{"id":1070,"date":"2023-05-04T10:00:36","date_gmt":"2023-05-04T04:30:36","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=1070"},"modified":"2023-05-05T09:16:53","modified_gmt":"2023-05-05T03:46:53","slug":"heights-and-distances","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/heights-and-distances\/","title":{"rendered":"Heights And Distances"},"content":{"rendered":"<h2><strong>Heights And Distances<\/strong><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-118836\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-1.png\" alt=\"Heights And Distances 1\" width=\"713\" height=\"349\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-1.png 713w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-1-300x147.png 300w\" sizes=\"(max-width: 713px) 100vw, 713px\" \/> <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-118838\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-2.png\" alt=\"Heights And Distances 2\" width=\"295\" height=\"501\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-2.png 295w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-2-177x300.png 177w\" sizes=\"(max-width: 295px) 100vw, 295px\" \/><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-118839\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-3.png\" alt=\"Heights And Distances 3\" width=\"722\" height=\"466\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-3.png 722w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-3-300x194.png 300w\" sizes=\"(max-width: 722px) 100vw, 722px\" \/><\/p>\n<p><strong>Angle Of Elevation<\/strong><br \/>\nThe 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 level, i.e. the case when we raise our head to look at the object. (see fig.)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134762\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-1-1.png\" alt=\"Heights And Distances 1\" width=\"227\" height=\"155\" \/><br \/>\n<strong>Angle Of Depression<\/strong><br \/>\nThe angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e. the case when we lower our head to look at the point being viewed. (See fig.)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134763\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-2-1.png\" alt=\"Heights And Distances 2\" width=\"288\" height=\"155\" \/><\/p>\n<p>More on\u00a0<a href=\"https:\/\/www.aplustopper.com\/lesson\/rs-aggarwal-class-10-solutions-height-and-distance\/\">RS Aggarwal Class 10 Height and Distance\u00a0Solutions<\/a><\/p>\n<h2><strong>Heights And Distances With Examples<\/strong><\/h2>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/-krQc0aA0w8?feature=oembed\" frameborder=\"0\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>The shadow of a building is 20 m long when the angle of elevation of the sun is 60\u00ba. Find the height of the building.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let AB be the building and AC be its shadow.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134764\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-3-1.png\" alt=\"Heights And Distances 3\" width=\"135\" height=\"167\" \/><br \/>\nThen, AC = 20 m and \u2220ACB = 60\u00ba.<br \/>\nLet AB = h.<br \/>\nThen, \\(\\frac { AB }{ AC }\\) = tan 60\u00ba =\u00a0\u221a3<br \/>\n\u21d2 h\/20 =\u00a0\u221a3<br \/>\n\u2234\u00a0h = (20 \u00d7 \u221a3 )m = (20 \u00d7 1.732) m<br \/>\n= 34.64 m.<\/p>\n<p><strong>Example 2: \u00a0 \u00a0<\/strong>If a vertical pole 6m high has a shadow of length 2\u00a0\u221a3 metres, find the angle of elevation of the sun.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let AB be the vertical pole and AC be its shadow.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134765\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-4.png\" alt=\"Heights And Distances 4\" width=\"148\" height=\"170\" \/><br \/>\nLet the angle of elevation be \u03b8. Then,<br \/>\nAB = 6 m, AC = 2\u00a0\u221a3 m<br \/>\nand \u2220ACB = \u03b8.<br \/>\n\\(\\text{Now, \u00a0tan }\\theta =\\frac{AB}{AC}=\\frac{6}{2\\sqrt{3}}=\\sqrt{3}=\\text{ tan }60{}^\\text{o}.\\)<br \/>\n\u2234\u00a0\u03b8 = 60\u00ba.<\/p>\n<p><strong>Example 3: \u00a0 \u00a0<\/strong>A ladder against a vertical wall makes an angle of 45\u00ba with the ground. The foot of the ladder is 3m from the wall. Find the length of the ladder.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let AB be the wall and CB, the ladder.