the shadow. Let the Sun makes an angle 0 from point A Given that, h=l (sun) am o B In right-angled triangle ABC p {+ sos-8) = (-h= (given) = tanBalatands* = 0-45" So, Assertion (A) is true. Reason (R): tis also true that tang = Perpendicular Base Hence, both Assertion (A) and Reason (R) are trueand Reason (R) isthe correct explanation of Assertion (A) 16. Altitude of an equilateral triangle divides the base into two equal parts, (c) Assertion (A): Let BC = h units be the height of tower and AB=b Units be the base of the tower. © Pro = BC ‘Then tan60°= pat > tan60°= 5 4) IF we tripled the height and base of tower Le. BC = 3h and AB = 3b, then angle will be ot 3h h tano= FF oS > tandee = tan 0 = tan 60° (from eq, (0) > 0-60", which is not tripled the original angle. So, Assertion (A) is false Reason (R): Let ABC be an equilateral triangle. Then AB=BC=CA =3V3em ‘ SIX “ke” s Let hbe the altitude of an equilateral triangle. TiP In right-angled AADB. use Pythagoras theorem, 2 (er 32] = 27-22 - 8! -3- [oF -@ 2-asen So. Reason (R) Is true. Hence, Assertion (A) Is false but Reason (A) is true. AD = \|(A8)? -(80)? (2) Assertion (A): Given tan 2 _ ac ° = tano- 2 80 nos AB Let BC = 12k and AB = 5k ie where kis a constant. In rightangled ABC, use ,A a Pythagoras theorem, 5 AC = (AB)? +(6c)? = V(5K)? + (12k)? eV 25K? + 14442 = 169K? =13k=13 units (Consider k= 1) So, Assertion (A) is true.18. 1. Let BC. him be the height of the tower and C be the a. Reason (R): itis a true relation that Ina right-angled triangle, (Hypotenuse)? = (Side)? + (Base)? So. Reason (R) is true Hence, bath Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (a) Assertion (A): Let A be the position of observer eye and BC= fim be the height of the tower. . | oe P Let AB » 50 m be distance between observers eye and foat ofthe tower. In right-ongled ABC. ac 7B tana0° Lik = Eu 50,V3 _ 50 > beg pam ‘So, Assertion (A) is true. Reason (R): tis true to say that while solving the problem of angle of elevatian/depression, triangle should be a right-angled triangle. Hence. both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A), line depression position of Sun. The length of the (Sun) shadow will be AB= V3h m. eee Let the elevation of Sun from f point A ls ZCAB =6. A i Inright-angled SABC, ac ( tano= ro tan hoa BB = tan 30° = 0-30" Let 6C = hm be the height ¢ of the tower. Let A be the foot ofthe point such that ie AB = 20m Also angle of elevation is 4 ZBAC = 30". A 20m 8 22. 23. 24. In right-angted triangle. =1153m Hence, height of the tower Is 1153 m. In right-angled aACO). °. E- ZOAC + ZACO) + ZCO,A = 180° (by angle sum property) = ZO,AC + 90° + 60" » 180" = -ZO)AC = 180° ~ 150° = 30" Here, ZED)A = ZO,AC = 30° (alternate angles) Also. ££0,A = 20,A8 = 45° (alternate angles) Hence, angles of depressions from paints 0, and 0, are respectively 30° and 45" Let C be the position af Sun and BC = hm be the height of the tower. Let AB = 10 m be the length of the Sun. © (Sun) Tromer hn dl on 8 In right-angled ABC. ec nage = BC ten age Fe a 0 > h=10m Hence. given statement is true Let BC = hmetre be the height £ of the tower and AB = 20 m be the length of shadow of tower, Inright-angled ABC. BC AB hm (tower) aA dp sansa = Bee 0 = ne2%m loneghen exasmantioe25. Let C be the position of the kite and A be the position of the boy, Let AC= Lm be the length of the string and BC = 21m be the height of the kite. Then string of kite makes an angle ACAB = 30 In right-angled aABC. BC AC © (ete) Im, 4 Jat m acs le [m4 sin30"= 12 27 = (= 21x2=42m Hence, given statement Is false, 26. Let initially height of tower ¢ be h mand AB =x". In right angled AABC. P h A x x x If the height of tower is doubled Le. BC = 2h. then tang = 22 x tan30" Here we see that angle is not doubled when height Is, doubled Hence, given statement is false. 