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Ex 9.1, 13 - As observed from top of a 75 m high lighthouse - Ex 9.1

Ex 9.1, 13 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 2
Ex 9.1, 13 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 3
Ex 9.1, 13 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 4

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Ex 9.1 , 13 As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. Given that height of the lighthouse is 75 m Hence, AD = 75 m And angle of depression of first ship is 45° So, ∠ PAC = 45 ° And angle of depression of second ship is 30° So, ∠ PAB = 30 ° We need to find distance between the two ships, i.e. BC Now, Lines PA & BD are parallel And AB is the transversal ∴ ∠ ABD = ∠ PAB So, ∠ ABD = 30° Similarly, Line PA & BD are parallel And AC is the transversal ∴ ∠ ACD = ∠ PAC So, ∠ ACD = 45° Since lighthouse is perpendicular to ground ∠ ADB = 90° In right angled triangle ACD, tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 45° = 𝐴𝐷/𝐶𝐷 1 = (" " 75)/𝐶𝐷 CD = 75m Similarly, In a right angle triangle ABD, tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵) tan 30° = 𝐴𝐷/𝐵𝐷 (" " 1)/√3 = (" " 75)/𝐵𝐷 BD = 75√3 BC + CD = 75 √3 BC + 75 = 75 √3 BC = 75 √3 – 75 BC = 75(√3 – 1) m Hence, the distance between two ships = 75(√3 – 1)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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.