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134766\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-5.png\" alt=\"Heights And Distances 5\" width=\"131\" height=\"163\" \/><br \/>\nThen, AC = 3m and \u2220ACB = 45\u00ba<br \/>\n\\(\\text{Now, \u00a0}\\frac{CB}{AC}=\\sec \\text{ 45}{}^\\text{o}=\\sqrt{2}\\Rightarrow \\frac{CB}{3}=\\sqrt{2} \\)<br \/>\n\u2234 \u00a0Length of the ladder = CB = 3\u00a0\u221a2<br \/>\n= (3 \u00d7 1.41) m = 4.23 m<\/p>\n<p><strong>Example 4: \u00a0 \u00a0<\/strong>A balloon is connected to a meteorological station by a cable of length 200 m, inclined at 60\u00ba to the horizontal. Find the height of the balloon from the ground. Assume that there is no slack in the cable.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let B be the balloon and AB be the vertical height. Let C be the meteorological station and CB be the cable.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134767\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-6.png\" alt=\"Heights And Distances 6\" width=\"133\" height=\"168\" \/><br \/>\nThen, BC = 200 m and \u2220ACB = 60\u00ba<br \/>\n\\( \\text{Then,\u00a0 }\\frac{AB}{BC}=\\text{sin }60{}^\\text{o}=\\frac{\\sqrt{3}}{2} \\)<br \/>\n\\( \\Rightarrow \\frac{AB}{200}=\\frac{\\sqrt{3}}{2} \u00a0\\)<br \/>\n\\( \\Rightarrow AB=\\left( \\frac{200\\times \\sqrt{3}}{2} \\right)m=173.2\\text{ }m. \\)<\/p>\n<p><strong>Example 5: \u00a0 \u00a0<\/strong>The pilot of a helicopter, at an altitude of 1200m finds that the two ships are sailing towards it in the same direction. The angle of depression of the ships as observed from the helicopter are 60\u00ba and 45\u00ba respectively. Find the distance between the two ships.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let B the position of the helicopter and let C, D be the ships. Let AB be the vertical height.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134768\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-7.png\" alt=\"Heights And Distances 7\" width=\"214\" height=\"158\" \/><br \/>\nThen, AB = 1200 m,<br \/>\n\u2220ACB = 60\u00ba and \u2220ADB = 45\u00ba.<br \/>\n\\( \\text{Then, }\\!\\!~\\!\\!\\text{ }\\frac{AD}{AB}=\\text{cot 45 }\\!\\!{}^\\text{o}\\!\\!\\text{ = 1} \\)<br \/>\n\\( \\Rightarrow \\frac{AD}{1200}=1\\Rightarrow \\text{AD = 1200 m} \\)<br \/>\n\\( \\text{And, }\\!\\!~\\!\\!\\text{ }\\frac{AC}{AB}=\\text{cot 60 }\\!\\!{}^\\text{o}\\!\\!\\text{ }=\\frac{1}{\\sqrt{3}}\\text{ } \\)<br \/>\n\\( \\Rightarrow \\frac{AC}{1200}=\\frac{1}{\\sqrt{3}}\\text{ } \\)<br \/>\n\\( \\Rightarrow AC=\\frac{1200}{\\sqrt{3}}=400\\sqrt{3}\\text{ m}\\text{.} \\)<\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/bf8TMKZjDbU?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 6: \u00a0 \u00a0<\/strong>A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30\u00ba and that of the top of the flagstaff is 45\u00ba. Find the height of the tower.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let AB be the tower and BC be the flagstaff.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134784\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-8.png\" alt=\"Heights And Distances 8\" width=\"222\" height=\"196\" \/><br \/>\nThen, BC = 7 m. Let AB = h.<br \/>\nLet O be the point of observation.<br \/>\nThen, \u2220AOB = 30\u00ba and \u2220AOC = 45\u00ba.<br \/>\n\\( \\text{Now, }\\frac{OA}{AC}=\\text{cot 45 }\\!\\!{}^\\text{o}\\!\\!\\text{ }=1 \\)<br \/>\n\u21d2\u00a0\u00a0OA = AC = h + 7.<br \/>\n\\( \\text{And, }\\frac{OA}{AB}=\\text{cot }30{}^\\text{o}=\\sqrt{3} \\)<br \/>\n\\( \\Rightarrow \\frac{OA}{h}=\\sqrt{3}\\Rightarrow OA=h\\sqrt{3} \\)<br \/>\n\u2234 \u00a0\u00a0h + 7 = h\u221a3<br \/>\n\\( \\Rightarrow \\frac{7}{\\sqrt{3}-1}\\times \\frac{\\sqrt{3}+1}{\\sqrt{3}+1}=\\frac{7(\\sqrt{3}+1)}{2}=9.562\\text{ }m \\)<\/p>\n<p><strong>Example 7: \u00a0 \u00a0<\/strong>From the top of a building 30 m high, the top and bottom of a tower are observed to have angles of depression 30\u00ba and 45\u00ba respectively. Find the height of the tower.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let AB be the building and CD be the tower.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134793\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-9.png\" alt=\"Heights And Distances 9\" width=\"217\" height=\"200\" \/><br \/>\nThen, AB = 30 m. Let DC = x.<br \/>\nDraw DE \u22a5\u00a0AB. Then AE = CD = x.<br \/>\n\u2234 \u00a0\u00a0BE = (30 \u2013 x) m.<br \/>\n\\( \\text{Now, }\\frac{AC}{AB}=\\text{cot 45 }\\!\\!{}^\\text{o}\\!\\!\\text{ }=1 \\)<br \/>\n\\( \\Rightarrow \\frac{AC}{30}=1\\Rightarrow AC=30\\text{ }m \\)<br \/>\n\u2234 \u00a0DE = AC = 30 m.<br \/>\n\\( \\frac{BE}{DE}=\\text{tan }30{}^\\text{o}=\\frac{1}{\\sqrt{3}}\\Rightarrow \\frac{BE}{30}=\\frac{1}{\\sqrt{3}} \\)<br \/>\n\\( \\Rightarrow BE=\\frac{30}{\\sqrt{3}} \\)<br \/>\n\\( CD=AE=AB-BE=\\left( 30-\\frac{30}{\\sqrt{3}} \\right) \u00a0\\)<br \/>\n\\( =30\\left( 1-\\frac{1}{\\sqrt{3}} \\right)\\,m \\)<\/p>\n<p><strong>Example 8: \u00a0 \u00a0<\/strong>From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let AB be the cliff and CD be the tower.