27. Let CD = hm be the height of the chimney and ‘AB = 20 mbe the distance covered from A to B. Given, ZDAC = 30” and Z0BC = 45" In right-angled aACD, co © tan30° ne" AC” AB BC Lh = or D yom (:AC= AB + BC =20 + BC) ‘And right-angled ABCD, © tanasr=- 2 . h hm aC > ac-h nA From eq. (1). we get Sale Lt. ZB 2h > 20+h= 3h = (/3-1)=20 = te za Jm a1 Hence. given statement Is true. ~@ Case Study Based Questions y Case Study 1 A group of students of class—X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name is Delhi Memorial, originally called All-India War Memorial, Monumental Sandstone Arch in New Delhi. is dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 metres) in height. a Based on the above information, solve the following questions: ‘What is the angle of elevation, if they are standing ata distance of 42/3 maway from the monument? a0 bare 45th 60" They want to see the tower (monument) at an angle of 60°. So, they want to know the distance where they should stand and hence find the distance. (UseV3 = 1.732] a2424m b 212M ca2Zm — 4.2564m If the altitude of the Sun is at 30°, then the helght, of the vertical tower that will cast a shadow of length 30 mis: an Qa Q3. b Om B «2m 4. 20Y3 m B The ratio of the length of a rod and its shadow is 24:83. The angle of elevation of the Sun is: azo bho 45". 80" ‘The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level, is: a. angle of elevation € corresponding angle a. 13m a4. Qs. b. angle of depression complete angle Solutions 1. Let the angle of elevation be 0. Given that, Height of the monument (BC) = 42m and AB=42V3 mNow. in right-angled AABC, i 42m 0=30° 1 Students toke the value of BB 1an 60° in precocity. 1 uti 1s wrong. Te correct volue of = tan 30° @ Lette required astance be xm, Given, angle of elevation (8) = 60° and height of the monument (BC) = 42m. TiP Memorize the vole of trigonometric angles property ond do practice more. Now. in right-angled AABC, tant = x = BB 3 14 «1732 24.24 So. option (a) is correct 3. Let the height of the vertical tower be hm. Given angle of t elevation (¢) = 30° hm and length of the _¥< | shadow (AB)= 30m aS 55,58 Now. In right-angled ABC. ac tano=E B30 nad 7 BE Some students confused the values of tan 30° and tan 60°. They take wrong value in haste. 4. Given, the ratio of the length of a rod and its shadow is 24:0y3. Let AC= 24k A Theo and 8C= 83k where kisa positive oa integer. Now. let angle of elevation of the Sun, “V iso. a eaK In right-angled AACB: So, option (b) is correct. 5. The angle formed by the Une of sight with the horizontal. when the object viewed Is above the horizontal level. is angle of elevation So. optian (a) is correct. Case Study 2 A satellite flying at height A is watching the top of the two tallest mountains in Sikkim and Karnataka, which are in Kangchenjunga (eight 8586 m) and Mullayanagiri (height 1930 m). The angles of depression from the satellite, to the top of Kangchenjunga and Mullayanagiri are 30° and 60" respectively. If the distance between middle of the bottom of both mountains is 2046 km, and the satellite is vertically above the mid-point of the distance between the two mountains. 8 Do Ss Kangchanjunga Mulleyanagi Based on the above information, solve the following questions Ql. The distance of the satellite from the top of Kangchenjunga is: a, 1181.22 km. b. 57752 km ©. 1937 km d. 1025.36 km Q2. The distance of the satellite from the top of Mullayanagir a 139.4 km, b. 57752 km © 2046 km d. 102536 kmQ3. a4. Qs. The vertical distance of the satellite from the ground is: 2.11384 km b. 599.2 km 1937 km 4. 1025.36 km What is the angle of elevation of the top of Kangchenjunga mountain, if a man is standing at a distance of 8586 m from Kangchenjunga? 230 base fa stone very far away makes 45° to the top of Mullayanagiri mountain. So, find the distance of this stone from the mountain. a. 118327m b.566.976m ©1930m 4.102536 m, Solutions Let the distance of the satellite from the top of Kangchenjunga is x km. : x ‘30 t se nm Ko J Aa 1023 km—oS Given that. height of the Kangchenjunga. AD = 586m and the distance between middle of the bottom of both mountains (DS) = 2046 km. But the satelite Is vertically above the mid-point of the distance between the two mountains Ds: Now, in right-angled aAGF, _AG > 1023 cos0=AS 5 cos30 = B_1023 2 x sf ee aa 9682x1732 x=118122 km So. option (a) Is correct. i Let the distance of the satellite from the top of Mullayanagiriis yk Given that. Height of the Mullayanagiri 5 =1930 mand the distance Ly » between the middle of the bottom WS of both mountains (05) = 2046 km But the satellite is vertically above the mid-point of the distance between the two mountains Ds_2046 si=PH= 22-2086 0234, Zinn Now, in right-angled APHF, = cosgor= 1028 y 4_yo23 - ay y= 2 1023 » 2046 km So, option (c) is correct. Inright-angled AAGF. tanag? = £6 2G -1023%1732 _cog.¢4m 3 and GI= AD = 8586 m 586 km (2 Thm = 1000 m) The distance of the satellite from the ground (F)=FG +61 = 5906 +8586 = 599.2 km So, option (b) is correct. Let the angle of elevation be 6 Given that. height of the Kangchenjunga (AD) = 8586 m and distance (CD) = BS86 m. Now inright-angled AADC, | tand = tan 45° = 0-45" So, option (b) Is correct. Ds— S56m —e Let the distance of the stone fram the mountain be xm P Given. angle of elevation 1 of a stone from the top of Mullayanagiri 2 nas" 8 ‘and helght of the (P5) = 1930 mNowy in right-angled aQsP. x=1930m So. option (¢) is correct. Case Study 3 Qu Q2. Radio towers are used for transmitting a range of communication services including radio and television, The tower will either act as an antenna itself or support one or more antennas on its structure, On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O. Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation Of the top of the Section B is 30° and the angle of elevation of the top of Section A is 45°. Based on the above information, solve the following questions: [case 2023) Find the length of the wire from the point O to the top of Section B. Find the distance AB. Or Find the area of AOPB. Q3. Find the height of the Section A from the base of the tower. Solutions 1. Let the Length of the wire from the point O to the top of section 8, ie. OB » im. Given, OP =36 cm and Z80P=30° Now in right-angled ABPO. OP B36 cm cos30°=- SF 5 Oe q 2,3 TA Lb * Es = 24/3 P 36cm 0 So. required length is 24/3 em. 2 Let AB=xem Given, ZAOP = 45° and OP = 36 cm Now in right-angled AAPO, age = AP. tan 45s AP. = “= 8 = AP=36cm. ‘Again in right- angled tho tanaa* -z "5 2 B 36 3 _ 363 => oe 288 rade “BB AB=AP—BP. = 36 - 1293 = 12(3-Y3)cm So. required distance AB is 12(3-4/3)cm. or Since, ABPO is a right-angled triangle. DN sam reac oPe = !abarexhe 36 x12y3_ 1 = OP XBP = = 216V3 cm? 3. From part (2) The elght ofthe Secton A fram the base of the tone’ "36cm Case Study 4 Kite Festival Kite festival is celebrated in many countries at different times of the year. In India, every year ldth January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites The picture given below, three kites flying together. In figure, the angles of elevation of wo kites (Points A and B) from the hands of a man (Point ©) are found to be 30” and 60" respectively. Taking AD =50 mand BE = 60 m. Based on the above information, solve the following questions. [CBSE 2022 Tormtt}QL. Find the lengths of strings used (take them In right-angled AABF. straight) for kites A and B as shown in the figure. —— AB =¥ (AF) (BF Q2. Find the distance’d' between these two kites. cat - =¥(70V3)? +(10)? Solutions = 16700100 = J14800 1. Inright-angled AADC = 12166m. sinaa° AD Hence. distance between two kites Is 121.66 m. x Case Study 5 = 3-2 Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was built by the King of Jaisalmer - AC =100m and rebuilt by Gadsi Singh in 14th century. The andin right-angled aCe, lake has many Chhatris. One of them is shown - BE below: singo* = BE 5 i _s0 2 BC 5 2 i, = 2, FB =40\3m Hence. length of the strings fam kites A and B are 100m and 40V3 m respectively 2. Inright-angled AOC, ao tanaor AD me OC a 50) We 0c = 0c =50f3m 8. andi right-angled aCEB, . BE taneor = 60 = rey 60 v3 = CE =x: BE -228 205m From figure, BF = BE-E€F =BE-AD =60-50=10m and AF =DE = C+ CE =50)3 + 20/3 Aim Observe the picture. From a point A, h m above from water level, the angle of elevation of top of Chhatri (point B) is 45° and angle of depression of its reflection in water (point C) is 60°. If the height of Chhatri above water level is (approximately) 10 m, then. Based on the above information, solve the following questions: [C0SE2022 Term} QL Draw a well-labelled figure based on the above information. Q2. Find the height (h) of the point A above water level. [Usey5 = 1.73) Solutions 1. From the given information. a figure is shown below. 8 (chhawi) © (Reflection of igure)2. TiP The imoge of point B with respect to water is of equal distance at point C.le., BE = EC Given, BS—Ea-1 ih 8D =BE-ED=10-h and OC =DE+€C=h+10 Inright-angled AADB. - 80, tonase £0 =F ap-wen tl . oF ap=10-h (i In ight-angled AADC - = 0. tangor = lo+h = 8-5 tosh a AD =: Se ° From eqs. (1) and (2), we get lo+h W-h= Se a ¥3(10-h) =10+h 5 15-54 =10+h = 1o(y3 -1) = H{1+-¥3) . 1o(3 -1), (V3-1) Wa) We) 103-17. (P07 = 103+1-23) 3-1 104-203) 2 = 10x22-3) 2 = 10 (2-173) 10 027=27m Case Study 6 ‘One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of 45°. When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of 30° and the ball flying towards him at the same height at an angle of elevation of 60”. Qu 2. Qs. é Based on the above information, solve the following questions [cose sop 2025-24] ‘At what distance from the foot of the tree was he observing the bird sitting on the tree? How far did the bird fly in the mentioned time? or After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball? What Is the speed of the bird (in m/min) if it had flown 20(V3 +1) m? Solutions Given. height of tree (AB) = 80 m and ZACB = 45" Now in right-angled AABC. AB tangs =F > BC =80m So, required distance Is 80 m. Given in 2 sec. the bird flying at the same height at an angle of elevation of 30°. 20CE #30" and DE= 80m Now in right-angled ADEC. oe ce = n= 20 Ee = cE=80Y3m BE =Ce-BC = 80/3 -80 = 80(V3 -1)m So, required distance the bird flew is BO(Y3 —1)m_ or After hitting the tree, the ball travel from A to F. Then angle of elevation of the ball from C is 60 je, ZECG=60° and FG=AB= 80m, Now In right-angled AFGC. FG 80. wre FE =o VI-2 tan30° 80 = Ge a80 1 FA=86=8C-C6 =80- ~ 001 |m Ba G So, required distance the ball travelled after hitting 1 thetreels eof 1-}m B 3. Given. in 2 sec. the bird had flown 20(Y3 +1). Distance Time taken 203 +1) ~z Speed of the bird = mysec = 2008+) om/min 2 = 600 (Y3 +1) m/min So, the required speed of the bird is 600 (5 +1) m/min “@ Very Short Answer Type Questions y Q1. The Length of the shadow of a tower on the plane ground is ./3 times the height of the tower. Find the angle of elevation of the Sun. __(cas€2023} Q2. The ratio of the height of a tower and the length of its shadow on the ground is 3:1. What is the angle of elevation of the Sun? [case 2017] 03. What is the angle of depression of the object at E from the observation point A, ifAD = ED? (case 2017} Q4. The angle of elevation of the top of a tower from a point on the ground which is 50 m away from the foot of the tower, is 30°. Find the height of the tower. [case 2023} Q5. Abuilding casts a shadow of length 5V3 mon the ground, when the Sun's elevation is 60°. Find the height of the building. Q6. Akite is flying, attached to a thread which is 140 m long. The thread makes an angle of 30° with the ground, Find the height ofthe kite from the ground, assuming that there is no slack in the thread. (w.imp} Q7. Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation 0 of the Sun is such that tan = [case 2023} Q8. In figure, AB is a 6 mhigh pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m, find the length of the ladder (Use 5 = 1.73) (CBSE 2016) Short Answer type-i Questions y Q1. The length of a string between a kite and a point on the ground is 70 m. If the string makes an angle Ow the grand eve such ttn =, se the kite is at what height from the ground? Q2. The angle of depression of car parked on the road from the top of a 150 m high tower is 30°. Find the distance of the car from the tower. 3. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. INCERT EXERCISE; Imp.) 4. The top of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x:y. [case 2015) Short Answer type-ii Questions y |. Aboy 1.7 mtallis standing on a horizontal ground, 50 m away from a building, The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. [Take V3 = 1.73] [BSE Sap 2022 Term-tI] Q2. The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the Sun is 60°, Find the angle of elevation of the Sun at the time of the longer shadow. [c8SE 2017) Q3. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower. [CBSE 2023) Q4. A person walking 45m towards a tower in a horizontal line through its base observes that angle of elevation of the top of the tower changes from 45° to 60°. Find the height of the tower. [Use V3 = 1.732] [case 2017) Q5. As observed from the top of a 100 m high light- house from the sea level, the angles of depression of two ships are a and §. It is given that one ship is exactly behind the other on the same side of the Light house. Based on the following figure, answer the following questions: Dep](i) In the given figure, if sin (38 —a) + and 0s (24-38) = dc > fi then find the values of and p. (li) Find the distance between the two ships. Q6. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, then find the width of the river. [c0SE 2022 Tern] Q7. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 30°. Determine the height of the tower and distance of the tower from the building. [case 2023) Q8. Two vertical poles of different heights are standing 20m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60° and angle of elevation of the top of the second pole fram the foot of the first pole is 30°. Find the difference between the heights of two poles. [Take V3 = 1.73] [c8se Sop 2022 Term] 99. A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 min.Find the speed of the boat (in m/min). [cose 2019, 17) 10. The angle of elevation of a cloud from a point hm above a lake is « and the angle of depression of its reflection in the lake is ). Prove that the height of fh (tanp + tana) (tan —tana) the cloud is [NCERT EXEMPLAR; CBSE 2017] ~) Long Answer type Questions w QL From the top of a tower 100 m high, a man observes: ‘two cars on the opposite angles of depression 30° and 45° respectively. Find the distance between the two [Use 3 = 1.73] {CBSE 2023] les of the tower with cars. Q2 Q3. Q4. 