Question

Suraj suraj and arjun take a straight route to the same terminal. And travel with a constant speech at the initial moment the positions of 2 and terminal points from an equilateral triangle when aashan cover a distance of 80 km the triangle becomes right angles when are there was a distance of 120 kilometres from the terminal. The suraj arrived at the. Find the distance between them at the initial momentum fuming that there are integral distances throughout the moments described

Answer 1

Answer:

initial Distance = 240 km

Step-by-step explanation:

Let say initial Distance = D km

suraj  Covered  D  km   When  Arjun covere  D - 120 km

their Speed ratio Surah : arjun ::  D : D - 120

Let say their speeds DK  & (D - 120)K

When arjun covered 80 km

Arjun Distance remained = D - 80 km

arjun covered 80 km  => Suraj Covered  = 80D/(D - 120)

Suraj Distance Remained = D - 80D/(D - 120)

= (D² - 120D - 80D)/(D - 120)

= (D² - 200D)/(D - 120)

Cos 60  =   (Suraj remaining distance )/(arjun remaining distance)

=> 1/2  =   {(D² - 200D)/(D - 120)} / (D - 80)

=> (D - 80)(D - 120) = 2(D² - 200D)

=> D² - 200D + 9600 = 2D² - 400D

=> D² - 200D  - 9600 = 0

=> D² - 240D + 40D - 9600 = 0

=> D (D - 240) + 40(D - 240) = 0

=> D = 240

initial Distance = 240 km