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134794\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-10.png\" alt=\"Heights And Distances 10\" width=\"239\" height=\"260\" \/><br \/>\nThen, AB = 25 m. From B draw BE \u22a5\u00a0CD.<br \/>\nLet \u2220EBD = \u2220ACB = \u03b1.<br \/>\n\\( \\text{Now, }\\frac{\\text{DE}}{\\text{BE}}=\\text{tan }\\alpha \\text{\u00a0 and\u00a0 }\\frac{\\text{AB}}{\\text{AC}}=\\text{tan }\\alpha \\text{ } \u00a0\\)<br \/>\n\\( \\frac{DE}{BE}=\\frac{AB}{AC}\\text{\u00a0 }So,\\text{ }DE=AB \\) [\u00a0\u2235 BE = AC]<br \/>\n\u2234 \u00a0CD = CE + DE = AB + AB = 2AB = 50m<\/p>\n<p><strong>Example 9: \u00a0 \u00a0<\/strong>The altitude of the sun at any instant is 60\u00ba. Find the height of the vertical pole that will cast a shadow of 30 m.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let AB be the pole and AC be its shadow.<br \/>\nThen, \u03b8\u00a0= 60\u00ba and AC = 30 m.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134795\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-11.png\" alt=\"Heights And Distances 11\" width=\"170\" height=\"144\" \/><br \/>\n\\( \\frac{AB}{AC}=\\text{tan }60{}^\\text{o} \u00a0\\)<br \/>\n\\( \\Rightarrow \\frac{AB}{30}=\\sqrt{3}\\Rightarrow AB=30\\sqrt{3}\\,\\,m \\)<\/p>\n<p><strong>Example 10: \u00a0 \u00a0<\/strong>When the sun is 30\u00ba above the horizontal,\u00a0Find\u00a0the length of shadow cast by a building 50m high.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Let AB be the building and AC be its shadow.<br \/>\nThen, AB = 50 m and \u03b8\u00a0= 30\u00ba.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134796\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-12.png\" alt=\"Heights And Distances 12\" width=\"163\" height=\"145\" \/><br \/>\n\\( \\frac{AC}{AB}=\\text{cot 3}0{}^\\text{o}=\\sqrt{3} \\)<br \/>\n\\( \\Rightarrow \\frac{AC}{50}=\\sqrt{3} \\)<br \/>\n\u21d2\u00a0AC = 50\u221a3\u00a0cm.<\/p>\n<p><strong>Example 11: \u00a0 \u00a0<\/strong>If the elevation of the sun changed from 30\u00ba to 60\u00ba, then find the difference between the lengths of shadows of a pole 15 m high, made at these two positions.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>When AB = 15m, \u03b8\u00a0= 30\u00ba,<br \/>\n\\( \\text{then\u00a0 }\\frac{AC}{AB}=\\text{tan}30{}^\\text{o} \\)<br \/>\n\\( \\Rightarrow AC=\\frac{15}{\\sqrt{3}}m. \\)<br \/>\nWhen AB = 15m, \u03b8\u00a0= 60\u00ba,<br \/>\n\\( \\text{then\u00a0 }\\frac{AC}{AB}=\\text{tan6}0{}^\\text{o} \\)<br \/>\n\u21d2\u00a0\u00a0AC = 15\u221a3 m.<br \/>\n\u2234\u00a0Diff. in lengths of shadows<br \/>\n\\( =\\left( 15\\sqrt{3}-\\frac{15}{\\sqrt{3}} \\right) \\)<br \/>\n\\( =\\frac{30}{\\sqrt{3}}=10\\sqrt{3}\\,\\,m \\)<\/p>\n<p><strong>Example 12: \u00a0 \u00a0<\/strong>The heights of two poles are 80 m and 62.5 m. If the line joining their tops makes an angle of 45\u00ba with the horizontal, then find the distance between the poles.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Let AB and CD be the poles such that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134797\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-13.png\" alt=\"Heights And Distances 13\" width=\"155\" height=\"241\" \/><br \/>\nAB = 80 m and CD = 62.5 m.<br \/>\nDraw DE \u22a5\u00a0AB. Then,\u00a0\u2220EDB = 45\u00ba<br \/>\nNow, BE = AB \u2013 AE = AB \u2013 CD = 17.5<br \/>\n\\(\\frac{DE}{BE}=\\text{cot }45{}^\\text{o}=1\\)<br \/>\n\u21d2\u00a0DE = BE = 17.5 m.<\/p>\n<p><strong>Example 13: \u00a0 \u00a0<\/strong>If the angle of elevation of cloud from a point 200 m above a lake is 30\u00ba and the angle of depression of its reflection in the lake is 60\u00ba, then find the height of the cloud above the lake.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let C be the cloud and C&#8217; be its reflection in the lake.<br \/>\nLet CS = C&#8217;S = x.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134798\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-14.png\" alt=\"Heights And Distances 14\" width=\"204\" height=\"277\" \/><br \/>\n\\( \\text{Now, }\\frac{BC}{AB}=\\text{tan }30{}^\\text{o}=\\frac{1}{\\sqrt{3}} \\)<br \/>\n\\( \\Rightarrow x-200=\\frac{AB}{\\sqrt{3}} \\)<br \/>\n\\( \\text{Also, }\\frac{BC&#8217;}{AB}=\\text{tan }60{}^\\text{o}=\\sqrt{3} \\)<br \/>\n\\( \\Rightarrow ~~x+200=(AB)\\sqrt{3}. \\)<br \/>\n\\( \\sqrt{3}(x-200)=\\frac{x+200}{\\sqrt{3}}\\text{ or }x=\\text{400}. \\)<br \/>\n\u2234 \u00a0CS = 400 m.