06. a7. QB. Qo. From point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. INCERT EXERCISE; COSE 2020} Two ships are approaching a light house from opposite directions. The angles of depression of the two ships from the top of a lighthouse are 30° and 45°, If the distance between the two ships is 100 m, find the height of the lighthouse. [Use V3 = 1.732] A straight highway leads to the foot of a tower. ‘Aman standing on the top of the 75 m high tower observes two cars at angles of depression of 30° and 60°, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. [Use 3 = 1.73] [CBSE 2023] ‘As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use V5 =1.732] _[ncenrexencise; c0s€ 2018, 17) There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole. [CSE 2019] ‘A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30°. A girl standing on the roof of a 20 m high building, finds the elevation of the same bird to be 45°. The boy and the girl are on the ‘opposite sides of the bird. Find the distance of the bird from the girl. (Given /2 = 1.414) [cose 2019 The angle of elevation of an aeroplane from a point ‘on the ground is 60°, After a flight of 30 sec, the angle of elevation changes to 30°. Ifthe aeroplane is flying at a constant height of 3600 /3 m, find the speed of the plane (in km/h). [cose 2019) The angle of elevation of a cloud from a point 60 m above the surface of the water of a Lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of water. [cose 2017]Solut ions Very Short Answer Type Questions Assen 1. Let AB be the tower and BC be if its shadow. b Let angle of elevation of the er sun bea 4 4 In right-angled 4 ABC, sane = AB tano = tif | h tano= ~ ins = ee tan30? B = 0=30 Hence, the elevation of the Sun is 30° NJERROR Students take the value of $ 5 ton 60° in haste. But it is wrong. So, the correct value IF B 2. Given that, the ratio of the height of a tower and the n 30° - length of its shadow on the P ground is ¥3:1 | Let tower height (PQ)= WS We A d (sQ)=t — 1 —> Let 2°5Q=0 (angle of elevation of the Sun) ‘Now. in right-angled PQS, = tano=tan60" > 0-60" So. the angle of elevation of the Sun Is 60" 3. Let the angle of depression of the object at € from the observation point A is 0. i A In right-angled ADE, / waoeap [+e Hence. the angle of depression is 45° 4, Given, the angle of elevation of the top (T) of a tower TB froma point (P) on the ground which is 30 m away from the foot (B) of the tower, is ZBPT = 30°, Le. BP = 30 mand let BT = hm Now. in right-angled ATEP, tan30° =! Br Lin ~ G30 . 239 cB 28 Ba - hao So, height of the tower is 10V3 m. 5. Let BC= hm be theheightof the building, AB = 5V3 m. and ZCAB = 60" gisun) em hm ko _ | wan In right-angled AARC. Ret = V3=—= sham a3 Hence, height of the building is 15 m. 6. Let AC be the length of thread and AB be the vertical height of the kite. Students should do practice, use of trigonometricat ratios in the triangle Let AB = him, AC = 140 m and ZACB = 30° In right-angled 4 ABC. sin30" a 2 140Hence, the height of the kite is 70 m 7. Let the length of the shadow BS = xm an the ground of a pole BP of height 18 m Given, 8P =18 m F (Pole) 6 and tane== | 7 lem vowinrangnes. | Be og Shadow a5 ee 5 xT=21m So, the required length of shadow ts 21m 8. From figure. 8D = AB- AD =254=346m In right-angled DC, 80 cO - LB 346 2 co co 3 17 4m sin60” a2 Hence. length of the ladder is 4m. Short Answer Type-1 Questions 1 Let C be the position of kite (Kite) and AC = 70 m be the length of the string Let string makes an 79, angle of @ from the ground. Let height of the kite from the ground be BCs hm. a 8 4 Given tano== 3 TRICK Use identities:cosec @ = ¥1 + cot?@ and tand-cot0 3 = to-2 coto=F cosec0 = VI+cot7o Hy Now In right-angled ABC, AC cosec 8 = = Ble > hs 56m Hence, height of the kite is 56 m, 2. Let AB = 150 m be the height of the tower and angle of depression Is 2 DAC = 30°. A ie yt 60m B Then, ACB = 2 DAC = 30° In right-angled A ABC. AB tone = aC ( " 10 BC > Bc =150/3m Hence. the distance of car from the tower's 1503 m. ERROR ‘Some candidates are unable to draw the diagram as per {alternate angles) tan30° the given dota ond lose their marks. 