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www.sakshieducation.com 10. HEIGHTS AND DISTANCES LONG ANSWER QUESTIONS 1. From a point on the level ground the angle of elevation of the top of the hill is α . After walking through a distance ‘d’ meters towards the foot of the hill the angles of elevation of the top is found to be β . Find the height of the hill B Solution: AB be the hill of height h P be the first point of observation A x Q d Q be the second point of observation The angle of elevation of β From P and Q are α and β respectively Given PQ = d let AQ = x In ∆ le BAQ tan β = h ⇒ x = h cot β x In ∆le BAP tan α = h ⇒ d + x = h cot α d+x d + h cot β = h cot α d = h {cot α − cot β } h= 2. d cot α − cot β From the cliff of a mountain 120 meters above the sea level the angles of depression of two boats on the same side of the mountain are 450 and 300 respectively. Find the distance between the both boats {assume the both the boats lie along the line from foot of the hill on the same sides of the hill ? Solution: Let AB be the hill Ax be the horizontal through the foot of the hill. By parallel to B horizontal line passing through the observing. Y 300 Exp: Let P, Q be the two boats 0 45 Given ∠y BQ = 300 120m 1 www.sakshieducation.com 450 300 X Q P A P www.sakshieducation.com From the fig ∠y BP = ∠BPA = 450 BQ = BQA = 300 Let PQ = x In triangle PQAB tan 450 = 120 AB ⇒1= PA PA ⇒ PA = 120 In triangle BAQ tan 300 = x = 120 3. ( AB 1 120 ⇒ = 0 AC 3 x + 120 ) 3 −1 The angles of elevation of the top of a mountain from the top and bottom of a pillar AB of height 5 meters are 300 and 600 respectively. Find the height of the mountain? Solution: Let PQ be the mountain of height ‘h’ AB is the pillar of height 5m Angles of elevation of Q from A and B are respectively 600 and 300 Let AP = BC = x Since AB = 5 CQ = h − 5 h-5 h In triangle QPA tan 600 = → (1) x h−5 In triangle QCB tan 30 = → (2) x 30 h tan 60 3 h = x ⇒ = (1) ÷ (2) we have 0 h − 5 tan 30 1 h−5 x 3 2 www.sakshieducation.com h x 5m 600 0 0 0 A x P www.sakshieducation.com 3= h 15 ⇒ 3h − 15 = h ⇒ 2h = 15 ⇒ h = 2 h−5 ∴ Height of mountain = 7.5m 4. From the top of a tree on the bank of a lake an aeroplane in the sky makes an angle of elevation α and its image in the river makes an angle of depression β . If the height of the tree from the water surface in a and the height of the a sin (α + β ) aeroplane in h show that h = sin ( β − α ) B h-a Solution : h Let ' β ' the position of aeroplane C b the its image Q a PQ be the tree of height ‘a’ draw QR parallel the AP Angle of elevation of B from Q is α h Angle of depression of C from Q is β BR = h – a c CR= h + Q Let QR= PQ = x In triangle QRB tan α = h−a → (1) x In triangle QRC tan β = h+a → (2) x (1) ÷ (2) we have ∴ h+a P tan α h − a = tan β h + a h+a tan β = tan α h−a Using compounds and dividends we have sin β sin α + h+ a −h− a tan β + tan α h cos β cos α = ⇒ = h +a− h +a tan β − tan α a sin β − sin α cos β cos α a sin ( β + α ) h sin β cos α + cos β sin α = ⇒h= a sin β cos α − cos β sin α sin ( β − α ) 3 www.sakshieducation.com www.sakshieducation.com 5. one end of the ladder is in contact with a wall and the other end is in contact with level ground making an angle α when the foot of the ladder is moved through a distance ‘a’ cms away from the wall, the end in contact with the wall slides through ‘b’ cms along the wall and the angle made by the ladder with the level is α + β now β . Show that a = b tan 2 Solution : - Let l be the length of the ladder OA = x and OB’ = y where AB is the initial position of ladder and A’B’ is the final position of ladder