<\/p>\n<p><strong>Example 14: \u00a0 \u00a0<\/strong>A balloon of radius \u03b3\u00a0makes an angle \u03b1\u00a0at the eye of an observer and the angle of elevation of its centre is \u03b2. Then find the height of its centre from the ground level.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let C be the centre of the balloon and O be the position of the observer at the horizontal line OX. Let OA and OB be the tangents to the balloon so that \u2220AOB = \u03b1, \u2220XOC = \u03b2 \u00a0and<br \/>\nCA = CB = \u03b3.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134800\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-15.png\" alt=\"Heights And Distances 15\" width=\"212\" height=\"218\" \/><br \/>\nClearly, right angled triangles OAC and OBC are congruent.<br \/>\n\u2220AOC = \u2220BOC =\u00a0\u03b1\/2.<br \/>\nLet CN \u22a5\u00a0OX.<br \/>\n\\( \\text{Now,\u00a0 }\\frac{OC}{CA}=\\text{cosec }\\!\\!~\\!\\!\\text{ }\\frac{\\alpha }{2} \u00a0\\)<br \/>\n\\( \\Rightarrow OC=\\gamma \\text{ cosec }\\!\\!~\\!\\!\\text{ }\\frac{\\alpha }{2}\\text{\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 }&#8230;\\text{(i)} \\)<br \/>\n\\( \\text{Also,\u00a0 }\\frac{CN}{OC}=\\sin \\beta\u00a0 \\)<br \/>\n\\( \\Rightarrow CN=OC\\text{ sin }\\beta =\\gamma \\text{ cosec }\\!\\!~\\!\\!\\text{ }\\frac{\\alpha }{2}\\sin \\beta \\text{\u00a0\u00a0 }\\left[ \\text{Using}\\left( \\text{i} \\right) \\right] \\)<\/p>\n<p><strong>Example 15: \u00a0 \u00a0<\/strong>The banks of a river are parallel.\u00a0 A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 45\u00ba and reaches the opposite bank at a point 20 m from the point opposite to the starting point. Find the breadth of the river.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0\u00a0Let A be the starting point and B, the end point of the swimmer.\u00a0 Then AB = 20 m and \u2220BAC = 45\u00ba.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134801\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-16.png\" alt=\"Heights And Distances 16\" width=\"212\" height=\"138\" \/><br \/>\n\\( \\text{Now, }\\frac{BC}{AB}=\\text{sin}45{}^\\text{o}=\\frac{1}{\\sqrt{2}} \\)<br \/>\n\\( \\Rightarrow \\frac{BC}{20}=\\frac{1}{\\sqrt{2}} \\)<br \/>\n\\( \\Rightarrow BC=\\frac{20\\times \\sqrt{2}}{2}=14.14\\text{ }m. \\)<\/p>\n<p><strong>Example 16: \u00a0 \u00a0<\/strong>A man on a cliff observes a fishing trawler at an angle of depression of 30\u00ba which is approaching the shore to the point immediately beneath the observer with a uniform speed. 6 minutes later, the angle of depression of the trawler is found to be 60\u00ba. Caliculate the time taken by the trawler to reach the shore.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0\u00a0Let AB be the cliff and C and D be the two positions of the fishing trawler.<br \/>\nThen, \u2220ACB = 30\u00ba and \u2220ADB = 60\u00ba<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134802\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-17.png\" alt=\"Heights And Distances 17\" width=\"242\" height=\"169\" \/><br \/>\nLet AB = h.<br \/>\n\\( \\text{Now, }\\frac{AD}{AB}=\\text{cot }60{}^\\text{o}=~\\frac{1}{\\sqrt{3}} \\)<br \/>\n\\( \\Rightarrow AD=\\frac{h}{\\sqrt{3}} \\)<br \/>\n\\( \\text{And, }\\frac{AC}{AB}=\\text{cot 3}0{}^\\text{o}=~\\sqrt{3} \\)<br \/>\n\u21d2 AC = \u221a3 h<br \/>\n\\( CD=ACAD=\\left( \\sqrt{3}\\,h-\\frac{h}{\\sqrt{3}} \\right)=\\frac{2h}{\\sqrt{3}} \\)<br \/>\nLet u m\/min be the uniform speed of the trawler.<br \/>\nDistance covered in 6 min = 6u metres.<br \/>\n\\( CD=6u~\\Rightarrow \\frac{2h}{\\sqrt{3}}=6u\\Rightarrow h=3\\sqrt{3}\\,\\,u \\)<br \/>\n\\( Now,\\text{ }AD=\\frac{h}{\\sqrt{3}}=\\frac{3\\sqrt{3}\\,\\,u}{\\sqrt{3}}=3u \\)<br \/>\nTime taken by trawler to reach A<br \/>\n\\( =\\frac{dis\\tan ce\\,AD}{speed}\\Rightarrow A=\\frac{3u}{u}=3\\min \\)<\/p>\n<p><strong>Example 17: \u00a0 \u00a0<\/strong>A boat is being rowed away from a cliff 150m high. At the top of the cliff the angle of depression of the boat changes from 60\u00ba to 45\u00ba in 2 minutes. Find the speed of the boat.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0\u00a0Let AB be the cliff and C and D be the two positions of the ship. Then, AB = 150 m,<br \/>\n\u2220ACB = 60\u00ba and \u2220ADB = 45\u00ba.