3. Let AC was the original tree. Due to storm, it wes broken into two parts. The broken part AB Is making an angle of 30° with the ground. Cin x In right-angled a A’CB. Height of the tree AB. BC=A’B BC (8 A8) i 8 (3h m4 8 -Him-Hacfim 20 Sem -em Hence, the helght of the tree Is 83m4, The base is same for both towers and their heights are given, ie. xand y respectively. eke eo Let the base of towers be BC = CO=k In right-angled ABC. U} In right-angled APQC. taneooE0 -¥ ck (2) wT > k = ya ky3 From eqs. (1) and (2), we get Kk ge y 3 2 3 Short Answer Type-lI Questions Let height ofthe tower be BE = hmand AD = L7 mbe the height af the boy E hm As Ao 1m am Also. given the angle of elevation of the boy to the top of the tower Is ZCDE = 60° Here eC 68a =ho17— (BCe ADs 17m) Aso. OC = AB = 50 m. Inight-angled triangle DCE E 60° -E tango = ho = hat Ba = 503 =h-17 = f=50x173+17 =8650+17 > © 88.20 m Hence. height of the bullding is 88.20 m. ‘Suppose B be the position of the Sun, Let the height of the tower be him and the angle between the Sun and the ground at the time of longer shadow be 0. AC and AD are the lengths of the shadow of the tower when the angle between the Sun and the ground are 60" and 0, respectively. 3 (Sun) A A 0 cx — & —— Let AC = xunit. then Given, AD=3AC => AD=3x In right-angled ABAC. a8 60° = AB tanso° AB = gee x = haw 0) In right-angled aBAD, AB_h tone ABP =» hs 3xxtan 0 Q) From eqs.) and (2). we get x = 3x-tane = tan = Je =tan30" = 0-30" 1. Let AB = 30 m be the height of the tower and CO = hm be the height of the another tower. Then. 8 30m ZCAD = 30 and ZACB = 60". In right-angled ACB, 30 = 4-2 5 @ i ald 5 BO dain Bag wo rightangled ACAD tanar =2 c= nage om Hence, the distance between the two towers and height of the other tower are 10V3m and 10 m respectively 4. Let AB= hmbe the height of the tower Let BC = xmand DC = 45 m In right-angled A ABC, 6 iy 45 mee In right-angled AABD. AB tangs. AB__AB 8D BC+CD < =i 5 ” heeds aL. . = has [using eq. (1) a = baa = has 45V3 , v3+1 = Eipaeniliners (rationalising the denominator) -8BHD (.(6-4) (0+ b= 0-B) (3P-1 _ 45(3+1732 3-1 = AB XA 122 — 196.47m Hence, the helght of the tower is 106.47 m. fi 5. (i) Given, sin (38—a) a ¢ ) are = sin (38 ~a! > 3p a= 45" 0 Also. cos(2a— 38) = cos(2a— 38) = 20-38 = 0° (2 On adding eqs. (1) and (2). we get a= 45° Put a = 45°in eq, (1), we get ‘com > S=100/3m 2% 2) In ightangled APSR PS 100 tana5 oF wo = 1-0 > RS=100m —(» tan 45° =1) ..(4) QR=QS—RS = 100\3-100 [from eg. (3)) =100(/3-1) m Hence, clstance between the two ships I 100(V/3-1)m. {EOMMONJERROR Sometime students get confused with the values of trigonometric angles. They substitute wrong values which leads to the wrong result. 6. Let A be the point of the bridge and 8 and D be the position of the opposite sides of the bridge. EA F Also, given the angle of depressions are ZEAB = 45° and ZFAD = 30° Then. ZABL = ZEAB = 45" and ZADC = ZFAD = 30° In right-angled ABCA. (by alternate angles) (by alternate angles) e AC tanas: =F AC = 1 = 8C=3m In right-angled aDCA. AC 30° = AC £2020"= eo = W-N3on a0 -8C+cD 34303 =3(1+¥3)m Hence, width of the rlver is 3(1+V3)m. Let ABs hm be the height of the tower and CD=7 cm be the height of the building Oem 38 Here, ZECB = ZDBC = 30° Let 8D=xm,CO=7m In right-angled aBOC. {alternate angles) 5 $ . = xeW3m a In right-angled a AEC. pA tango: = AE ¢ = 3 fax=h-7 = ¥Bx743eh-7 (using eq. (1)) = h=2147=28m Hence, the distance of the tower from the building is 73 m and height of the tower is 30 m. Let heights of two different poles be 8C = hy m and AD = hy m. é © by "led ald Ns —aa— ZBAC = 60" ZABD «= 30° and AB = 20m In right-angled 4 ABC, Aliso given, 3) > = BC tano = = ott Bax = hy = 203m In right-angled BAO, me $4 0 3 > np ae 208, 3 The difference between the height of two poles = y= hy = 20Y3-2245 s0y3-20)3 _40y5 3 3 40x173 692 3 3 =2307m 9. In the figure, AB represents the 150 m high cliff Initially. the boats at point Cand it moves to point D in 2 min and as itis given that the angle of depression of the boat changes fram 60° to 45°. x ee So, cDAK= 08° and ZCAK = 60° TiP {fo tonsvesol intersects two parcel tne, then each pair of alternate interior angles is equal. ‘ZOAX = ZADB = 45" and ZCAX = ZACB = 60" (alternate interior angles) In right-angled SABC, 2 8B tan 60° ac 150 = B “(> AB=150.m and BC= ye) 5 yo $. 