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Example 6 - The angles of depression of top and bottom - Questions easy to difficult

  1. Chapter 9 Class 10 Some Applications of Trigonometry
  2. Serial order wise
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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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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