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134803\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-18.png\" alt=\"Heights And Distances 18\" width=\"131\" height=\"162\" \/><br \/>\n\\( Now,\\frac{AD}{AB}=\\text{cot }45{}^\\text{o}=1 \\)<br \/>\n\\( \\Rightarrow \\frac{AD}{150}=1\\Rightarrow AD\\text{ }=\\text{ }150\\text{ }m. \\)<br \/>\n\\( \\frac{AC}{AB}=\\text{cot 60}{}^\\text{o}=\\frac{1}{\\sqrt{3}}\\Rightarrow \\frac{AC}{150}=\\frac{1}{\\sqrt{3}} \\)<br \/>\n\\( \\Rightarrow AC=\\frac{150}{\\sqrt{3}}=50\\sqrt{3}=\\text{ }86.6\\text{ }m \\)<br \/>\n\u2234 CD = AD \u2013 AC = (150 \u2013 86.6) m = 63.4 m<br \/>\nThus, distance covered in 2 min. = 63.4 m<br \/>\n\u2234 Speed of the boat<br \/>\n\\( =\\left( \\frac{63.4}{2}\\times \\frac{60}{1000} \\right)\\,\\,km\/hr.=\\text{ }1.9\\text{ }km\/hr. \\)<\/p>\n<p><strong>Example 18: \u00a0 <\/strong>\u00a0A tower is 100\u221a3 metres high. Find the angle of elevation of its top from a point 100 metres away from its foot.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let AB be the tower of height 100\u221a3 metres, and let C be a point at a distance of 100 metres from the foot of the tower.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134804\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-19.png\" alt=\"Heights And Distances 19\" width=\"208\" height=\"174\" \/><br \/>\nLet \u03b8 be the angle of elevation of the top of the tower from point C.<br \/>\nIn \u2206CAB, we have<br \/>\n\\( \\tan \\theta =\\frac{AB}{AC} \\\\ \u00a0\\)<br \/>\n\\( \\Rightarrow \\tan \\theta =\\frac{100\\sqrt{3}}{100}=\\sqrt{3} \\)<br \/>\n\u21d2\u00a0\u03b8 = 60\u00ba<br \/>\nHence, the angle of elevation of the top of the tower from a point 100 metres away from its foot is 60\u00ba.<strong>\u00a0<\/strong><\/p>\n<p><strong>Example 19: \u00a0 \u00a0<\/strong>From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30\u00ba. The angle of elevation of the top of a water tank (on the top of the tower) is 45\u00ba. Find the (i) height of the tower (ii) the depth of the tank.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let BC be the tower of height h metre and CD be the water tank of height h<sub>1<\/sub>\u00a0metre.<br \/>\nLet A be a point on the ground at a distance of 40 m away from the foot B of the tower.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134805\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-20.png\" alt=\"Heights And Distances 20\" width=\"159\" height=\"186\" \/><br \/>\nIn \u2206ABD, we have tan 45\u00ba = \\(\\frac { BD }{ AB }\\)<br \/>\n\\( \\Rightarrow 1=\\frac{h+{{h}_{1}}}{40}\\Rightarrow ~~h+{{h}_{1}}=40\\text{ }m~\\text{ }\\text{&#8230;&#8230;}\\left( \\text{i} \\right) \\)<br \/>\nIn \u2206ABC, we have<br \/>\n\\( \\text{tan }30{}^\\text{o}\\text{ }=\\frac{BC}{AB}\\Rightarrow \\frac{1}{\\sqrt{3}}=\\frac{h}{40} \\)<br \/>\n\\( \\Rightarrow h=\\frac{40}{\\sqrt{3}}=\\frac{40\\sqrt{3}}{3}=23.1\\text{ m} \\)<br \/>\nSubstituting the value of h in (i), we get<br \/>\n23.1 + h<sub>1<\/sub>\u00a0= 40<br \/>\n\u21d2 h<sub>1<\/sub>\u00a0= (40 \u2013 23.1)m = 16.9 m<\/p>\n<p><strong>Example 20: \u00a0 \u00a0<\/strong>Two stations due south of a leaning tower which leans towards the north are at distance a and b from its foot. If \u03b1, \u03b2 be the elevations of the top of the tower from these stations, prove that its inclination \u03b8 to the horizontal is given by<br \/>\n\\(\\text{cot }\\theta =\\frac{b\\,\\,\\cot \\,\\,\\alpha -a\\,\\,\\cot \\,\\,\\beta }{b-a} \\)<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let AB be the leaning tower and let C and D be two given stations at distances a and b respectively from the foot A of the tower.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134806\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-21.png\" alt=\"Heights And Distances 21\" width=\"261\" height=\"248\" \/><br \/>\nLet AE = x and BE = h<br \/>\nIn \u2206ABE, we have<br \/>\n\\( \\tan \\theta =\\frac{BE}{AE}\\Rightarrow \\tan \\theta =\\frac{h}{x} \\)<br \/>\n\u21d2 x = h cot \u03b8 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nIn \u2206CBE, we have<br \/>\n\\( \\tan \\alpha =\\frac{BE}{CE} \\)<br \/>\n\\( \\Rightarrow \\tan \\alpha =\\frac{h}{a+x} \\)<br \/>\n\u21d2 a + x = h cot \u03b1<br \/>\n\u21d2 x = h cot \u03b1 \u2013 a \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nIn \u2206DBE, we have<br \/>\n\\( \\tan \\beta =\\frac{BE}{DE} \\)<br \/>\n\\( \\Rightarrow \\tan \\beta =\\frac{h}{b+x} \\)<br \/>\n\u21d2 b + x = h cot \u03b2<br \/>\n\u21d2 x = h cot \u03b2\u2013 b \u00a0 \u00a0 \u00a0 &#8230;.(iii)<br \/>\nFrom equations (i) and (ii), we have<br \/>\nh cot \u03b8= h cot \u03b1\u2013 a<br \/>\n\u21d2 h (cot \u03b1\u2013 cot \u03b8) = a<br \/>\n\\( \\Rightarrow h=\\frac{a}{\\cot \\alpha -\\cot \\theta } \\)<br \/>\nFrom equation (i) and (iii), we get<br \/>\nh cot \u03b8= h cot \u03b2\u2013 b<br \/>\n\u21d2 h (cot \u03b2\u2013 cot \u03b8) = b<br \/>\n\\( \\Rightarrow h=\\frac{b}{\\cot \\beta -\\cot \\theta } \\)<br \/>\nEquating the values of h from equations (iv) and (v), we get<br \/>\n\\( \\frac{a}{\\cot \\alpha -\\cot \\theta }=\\frac{b}{\\cot \\beta -\\cot \\theta } \\)<br \/>\n\u21d2 a(cot \u03b2\u2013 cot \u03b8) = b(cot \u03b1\u2013 cot \u03b8)<br \/>\n\u21d2 (b \u2013 a) cot \u03b8= b cot \u03b1\u2013 a cot \u03b2<br \/>\n\\( \\cot \\theta =\\frac{b\\,\\,\\cot \\alpha -a\\cot \\beta }{b-a} \\)<\/p>\n<p><strong>Example 21: \u00a0 \u00a0<\/strong>If the angle of elevation of a cloud from a point h metres above a lake is \u03b1 and the angle of depression of its reflection in the lake is \u03b2, prove that the height of the cloud is \\(\\frac{h(\\tan \\alpha +\\tan \\beta \\,)}{\\tan \\beta -\\tan \\alpha } \\).<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let AB be the surface of the lake and let P be a point of observation such that AP = h metres. Let C be the position of the cloud and C\u00b4 be its reflection in the lake. Then, CB = C\u00b4B. Let PM be perpendicular from P on CB. Then,<br \/>\n\u2220CPM = \u03b1 and \u2220MPC\u00b4= \u03b2 Let CM = x.<br \/>\nThen, CB = CM + MB = CM + PA = x + h.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134808\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-22.png\" alt=\"Heights And Distances 22\" width=\"230\" height=\"276\" \/><br \/>\nIn \u2206CPM, we have<br \/>\n\\( \\tan \\alpha =\\frac{CM}{PM} \\)<br \/>\n\\( \\Rightarrow \\tan \\alpha =\\frac{x}{AB} \\)<br \/>\n\u21d2 AB = x cot \u03b1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\nIn \u2206PMC\u00b4, we have<br \/>\n\\( \\tan \\beta =\\frac{C\\acute{\\ }M}{PM} \\)<br \/>\n\\( \\Rightarrow \\tan \\beta =\\frac{x+2h}{AB} \\)<br \/>\n[\u2235 C\u00b4M = C\u00b4B + BM = x + h + h]<br \/>\n\u21d2 AB = (x + 2h) cot \u03b2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nFrom (i) and (ii), we have<br \/>\nx cot \u03b1= (x + 2h) cot \u03b2<br \/>\n\u21d2 x(cot \u03b1\u2013 cot \u03b2) = 2h cot \u03b2<br \/>\n\\( \\Rightarrow x\\left( \\frac{1}{\\tan \\alpha }-\\frac{1}{\\tan \\beta } \\right)=\\frac{2h}{\\tan \\beta } \\)<br \/>\n\\( \\Rightarrow x\\left( \\frac{\\tan \\beta -\\tan \\alpha }{\\tan \\alpha \\tan \\beta } \\right)=\\frac{2h}{\\tan \\beta } \\)<br \/>\n\\( \\Rightarrow x=\\frac{2h\\tan \\alpha }{\\tan \\beta -\\tan \\alpha } \\)<br \/>\nHence,<br \/>\nHeight of the cloud = x + h<br \/>\n\\( =\\frac{2h\\tan \\alpha }{\\tan \\beta -\\tan \\alpha }+h \\)<br \/>\n\\( =\\frac{2h\\tan \\alpha +h\\tan \\beta -h\\tan \\alpha }{\\tan \\beta -\\tan \\alpha } \\)<br \/>\n\\( =\\frac{h(\\tan \\alpha +\\tan \\beta \\,)}{\\tan \\beta -\\tan \\alpha } \\)<\/p>\n<p><strong>Example 22: \u00a0 \u00a0<\/strong>There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively 30\u00ba and 45\u00ba, find the height of the tree.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0Let OA be the tree of height h metre.<br \/>\nIn triangles POA and QOA, we have<br \/>\n\\( \\text{tan }30{}^\\text{o}=\\frac{OA}{OP}\\text{ and tan }45{}^\\text{o}=\\frac{OA}{OQ} \\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134810\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-23.png\" alt=\"Heights And Distances 23\" width=\"252\" height=\"197\" \/><br \/>\n\\( \\Rightarrow \\frac{1}{\\sqrt{3}}=\\frac{h}{OP}\\text{ and }1=\\frac{h}{OQ} \u00a0\\)<br \/>\n\u21d2 OP = \u221a3 h and OQ = h<br \/>\n\u21d2 OP + OQ = \u221a3 h + h \u21d2 PQ = (\u221a3+ 1)h<br \/>\n\u21d2 100 = (\u221a3 + 1)h [\u2235 PQ = 100 m]<br \/>\n\\( \\Rightarrow h=\\frac{100}{\\sqrt{3}+1}\\Rightarrow h=\\frac{100(\\sqrt{3}-1)}{2}\\text{ m} \\)<br \/>\n\u21d2 h = 50(1.732 \u2013 1) m = 36.6 m<br \/>\nHence, the height of the tree is 36.6 m<\/p>\n<p><strong>Example 23: \u00a0 \u00a0<\/strong>The angle of elevation of a cliff from a fixed point is \u03b8. After going up a distance of k metres towards the top of cliff at an angle of \u03c6, it is found that the angle of elevation is \u03b1. Show that the height of the cliff is metres<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0Let AB be the cliff and O be the fixed point such that the angle of elevation of the cliff from O is \u03b8 i.e.\u00a0\u2220AOB = \u03b8. Let \u2220AOC = \u03c6 and<br \/>\nOC = k metres. From C draw CD and CE perpendiculars on AB and OA respectively.<br \/>\nThen, \u2220DCB = \u03b1.