Given AA = a BB’ = b B b AB = A’B’ = l B' In triangle B OA y y+b sin α = ⇒ y + b = l sin α → (1) l cos α = x ⇒ x = l cos α → (2) 2 In triangle B’OA’ sin β = y ⇒ y = l sin β → (3) l cos β = a+x ⇒ a + x = l cos β → (4) l (1) - (3) we have b = l ( sin α − sin β ) → ( 5) (4) – (2) we have a = l ( cos β − cos α ) → ( 6 ) (5) ÷ (6) we have b l ( sin α − sin β ) = a l ( cos β − cos α ) 4 www.sakshieducation.com O x A a A' www.sakshieducation.com b = a α+β α −β l 2 cos sin 2 2 α + β α − β 2 sin sin 2 2 b α + β = cot a 2 α + β a = b tan 2 6. The angles of depression of two mile stones along a road when viewed from an tan α tan β areoplane are α , β . Show that the aeroplane is flying at a height of tan α + tan β from the road level. Taking the two mile stones to be on opposite side? Solution: - Let A be the position of the aeroplane P, Q be two mile stones Given PQ = 1 mile Draw XAY parallel to PQ through the observes eye Given ∠ × AP = ∠APQ = α ∴ ∠XAP = ∠APQ = α ∠YAQ = β ∠YAQ = ∠PQA = β In triangle PMA A x y h tan α = ⇒ PM = h cot α PM In triangle tan β = h ⇒ QM = h cot β QM Given PQ = 1 PM + QM = 1 ⇒ h cot α + h cot β = l h= 1 tan α tan β = cot α + cot β tan α + tan β 5 www.sakshieducation.com h P M Q www.sakshieducation.com 7. From the point on a bridge across a river the angles of depression of the banks on opposite sides of the river are 300 and 450 respectively. If the bridge is at a height of 4 meters from its banks find the width of the river? Solution : - Let A, B represent points on the bank on opposite sides of the river so that AB is the width of the river. Let P be a point on the bridge at a height of 4 meters i.e., PD = 4 mt A In triangle PDA tan 300 = 4 ⇒ AD = 4 3 AD 300 h 4 In triangle PDB tan 45 = ⇒ = 1 ⇒ DB = 4 DB DB 8. ( 450 A 0 AB = AD + DB = 4 3 + 4 = 4 450 300 B D ) 3 +1 From the top of a light house 300 mts high the angle of dipressino made by two boats situated in a horizontal line along the foot of the light house are 300 and 450 find the distance between the two boats (i) when they are on the same side (ii) Q when they are on either side of the light house? Y 450 30 Solution : - (i) 300mt Q Refer to problem ‘2’ 30 450 300 B And try your self 0 450 P A 300mt Solution (ii) Refer to problem 7 A P B And try your self 9. From the top of a hill 200 meters high the angle of depression of the top and bottom of a pillar on the level ground are 300 and 600 respectively. Find the height of the pillar.? Solution : - Let PQ be the full of height 200 mt AB be the pillar of height h Q 600 300 Draw QY parallel to horizontal B Angle of depression of B from Q is 30 300 h 200 200mt C 0 h 6 www.sakshieducation.com 600 A P www.sakshieducation.com Angle of depression of A from Q is 600 Draw BC parallel to Qy Since AB = h QC = 200 – h Let BC = AP = x In ∆ le QCB tan 300 = 200 − h x -------(1) In ∆ le QPA tan 600 = 100 x --------(2) (1) ÷ (2) we have tan 300 200 − h 1 200 − h = ⇒ = 0 tan 60 100 3 200 200 = 600 − 3h ⇒ h = 10. 400 3 A pillar of 10 meter height is mounted on a tower from a point on the level ground, the angles of elevation of the top of the foot of the pillar are 750 and 450 respectively find the height of the tower.? Solution : - Let AB be the tower and BC be the pillar of height 10mt Let P be the point of observation Angles of elevation from P to B and C are respectively 450 and 750 In triangle BAP tan 450 = h -------(1) x In triangle CAP tan 750 = h + 10 x -------(2) tan 450 h = 0 tan 75 h + 10 C 10mt B 1 h = 2 + 3 h + 10 ( h 450 ) h + 10 = 2 + 3 h A 7 www.sakshieducation.com x 750 P www.sakshieducation.com ( ) 10 =5 3 +1 10 = 1 + 3 h ⇒ h = 11. ( ) 3 −1 From the top of a pillar of a height 80 meters the angles of depression of the top and foot of the another pillar are 300 and 450 respectively. Find the distance between the two pillars and the height of the second pillar ? Solution : - Let PQ be the pillar of height 80 mts AB be the second pillar of height h y 300 450 Draw BX parallel to AP B 300 QY parallel to AP 0 From Q are 30 and 45 h 450 A In triangle QXB tan 300 = 80m x x Angles of depression of B and A 0 80-h x P 80 − h -------(1) x In triangle QPA tan 450 = 80 h ----------(2) (1) ÷ (2) when tan 300 80 − h = tan 450 80 80 3 − h 3 = 80 ⇒ 80 12. ( ) 3 −1 / 3 From a point A on the level ground away from the foot of the tower the angle of elevation of the top of the tower is 300 . From a point at a height of h meters vertically above A the angle of elevation of the foot of the tower is 600 . Find the height of tower in term of h ? Solution: - Let PQ be the tower of height ‘x’ Q A be the first point of observation 600 y B be the point of a height of ‘h’ h 600 From the point A P 8 www.sakshieducation.com B 300 A www.sakshieducation.com Angle of elevation of Q from A is 300 Angle of depression of P from B is 600 In triangle QPA tan 300 = x y In triangle BAP tan 600 = h y Let AP = 4 x y tan 30 ⇒1= x⇒x=h = 0 h tan 60 3 h 3 y 0 13. The angle of elevation of the top of a tower from the foot of the building is twice the angle of elevation from the top of the building. The height of the building is 50 meters and the height of the tower is 75 meters. Find the angle of elevation of the top of the tower from the foot of the B building? 75m Let AB be the tower of height 75 mt PQ be the building of height 50m Q x 50m Draw QX parallel to AP A Angles of elevation of B from Q AP Angle of elevation of B from Q is α Angle of elevation of B from P is 2α In triangle BAP tan 2α = 75 --------------(1) x In triangle BXQ tan α = 25 -----------------(2) x (1) ÷ (2) we have tan 2α = 3 ⇒ tan 2α = 3 tan α tan α 2 tan α = 3 tan α ⇒ 2 = 3 − 3 tan 2 α 1 − tan α 3 tan 2 α = 1 ⇒ tan α = 1 3 α = 300 ⇒ 2 α = 600 9 www.sakshieducation.com x P www.sakshieducation.com 14. As observed from the top of a light house of height 75m from the sea level the angles of depressions of two ships are 300 and 450 . It one ship is exactly between the two ships? Solution : - X H 0 300 45 Refer to Problem 2s 300 And try your self 15. 450 P Q L The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary prove that the height of the tower is 6m B Solution: - AB be the tower f height h Angle of elevation of B from C and D Are respectively are α , β A 4m 9m Given α + β = 900 α = 900 − β ( tan α = tan 900 − β ) tan α = cot β ⇒ tan α = 1 ⇒ tan α tan β = 1 ----------------(1) tan β In triangle BAC tan α = h 4 In triangle BAQ tan β = h 4 tan α tan β = 1 ⇒ h2 =1⇒ h = 6 36 10 www.sakshieducation.com C D www.sakshieducation.com 16. A and B are two points on the line segment joining the feet of two equal pillars. AB = 30 meters. From A the angles of elevation of top of the pillars are 300 and 600 . From B they are 600 and 300 respectively in the same order find the height of the pillars and the distance between them? Solution : - tan 600 = S B In triangle QPB h ⇒h=x 3 x h h In ∆le QPA tan 300 = P h x + 30 x + 30 = h 3 But h = x 3 ∴ x + 30 = 3 x ⇒ x = 15 h = 15 3 In triangle SRY tan 600 = 3= 300 300 600 h y 15 3 ⇒ y = 15 y Distance between the poles = x + 30 + y = 60m Height of pillar = 15 3 11 www.sakshieducation.com x B 30 A 600 y R www.sakshieducation.com 16. Over a tower AB of height h mt there is a flat staff BC. AB and BC are making equal angles at a point