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Ex 9.1, 12 - From the top of a 7 m high building, angle - Questions easy to difficult

Ex 9.1, 12 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 2
Ex 9.1, 12 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 3 Ex 9.1, 12 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 4

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Ex 9.1 , 12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. Let building be AB & tower be CE Given height of building = AB = 7m From the top of building, angle of elevation of top of tower = 60°. Hence, ∠EAD = 60° Angle of depression of the foot of the tower = 45° Hence, ∠CAD = 45° We need to find height of tower i.e. CE Since AB & CD are parallel, CD = AB = 7 m Also, AD & BC are parallel So, AD = BC Since tower & building are vertical to ground ∠ ABC = 90° & ∠ EDA = 90° Now, AD & BC are parallel, taking AC as transversal ∠ ACB = ∠ DAC ∠ ACB = 45° In right angle triangle ABC, tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 45° = 𝐴𝐵/𝐵𝐶 1 = 𝐴𝐵/𝐵𝐶 1 = (" " 7)/𝐵𝐶 BC = 7m Since BC = AD So, AD = 7m Now, In a right angle triangle ADE, tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan 60° = 𝐸𝐷/𝐴𝐷 √3 = 𝐸𝐷/𝐴𝐷 √3 = 𝐸𝐷/7 7√3 = ED ED = 7√3 Height of tower = ED + DC = 7√3 + 7 =7(√3 + 1)m

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.