<br \/>\nLet h be the height of the cliff AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134811\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-24.png\" alt=\"Heights And Distances 24\" width=\"262\" height=\"183\" \/><br \/>\nIn \u2206OCE, we have<br \/>\n\\( \\sin \\phi =\\frac{CE}{OC} \\)<br \/>\n\\( \\Rightarrow \\sin \\phi =\\frac{CE}{k} \\)<br \/>\n\u21d2 CE = k sin \u03c6 \u00a0 \u00a0 \u00a0 &#8230;(i) \u00a0 \u00a0 [\u2235 CE = AD]<br \/>\n\u21d2 AD = k sin \u03c6<br \/>\n\\( \\text{And, cos}\\phi =\\frac{OE}{OC} \\)<br \/>\n\\( \\Rightarrow \\text{cos}\\phi =\\frac{OE}{k} \\)<br \/>\n\u21d2 OE = k cos \u03c6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nIn \u2206 OAB, we have<br \/>\n\\( \\text{tan }\\theta =\\frac{AB}{OA} \\)<br \/>\n\\( \\Rightarrow \\text{tan }\\theta =\\frac{h}{OA} \\)<br \/>\n\u21d2 OA = h cot \u03b8 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(iii)<br \/>\nCD = EA = OA \u2013 OE<br \/>\n= h cot \u03b8 \u2013 k cos \u03c6 \u00a0 \u00a0 \u00a0 \u00a0&#8230;..(iv) \u00a0 \u00a0 \u00a0[Using eqs.(ii) and (iii)]<br \/>\nand, BD = AB \u2013 AD = AB \u2013 CE<br \/>\n= (h \u2013 k sin \u03c6) \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(v) \u00a0 \u00a0 \u00a0 [Using equation (i)]<br \/>\nIn \u2206BCD, we have<br \/>\n\\( \\text{tan}\\alpha =\\frac{BD}{CD}\\Rightarrow \\text{ tan}\\alpha =\\frac{h-k\\sin \\varphi }{h\\cot \\theta -k\\cos \\varphi } \\) [Using equations (iv) and (v)]<br \/>\n\\( \\Rightarrow \\frac{1}{\\cot \\alpha }=\\frac{h-k\\sin \\varphi }{h\\cot \\theta -k\\cos \\varphi} \\)<br \/>\n\u21d2 h cot \u03b1 \u2013 k sin \u03c6 cot \u03b1 = h cot \u03b8 \u2013 k cos \u03c6<br \/>\n\u21d2 h(cot \u03b8 \u2013 cot \u03b1) = k(cos \u03c6 \u2013 sin \u03c6cot \u03b1)<br \/>\n\\( \\Rightarrow h=\\frac{k(\\cos \\varphi -\\sin \\varphi \\cot \\alpha )}{\\cot \\theta -\\cot \\alpha } \\)<\/p>\n<p><strong>Example 24: \u00a0 \u00a0<\/strong>At the foot of a mountain the elevation of its summit is 45\u00ba; after ascending 1000 m towards the mountain up a slope of 30\u00ba inclination is found to be 60\u00ba. Find the height of the mountain.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let F be the foot and S be the summit of the mountain FOS. Then, \u2220OFS = 45\u00ba and therefore \u2220OSF = 45\u00ba. Consequently,<br \/>\nOF = OS = h km(say).<br \/>\nLet FP = 1000 m = 1 km be the slope so that<br \/>\n\u2220OFP = 30\u00ba. Draw PM \u22a5OS and PL \u22a5OF.<br \/>\nJoin PS. It is given that \u2220MPS = 60\u00ba.<br \/>\nIn \u2206FPL, We have<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134812\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-25.png\" alt=\"Heights And Distances 25\" width=\"225\" height=\"219\" \/><br \/>\n\\( \\text{sin }30{}^\\text{o}=\\frac{PL}{PF} \\)<br \/>\n\\( \\Rightarrow PL=PF\\text{ sin }30{}^\\text{o}=\\left( 1\\times \\frac{1}{2} \\right)=\\frac{1}{2}km. \\)<br \/>\n\\( OM=PL=\\frac{1}{2}km \\)<br \/>\n\\( \\Rightarrow MS=OS-OM=~\\left( h-\\frac{1}{2} \\right)km~~~~\\ldots .\\left( i \\right) \\)<br \/>\n\\( \\text{Also, cos }30{}^\\text{o}=\\frac{FL}{PF} \\)<br \/>\n\\( \\Rightarrow FL=PF\\text{ cos }30{}^\\text{o}=\\left( 1\\times \\frac{\\sqrt{3}}{2} \\right)=\\frac{\\sqrt{3}}{2}km \\)<br \/>\nNow, h = OS = OF = OL + LF<br \/>\n\\( \\Rightarrow h=OL+\\frac{\\sqrt{3}}{2} \\)<br \/>\n\\( \\Rightarrow OL=\\left( h-\\frac{\\sqrt{3}}{2} \\right)km \\)<br \/>\n\\( \\Rightarrow PM=\\left( h-\\frac{\\sqrt{3}}{2} \\right)km \\)<br \/>\nIn \u2206PSM, we have<br \/>\n\\( \\text{tan }60{}^\\text{o}=\\frac{SM}{PM} \\)<br \/>\n\u21d2 SM = PM. tan 60\u00ba \u2026..(ii)<br \/>\n\\( \\Rightarrow \\left( h-\\frac{1}{2} \\right)=\\left( h-\\frac{\\sqrt{3}}{2} \\right)\\sqrt{3} \\)\u00a0 \u00a0\u00a0[Using equations (i) and (ii)]<br \/>\n\\( \\Rightarrow h-\\frac{1}{2}=h\\sqrt{3}-\\frac{3}{2} \\)<br \/>\n\u21d2 h(\u221a3 \u2013 1) = 1<br \/>\n\\( \\Rightarrow h=\\frac{1}{\\sqrt{3}-1} \\)<br \/>\n\\( \\Rightarrow h=\\frac{\\sqrt{3}+1}{(\\sqrt{3}-1)\\,(\\sqrt{3}+1)}=\\frac{\\sqrt{3}+1}{2} \\)<br \/>\n\\( =\\frac{2.732}{2}=1.336\\text{ }km \\)<br \/>\nHence, the height of the mountain is 1.366 km.<\/p>\n<p><strong>Example 25: \u00a0 \u00a0<\/strong>The angle of elevation of the top of a tower from a point A due south of the tower is \u03b1 and from B due east of the tower is \u03b2. If AB = d, show that the height of the tower is \u00a0\\(\\frac{d}{\\sqrt{{{\\cot }^{2}}\\alpha +{{\\cot }^{2}}\\beta }} \\)<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong>Let OP be the tower and let A and B be two points due south and east respectively of the tower such that \u2220OAP = \u03b1and \u2220OBP = \u03b2.<br \/>\nLet OP = h. In \u2206OAP, we have<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134814\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-26.png\" alt=\"Heights And Distances 26\" width=\"281\" height=\"228\" \/><br \/>\n\\( \\tan \\alpha =\\frac{h}{OA} \\)<br \/>\n\u21d2 OA = h cot \u03b1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nIn \u2206OBP, we have<br \/>\n\\( \\tan \\beta =\\frac{h}{OB} \\)<br \/>\n\u21d2 OB = h cot \u03b2. