Question 34 (A) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Basic - Solutions of Sample Papers for Class 10 Boards

Transcript

Question 34 (A) The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 meters, find the height of the chimney. Also, find the length of the wire tied from the top of the chimney to the top of tower.Let tower be AB & chimney be CD Given Height of the tower = 40 m ∴ AB = 40 m Angle of elevation of top of chimney to foot of tower = 60° ∴ ∠ DBC = 60° Angle of depression of foot of chimney from top of tower= 60° ∴ ∠ ACB = 30° We need to find height of building i.e. AB And, length of the wire tied from the top of the chimney to the top of tower i.e. AD In a right angle triangle ABC, tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 30° = 𝐴𝐵/𝐵𝐶 tan 30° = 𝟒0/𝐵𝐶 1/√3 = (" " 40)/𝐵𝐶 BC = 𝟒𝟎√𝟑 In a right angle triangle DBC, tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵) tan B = 𝐷𝐶/𝐵𝐶 tan 60° = 𝑫𝑪/(𝟒𝟎√𝟑) √3 = 𝐷𝐶/(40√3) √3 × 40√3 = DC DC = √𝟑 × 𝟒𝟎√𝟑 DC = 40 × √3 × √3 DC = 40 × 3 DC = 120 m Now, we need to find AD Since ∠ E = 90° And, DE = CD – EC DE = CD – AB DE = 120 – 40 DE = 80 m And, AE = BC = 𝟒𝟎√𝟑 m Now, In right triangle ∆ AED By Pythagoras Theorem AD2 = AE2 + DE2 AD2 = (𝟒𝟎√𝟑)^𝟐+〖𝟖𝟎〗^𝟐 AD2 = (40)^2 ×(√3)^2+〖(40 × 2) 〗^2 AD2 = (40)^2 ×(√3)^2+〖(40) 〗^2 × 2^2 AD2 = (40)^2 ×〖3(√3)〗^2+〖(40) 〗^2 × 2^2 AD2 = (40)^2 × 3+〖(40) 〗^2 × 4 AD2 = (40)^2 [3+4] AD2 = (40)^2 × 7 AD = √((40)^2 × 7) AD = 𝟒𝟎√𝟕 m Thus, length of wire tied from the top of the chimney to the top of tower is 40 √7 𝑚.