1503 505m Inrightangled aas0, AB AB ‘BD BC+CD (- y=50V3m)= x=150-50V3m The boat covers CD distance in 2 min. = 4(150-50¥3) evemin =25(3-V3) m/min 253 (3 — 1) m/min 10. Let P be the position of cloud, S be the point of observation and MR is the surface of lake. TiP The concept of angle of depression and angle of elevation must be understood deeply and clearly. Let PR=RQ=xm (height of cloud above the take) SM=NR= hm PN=PR-NR =x-h ON=QR+RN =xth and MR=SN=ym In right-angled APNS. =PN tan a= BN 2 tani () In right-angled ASNQ, = (2) Dividing eq. (1) by eq. (2). we get ach tana __¥ tanB xh y = fan. tanp xh = (x+Atana = (x—htan p > xtana+htana=xtan p-htanp = h (tan a + tan B) = x (tan p-tana) (tana-+ tang) tanB=tana The height of the cloud is h(tana+tanB) (tanB=tana) Hence proved. Long Answer Type Questions 1. Lethe top of the tawer AO, aman at point A. observed the angle of depression 30° of car C and angle of depression 45° of car B on both sides of a tower. a x 45° 30" ¥ 45° rf vw TiP Ifo transversal intersects two parallel lines, then each poi of alternate interior angles sequal ZXAB = ZAB0 = 45° ANAC = ZACO = 30° OA =100 m and (alternate angles) Given, geTip The concept of angle of depression must be understood deeply ond clearly. In right-angled AAOB, 0A wands = (let OB = xm) = 1 x = x-100m 4 In right-angled aAOC. OA _ 100 peal Oana (let OC = ym) = 1 100 Boy » y=100¥3 m @ Therefore, widthiof therriver sixty =100 + 100y3 = 100 + 100 « 173 = 100 + 173 273m Hence, distance between two cars is 273 m. A 2. Let AB be the tower and BC be the building Let OC = xm. AB = hm and c= 20m. In ight-angled 49¢0. ec tangs = 86 mecIn right-angled aACo, AC _ AB+BC tangor=AE = A6~4 +20 > ete = Vix=h+20 => ¥3x20=h+20 (:x20m) = h=20Y3-20 =20(Y3-1)m Hence. the height of tower is 20 (v3 —1)m. 3. Let PQ be the lighthouse. and A and Bare the position of two ships. TiP The concept of angle of depression must be understood deeply and clearly. a3 Let PQ = hm, AQ = xm and QB = ym. The distance between two ships (AB) = x+y=100m_ (Given) In right-angled a PQA ton 450 F2 Fo eon ) In right-angled aPQ8, SS _ Haas fan 30°98 > Sy = y=vah 2) Adding eqs. (I) and (2). we get xtysntyah 5 h+yBh=100 (-x+ y= 100 m) > V3+1)h=100 _ 00, 3-1 = nee 31 (rationalising the denominator) 100(V3-1) _100(1732-1) 3-1 2 (0+b) (o-b)=o?-) 250 «0.732=366m Hence. the height of the lighthouse is 36.6 m. 4, Given, the angles of depression of two cars A and B from the man standing on the top 0 of the tower CD (say) with height 75 m are 60° and 30° respectively 18m If @ transversal intersects two parallel lines, then each pair of alternate intesfor angles is equal. ‘ZXDB = ZC80 = 30" and ZXDA= ZCAD= 60" —_(alternate angles) Let the distance between the cars, AB= ym. In right-angled ACD. co . 6 tango"=F = 8-72 5 ss cA= E ( In right-angled 4BCD. Oo o BC” CA+AB CA+y 175 tan 30° =. (from eq. ()) vy 7545-73-73 (5-1 150,83 = yet EEO EE 50v3, 3 BOM = 50V3 =50x173= 865 So, required distance between two cars Is 86.5m, 5. Let AB be the light house. C and D be the position of the ships. TiP {fe tronsvesol intersects two poral tines, then each par of alternate interior ongles Is equal ZEAD = ZADB = 30" and ZEAC = ZACB=45° (alternate interior angles) e