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nSince OAB is a right angled triangle. Therefore,<br \/>\nAB<sup>2<\/sup> = OA<sup>2<\/sup> + OB<sup>2<\/sup><br \/>\n\u21d2 d<sup>2<\/sup> = h<sup>2<\/sup> cot<sup>2<\/sup>\u00a0\u03b1 + h<sup>2<\/sup> cot<sup>2<\/sup>\u00a0\u03b2<br \/>\n\\( \\Rightarrow h=\\frac{d}{\\sqrt{{{\\cot }^{2}}\\alpha +{{\\cot }^{2}}\\beta }} \\) \u00a0 \u00a0[Using (i) and (ii)]<\/p>\n<p><strong>Example 26: \u00a0 \u00a0\u00a0<\/strong>The elevation of a tower at a station A due north of it is \u03b1 and at a station B due west of A is \u03b2. Prove that the height of the tower is \u00a0\u00a0\\(\\frac{AB\\sin \\alpha \\sin \\beta }{\\sqrt{{{\\sin }^{2}}\\alpha -{{\\sin }^{2}}\\beta }} \\)<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong>Let OP be the tower and let A be a point due north of the tower OP and let B be the point due west of A. Such that \u2220OAP = and \u2220OBP = Let h be the height of the tower.<br \/>\nIn right angled triangles OAP and OBP, we have<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134815\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-27.png\" alt=\"Heights And Distances 27\" width=\"183\" height=\"187\" \/><br \/>\n\\( \\tan \\alpha =\\frac{h}{OA}\\text{ and }\\tan \\beta =\\frac{h}{OB} \\)<br \/>\n\u21d2 OA = h cot \u03b1 and OB = h cot \u03b2.<br \/>\nIn \u2206OAB, we have<br \/>\nOB<sup>2<\/sup> = OA<sup>2<\/sup> + AB<sup>2<\/sup><br \/>\n\u21d2 AB<sup>2<\/sup> = OB<sup>2<\/sup> \u2013 OA<sup>2<\/sup><br \/>\n\u21d2 AB<sup>2<\/sup> = h<sup>2<\/sup> cot<sup>2<\/sup> \u03b2\u2013 h2 cot<sup>2<\/sup> \u03b1<br \/>\n\u21d2 AB<sup>2<\/sup> = h<sup>2<\/sup> [cot<sup>2<\/sup> \u03b2\u2013 cot<sup>2<\/sup> \u03b1]<br \/>\n\u21d2 AB<sup>2<\/sup> = h<sup>2<\/sup>[(cosec<sup>2<\/sup> \u03b2\u2013 1) \u2013 (cosec<sup>2<\/sup> \u03b1\u2013 1)]<br \/>\n\u21d2 AB<sup>2<\/sup> = h<sup>2<\/sup>(cosec<sup>2<\/sup> \u03b2\u2013 cosec<sup>2<\/sup> \u03b1)<br \/>\n\\( \\Rightarrow A{{B}^{2}}={{h}^{2}}\\left( \\frac{{{\\sin }^{2}}\\alpha -{{\\sin }^{2}}\\beta }{{{\\sin }^{2}}\\alpha {{\\sin }^{2}}\\beta } \\right) \\)<br \/>\n\\( \\Rightarrow h=\\frac{AB\\sin \\alpha \\sin \\beta }{\\sqrt{{{\\sin }^{2}}\\alpha -{{\\sin }^{2}}\\beta }} \\)<\/p>\n<p><strong>Example 27: \u00a0 \u00a0\u00a0<\/strong>An aeroplane when flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60\u00ba and 45\u00ba respectively. Find the vertical distance between the aeroplanes at that instant.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let P and Q be the positions of two aeroplanes when Q is vertically below P and OP = 4000 m. Let the angles of elevation of P and Q at a point A on the ground be 60\u00ba and 45\u00ba respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134816\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-28.png\" alt=\"Heights And Distances 28\" width=\"201\" height=\"200\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-28.png 201w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-28-150x150.png 150w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2019\/04\/Heights-And-Distances-28-100x100.png 100w\" sizes=\"(max-width: 201px) 100vw, 201px\" \/><br \/>\n\\( \\text{tan }60{}^\\text{o}=\\frac{OP}{OA}\\text{ and tan 45}{}^\\text{o}=\\frac{OQ}{OA} \\)<br \/>\n\\( \\Rightarrow \\sqrt{3}=\\frac{4000}{OA}\\text{ and 1=}\\frac{OQ}{OA} \\)<br \/>\n\\( \\Rightarrow OA=\\frac{4000}{\\sqrt{3}}\\text{ and }OQ=OA \\)<br \/>\n\\( \\Rightarrow OQ=\\frac{4000}{\\sqrt{3}}m \\)<br \/>\nIn triangles AOP and AOQ, we have<br \/>\n\u2234 Vertical distance between the aeroplanes<br \/>\n= PQ = OP \u2013 OQ<br \/>\n\\(=\\left( 4000-\\frac{4000}{\\sqrt{3}} \\right)=4000\\frac{(\\sqrt{3}-1)}{\\sqrt{3}}m \\)<br \/>\n= 1690.53 m<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Heights And Distances Angle Of Elevation 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 level, i.e. the case when we raise our head to look at the object. (see fig.) Angle Of Depression The [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[168,169,170,171,146],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Heights And Distances - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/heights-and-distances\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Heights And Distances\" \/>\n<meta property=\"og:description\" content=\"Heights And Distances Angle Of Elevation 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 level, i.e. the case when we raise our head to look at the object. 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