# EX 9.1, 9

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.1 , 9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Let building be AB & tower be CD Given Height of the tower = 50 m Hence, CD = 50m Angle of elevation of top of building from foot of tower = 30° Hence, ∠ACB = 30° Angle of elevation of top of tower from foot of building = 60° Hence, ∠DBC = 60° We need to find height of building i.e. AB Since building & tower are perpendicular to ground ∠ ABC = 90° & ∠ DCB = 90° Similarly, In a right angle triangle ABC, tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 30° = (" " 𝐴𝐵)/𝐵𝐶 (" " 1)/√3 = (" " 𝐴𝐵)/((" " 50)/√3) (" " 1)/√3 = AB × √3/50 (" " 1)/√3 × 50/√3 = AB AB = (" " 1)/√3 × 50/√3 AB = (" " 50)/3 m Hence, height of building = (" " 50)/3 m

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .