Ex 6.5, 11
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 11/2 hours?
Given :
Speed of north flying aeroplane = 1000 km/hr.
Speed of west-flying aeroplane = 1200 km/hr.
To find: Distance between the two planes
after 1.5 hours , i.e. , BC
Solution:
We know that
Speed = (π·ππ π‘ππππ )/ππππ
Distance = speed Γ time
Now, we have to find distance BC
Since North and West are perpendicular,
β BAC = 90Β°
So, Ξ ABC is a right triangle
Using Pythagoras theorem in right angle triangle ACB
(Hypotenuse)2 = (Height)2 + (Base)2
(BC)2 = (AB)2 + (AC)2
(BC)2 = (1500)2 + (1800)2
(BC)2 = 2250000 + 3240000
(BC) = β5490000
= β(3Γ3Γ61Γ100Γ100)
= β((3)^2Γ(100)^2 )Γβ61
= 3Γ100β61
= 300 β61 km
Hence, two planes would be 300β61 km from each other.

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.