 Hello and welcome to the session, let's work out the following problem. It says two men on either side of a cliff 80 meter high observe the angles of elevation of the top of the cliff to be 30 degree and 60 degree respectively. Find the distance between two men. So let's now move on to the solution. We are given that two men on the either side of a cliff 80 meter high observe that angle of elevation is 30 meter, 30 degrees and 60 degrees respectively. So suppose let cd be the cliff and this is 80 meter high and distance of first men from the cliff be x meter, let ac be x meter and distance of the second man from the cliff is y meter that is bc is y meter. Now in triangle acd we have cd is equal to 80 meter, ac is x meter and this is a right angles triangle. So perpendicular upon base is equal to 10 degrees perpendicular is cd bases ac is equal to 10 30 degrees a cd is 80 meter this is x and and 30 is 1 by root 3. So x is 80 root 3 meter. Now similarly in triangle bcd we have cd is 80 meter bc is y meter and cd upon bc is equal to tan 60 degrees because here the angle of elevation is 60 degrees. So this is tan 60 degrees a cd is 80 vc is y and tan 60 degrees is root 3. So this implies y is equal to 80 by root 3 meter. So this much distance is 80 root 3 and this distance is 80 by root 3. Now we have to find the distance between two men. So therefore distance between two men is equal to ac plus cb that is x plus y meter. Now x is 80 root 3 y is 80 by root 3 meter. So now taking LCM we have 80 into root 3 into root 3 is 3 plus 80 upon root 3 this is again equal to taking 80 common we have 80 into 3 plus 1 upon root 3 80 into 4 is 3 20 upon root 3. Now multiplying and dividing by root 3 we have 320 upon root 3 into root 3 by root 3 we have 320 into root 3 upon 3 which is equal to 184.5 approximately. So the distance between two men is 184.5 meter approximately. So this completes the question and the session. Bye for now good. Have a good day.