distance d mt from the foot A of the tower. Show that the (d + h ) h 2 height of the flag staff is 2 d 2 − h2 Solution : - AB is the tower of height h BC is the flag staff C Let the height of the flat staff be ‘x’ X P be any point let Ap = d B In triangle BAD h tan α = h d A In triangle CAD tan 2α = h+x d 2 tan α h+x = 2 1−tan α d h 2 h+x d = h + x ⇒ 2h × d = 2 2 2 h d d d −h d 1− 2 d 2 2hd 2 2hd 2 − hd 2 + h3 − h = x ⇒ x = d 2 − h2 d 2 − h2 x= ( h d 2 + h2 hd 2 + h3 ⇒ x = d 2 − h2 d 2 − h2 ) 12 www.sakshieducation.com d P www.sakshieducation.com 17. ABCD is a monument built vertically on the ground from a point P away from the moment on the level ground the angles of elevation of AB, AC, AD are α , β and γ respectively. If AB= a, AC = b , AD = c AP = x and α + β + γ = 1800 then show that ( a + b + c ) x 2 = abc Solution : - ABCD is movement A is on the ground Given AB = a AC = b AD = C P be a point of observation Given AP = x α , β , γ are the angles of elevation of B, C, D from P In triangle BAP tan α = a x D C C B In triangle CAP tan β = In triangle DAP tan γ = b x a A c x Given that α + β + γ = 1800 ⇒ α + β = 1800 − γ tan α + tan β = − tan γ ⇒ tan α + tan β + tan γ = tan α tan β tan γ 1 − tan α tan β a b c abc + + = 3 ⇒ ( a + b + c ) x 2 = abc x x x x 13 www.sakshieducation.com a P www.sakshieducation.com 18. An aeroplane is moving one kilometers high from west to east horizontally. From a point on the ground the angle of elevation of the aeroplane is 600 and after 60 seconds the angles of elevation of the aeroplane is observed as 300 . Find the speed of the aeroplane in km/hr ? Solution : - Let C, E be the positions of aeroplane P be the point of observation C Angle of elevation of C from P is 600 E 1km 600 Angle of elevation of E from P is 300 300 P Draw CD and EF perpendicular to horizontal line PX In triangle CDP tan 600 = ⇒ CD = 1 PD 1 3 In triangle EFP tan 300 = 1 PF PF = 3 Distance travelled by aeropalne in 10 seconds = CE = DF CE = DF = PD − CD = 3 − 1 2 = 3 3 2 2 720 dis tan ce 3 = = × 360 = = 240 3 km / h ∴ Speed = time 10 3 3 360 14 www.sakshieducation.com D F x www.sakshieducation.com 19. A pillar is leaning towards east and α , β are the angles of elevation of the top of the pillar from points due west of the pillar at distance a and b respectively. Show that the angle between the pillar and the horizontal is b−a tan −1 b cot α − a cot β Solution : B Let AB be the pole inclined towards east P, Q are the points of observations h At a distance of a and b from A respectively Angles of elevation from P, Q to B are α , β Draw BM perpendicular to AQ let BM = h AM = x In triangle AMB tan θ = h ⇒ x = h cot θ → (1) x In triangle PMB tan α = h ⇒ a + x = h cot α → (2) a+x In triangle QMB tan β = h ⇒ b + x = h cot β → (3) b+x (2) ------(1) we have a + x − x = h {cot α − cot θ } → (4) (3) -------(1) we have b + x − x = h {cot β − cot θ } → (5) (4) ÷ (5) we have a cot α − cot θ = b cot β − cot θ a cot β −a cot θ = b cot α − b cot θ ( b − a ) cot θ = b cot α − a cot β cot θ = b cot α − a cot β b−a ⇒ tan θ = b−a b cot α − a cot β b−a b cot α −a cot β θ = tan −1 15 www.sakshieducation.com Q P P a A x M www.sakshieducation.com 20. From a point B on the level ground away from the foot of the hill AD, the top of the hill makes an angles of elevation α from the point B, the point C is reached by moving a distance d along a slant/slope which makes an angle ϑ with the horizontal. If B is the angle of elevation of the top of the hill from C, find the height of the hill ? Solution : - AD be the hill b is the point of observation A BC is the inclined plane Draw