send mail to support@abhimanu.com mentioning your email id and mobileno registered with us! if details not recieved
Resend Opt after 60 Sec.
By Loging in you agree to Terms of Services and Privacy Policy
Claim your free MCQ
Please specify
Sorry for the inconvenience but we’re performing some maintenance at the moment. Website can be slow during this phase..
Please verify your mobile number
Login not allowed, Please logout from existing browser
Please update your name
Subscribe to Notifications
Stay updated with the latest Current affairs and other important updates regarding video Lectures, Test Schedules, live sessions etc..
Your Free user account at abhipedia has been created.
Remember, success is a journey, not a destination. Stay motivated and keep moving forward!
Refer & Earn
Enquire Now
My Abhipedia Earning
Kindly Login to view your earning
Support
Type your modal answer and submitt for approval
A man is watching from the top of a tower, a boat speeding away from the tower. The angle of depression from the top of the tower to the boat is 600 when the boat is 80m from the tower. After 10 seconds, the angle of depression becomes 300. What is the speed of the boat? (Assume that the boat is running in still water).
20 m/sec
10 m/sec
16 m/sec
18 m/sec
Let AB is the tower and boat is at points C and D when the angles of depression are 600 and 300 respectively In ΔABC, tan 600= AB/AC ⇒ √3 = AB/80 ⇒ AB = 80√3m Again in ΔBAD, tan 300 = AB/AD ⇒ 1/√3 = 80√3/AD ⇒ AD = 80√3 × √3 = 240m ∴ CD = 240 – 80 = 160m The boat took 10 seconds to cover 160m ∴ The speed of the boat = 160/10 = 16m/s
By: Amit Kumar ProfileResourcesReport error
Access to prime resources
New Courses