The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

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Question 28 The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower. Let building be AB & tower be CD Height of building = AB = 50 m Let, Height of tower = PC And, distance between tower and building = AC Angle of depression to top of building = ∠ QDB = 30° Angle of depression to bottom of building = ∠ QDA = 60° We have to find height of tower i.e. CD And distance between tower &building i.e. AC Draw BE parallel to AC & DQ Lines DQ & BE are parallel, And BD is the transversal ∠ DBE = ∠ QDB ∠ DBE = 30° Now, AC and BE are parallel lines. So, AC = BE Lines DQ & AC are parallel, And AD is the transversal ∠ DAC = ∠ QDA ∠ DAC = 45° Similarly, AB and CE are also parallel lines. So, CD = AB CD = 50 m Also, Since DC is perpendicular AC ∠ DEB = ∠ DCA = 90° In right angle triangle DBE tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐵)/(Side 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵) tan B = (" " 𝐷𝐸)/𝐵𝐸 tan 30° = 𝐷𝐸/𝐵𝐸 1/√3 = (" " 𝐷𝐸)/𝐵𝐸 BE = DE √3 In right angle triangle PAC tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan 60° = (" " 𝐷𝐶)/𝐴𝐶 √3 = (" " 𝐷𝐶)/𝐴𝐶 AC = (" " 𝐷𝐶)/√3 "BE = " (" " 𝐷𝐶)/√3 From (1) & (2) DE √3 = (" " 𝐷𝐶)/√3 DE√3 × √3 = DC 3DE = DC 3DE = DE + EC 3DE – DE = EC 2DE = EC 2DE = 50 DE = 50/2 DE = 25 m ∴ Height of tower = DC = DE + EC = 25 + 50 = 75 m From (1) BE = DE √3 Putting DE = 25 BE = 25√3 m So, AC = BE = 25 √3 m ∴ Distance between two buildings = AC = 25√𝟑 m

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