Question 27 (OR 1 st question)

A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°,  how long will the car take to reach the observation tower from this point?     1. Class 10
2. Solutions of Sample Papers for Class 10 Boards
3. CBSE Class 10 Sample Paper for 2019 Boards

Transcript

Question 27 (OR 1st question) A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point? Let the man & tower height be AD Given that man sees car first at an angle of depression of 30° So, ∠ PAB = 30° After 12 minutes, the man sees car at an angle of depression of 45° So, ∠ PAC = 45° Now, tower is vertical, So, ∠ ADB = 90° Also, Lines PA & BD are parallel And AB is the transversal ∠ ABD = ∠ PAB (Alternate Angles) So, ∠ ABD = 30° Lines PA & BD are parallel And AC is the transversal ∠ ACD = ∠ PAC (Alternate Angles) So, ∠ ACD = 45° In right angle triangle ACD tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 45° = AD/CD 1 = 𝐴𝐷/𝐶𝐷 CD = AD AD = CD In right angle triangle ABD tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵) tan 30° = (" " AD)/BD (" " 1)/√3 = 𝐴𝐷/𝐵𝐷 BD/√3 = AD AD = BD/√3 From (1) & (2) CD = 𝐵𝐷/√3 √3CD = BD √3CD = BC + CD √3CD – CD = BC CD (√3 – 1) = BC CD = 𝐵𝐶/((√3 − 1) ) CD = 𝐵𝐶/((√3 − 1) ) × ((√3 + 1))/((√3 + 1) ) CD = 𝐵𝐶(√3 + 1)/(((√3)^2 − 1^2 ) ) CD = 𝐵𝐶(√3 + 1)/((3 − 1) ) CD = 𝐵𝐶(√3 + 1)/2 Given that Time taken to cover BC = 12 minutes Time taken to cover 𝐵𝐶(√3 + 1)/2 = 12 × ((√3 + 1))/2 minutes = 6 (√3 + 1) minutes ∴ Time taken to cover CD = 6 (√3 + 1) minutes Hence, it takes 6 (√𝟑 + 1) minutes to reach to the foot of the tower.

CBSE Class 10 Sample Paper for 2019 Boards

Class 10
Solutions of Sample Papers for Class 10 Boards 