Examples

Chapter 9 Class 10 Some Applications of Trigonometry
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Example 6 The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30 and 45 , respectively. Find the height of the multi-storeyed building and the distance between the two buildings. Height of tall building (AB) = 8m Let, Height of the multi-storeyed building = PC And, distance between two building = AC Angle of depression to top of building = QPB = 30 Angle of depression to bottom of building = QPA = 45 We have to find height of multi-storeyed building i.e. PC And distance between two buildings i.e. AC Draw BD parallel to AC & PQ Lines PQ & BD are parallel, And BP is the transversal PBD = QPB PBD = 30 Now, AC and BD are parallel lines. So, AC = BD Similarly, AB and CD are also parallel lines. So, CD = AB CD = 8 metre Also, Since PC is perpendicular AC PDB = PCA = 90 From (1) & (2) PD 3 = PC PD 3 = PD + DC PD 3 = PD + 8 PD 3 PD = 8 PD ( 3 1) = 8 PD = 8/(( 3 1) ) Rationalizing PD = 8/(( 3 1) ) (( 3 + 1))/(( 3 + 1) ) PD = 8( 3 + 1)/( 3 1)( 3 + 1) PD = 8( 3 + 1)/(( 3)^2 1^2 ) PD = 8( 3 + 1)/(3 1) PD = 8( 3 + 1)/2 PD = 4( 3 + 1) Now, PC = PD + DC = 4( 3 + 1) + 8 = 4 3 + 4 + 8 = 4 3 + 12 = 4( 3 + 3) So, height of multi-storeyed building = PC = 4( 3 + 3) m From (1) BD = PD 3 BD = 4( 3 + 1) 3 BD = 4 3 ( 3 + 1) BD = 4 3 " " 3 + 4 3 " " 1 BD = 4(3) + 4 3 BD = 12 + 4 3 BD = 4(3 + 3) m So, AC = BD = 4(3 + 3) m Distance between two buildings = AC = 4(3 + 3) m

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.