If the angle of elevation from Shreya’s eye changes from 45° to 30°

Question 38 (iii) - Choice (A) If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back. Note: She moves back, so she is going from left to right We need to find distance HF Also, by symmetry HF = GE In right angle triangle AGC tan G = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐺)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐺) tan G = AC/𝐺𝐶 tan 30° = 41/GC 1/√3 = 41/GC GC = 41 √𝟑 In right angle triangle AEC tan E = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑆)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑆) tan E = 𝐴𝐶/𝐸𝐶 tan 45° = 𝐴𝐶/𝐸𝐶 1 = 41/𝐸𝐶 EC = 41 Now, GC = GE + EC Putting values 41√𝟑 = GE + 41 41√3 – 41 = GE 41 (√3 – 1) = GE GE = 41 (√𝟑 – 1) m ∴ The distance she moved back = 41 (√𝟑 – 1) m

Question 38 (iii) (Choice A) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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