Question 35 (b) - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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Question 35 (b) The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45°and 30°. If the ships are 200 m apart and one ship is exactly behind the other then find the height of lighthouse. Let’s draw a diagram B A D 200 m ? 30° P 30° C 45° 45° Let the lighthouse be AB Given that Distance between two ships = CD = 200 m Also, ∠ ADB = 30° & ∠ ACB = 45° We need to find Height of light house i.e. AB Now, in right triangle ∆ ABC tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐶) tan 45° = (𝑨𝑩)/(𝑩𝑪) 1=(𝐴𝐵)/(𝐵𝐶) BC = AB Also, in right triangle ∆ ABD tan D = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒 𝐷)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐷) tan 30° = (𝑨𝑩)/(𝑩𝑫) (1)/(sqrt(3))=(𝐴𝐵)/(𝐵𝐷) BD = BC = AB B A D 200 m ? 30° P 30° C 45° 45° 1=(𝐴𝐵)/(𝐵𝐶) BC = AB Also, in right triangle ∆ ABD tan D = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒 𝐷)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐷) tan 30° = (𝑨𝑩)/(𝑩𝑫) (1)/(sqrt(3))=(𝐴𝐵)/(𝐵𝐷) BD = sqrt(𝟑) AB Putting BD = BC + CD BC + CD = sqrt(3) AB BC + 200 = sqrt(𝟑) AB Putting BC = AB AB + 200 = sqrt(3) AB 200 = sqrt(3) AB – AB 200 = (sqrt(3)−1) AB 200 = (sqrt(3)−1) AB AB + 200 = sqrt(𝟑) AB B A D 200 m ? 30° P 30° C 45° 45° AB + 200 = sqrt(3) AB 200 = sqrt(3) AB – AB (200)/((sqrt(3) − 1)) = AB AB = (𝟐𝟎𝟎)/((sqrt(𝟑) − 𝟏)) Rationalizing AB = (200)/((sqrt(3) − 1)) ×((sqrt(3) + 1))/((sqrt(3) + 1)) AB = (200(sqrt(3) + 1))/((sqrt(3))^(2)− 1^(2)) AB = (200(sqrt(3) + 1))/(3 − 1) AB = (200(sqrt(3) + 1))/(2) AB = 100(sqrt(𝟑) + 𝟏) m ∴ Height of the lighthouse is 100(sqrt(𝟑) + 𝟏) m 200 = (sqrt(3)−1) AB AB + 200 = sqrt(𝟑) AB B A D 200 m ? 30° P 30° C 45° 45° ∴ Height of the lighthouse is 100(sqrt(𝟑) + 𝟏) m