CE parallel to BD Angle of elevation of A from B is α E Angle of elevation of A from C is β Angle of inclination of BC with BD is ϑ In ∆ le ADB ∠BAD 900 − α In ∆le CEA ∠CAE = 900 − β In ∆le BCA ∠ABC = α − ϑ ∠BAC = ( 90 − α ) − ( 90 − β ) = β − α ∠BAC = 1800 − {∠BAC + ∠ABC} = 1800 − { α − ϑ + β − α } = 1800 − ( β − ϑ ) Using since rule BC AB = 0 sin ( β − α ) sin180 − ( β − ϑ ) d sin ( β − α ) = d sin ( β − ϑ ) AB ⇒ AB = sin ( β − ϑ ) sin ( β − α ) In triangle BDA sin α = AD ⇒ AD = AB sin α AB 16 www.sakshieducation.com C B D www.sakshieducation.com ∴ AD = 21. d sin α sin ( β − ϑ ) sin ( β − α ) From a point on the slope of a hill, the angle of elevation of the top of the hill is 450 . After walking 200 mt from that point on a slope of 150 angle with horizontal the same point on the top of the hill makes an angle of elevation 600 show that the height of the hill is 100 ( 6+ 6 ) A Solution : 600 Do the problem again by taking α = 450 , ϑ = 150 , β = 600 200mt E C and d = 200 m in the above problem 0 450 15 B 22. D A church tower AB standing on a level plane is surrounded by a spire BC of the 1 same height as the tower D is a point in AB such that AD = AD. At a point on 3 the plane 100 mt from the foot of the tower the angles subtended by AD and BC are equal. Find the height of tower ? Solution :- In triangle OQP C h tan θ = 300 h B D In triangle BAP tan (θ + Q ) = h 100 P In triangle CAP tan ( 2θ + Q ) = 2h h = 100 50 tan ( 2θ + Q ) = h h ⇒ tan {θ + θ + Q} = 50 50 tan θ + tan (θ + Q ) 1 − tan θ tan (θ + Q ) = h 50 h h + 300 100 = h ⇒ 400 h × 30000 = h h2 50 30, 000 30000 − h 2 50 1− 30, 000 17 www.sakshieducation.com 100mt A h www.sakshieducation.com 20, 000 = 30, 000 − h 2 ⇒ h 2 = 10, 000 ⇒ h = 100mt 23. A man observes a tower AB of height h from a point P on the ground the moves a distance ‘d’ towards the foot of the tower and finds that the angle of elevation is 3d in the same direction and the angle of doubled. He further moves a distance 4 elevation is three times that at P. Prove that 36h 2 = 35d 2 Solution : In triangle PQB ∠PQB = 180 − 2α B ∠PBQ = 180 − {α + 180 − 2α } =α d 180-R 180-k ∴∠BPQ = ∠PBQ = α ∴ BQ = BQ = d {sides opposite to equal angles are equal} In triangle BQR ∠QRB = 180 − 3α ∠QBR = 1800 − {2α + 180 − 3α } = α Using since rule in this triangle BQR BQ QR d sin 3α = ⇒ = 3d sin (180 − 3α ) sin α sin α 4 4 4 = 3 − 4 sin 2 α ⇒ 4sin 2 α = 3 − 3 3 sin 2 α = 5 12 cos 2 α = 7 12 In ∆ le QAB sin 2α = AB h ⇒ sin 2α = BQ d h = 2d sin α cos α ⇒ h 2 = 4d 2sin 2 α cos 2 α h2 = 4 d 2 × 5 7 × ⇒ 36h 2 = 35d 2 12 3 12 18 www.sakshieducation.com P d Q R A www.sakshieducation.com 24. An observer finds that the angle of elevation of the tower is θ , an advancing ‘a’ meters towards the tower, the elevation is 450 and an advancing b meter near the elevation is 900 − θ . Find the height of the B tower.? Solution : h AB be the tower of height h 900 - Q 450 P the first point of observation P a Q b From which the angle of elevation is 'θ ' R x A Q be the second point such that PQ = a and angle of elevation from Q is 450 R be the third pant such that QR = b and angle of elevation is 900 − θ let AR = x In triangle BAR tan ( 900 − θ ) = In triangle QAB tan 450 = In triangle PAB tan θ = h h ⇒ cot θ = ⇒ x = h tan θ -------(1) 2 x h ⇒ h = b + x ---------(2) b+x h ⇒ a + b + x = h cot θ -------(3) a+b+ x (2) – (1) we have (3) – (2) we have b = h (1 − tan θ ) ------(4) a = h {cot θ − 1} -------(5) ( 4) (5) we have b h (1− tan θ ) = a h ( cot θ − 1) h {1 − tan θ } b b ⇒ = tan θ = a a 1 − 1 h tan θ ab b Sub tan θ is (3) we have b = h 1 − ⇒ h = a−b a 19 www.sakshieducation.com