 Hello and welcome to the session. Let us discuss the following question, question says, as observed from the top of a 75 meter high lighthouse, from the sea level, the angles of depression of two ships are 30 degrees and 45 degrees. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. First of all let us understand what is angle of depression. Angle of depression of an object viewed is the angle formed by the line of sight with the horizontal when it is below the horizontal level. Now here in this case angle of depression of r from q is equal to angle pqr and this angle is equal to angle qrs that is the angle of elevation of q from r. These two lines are parallel and this is transversal so angle pqr is equal to angle qrs since they are alternate angles. So we can write angle of depression of r from q that is angle pqr is equal to angle qrs that is angle of elevation of q from r. This is the key idea to solve the given question. Let us now start the solution. First of all let us draw a simple diagram to represent the problem. Let here AB denotes the lighthouse. We know height of the lighthouse is equal to 75 meters so we can write let AB is equal to 75 meters denotes the lighthouse. Now C and D represent the position of two ships in the sea and angle of depression of these two ships from top of the lighthouse are 30 degrees and 45 degrees. Now ship at point D has angle of depression from point A 30 degrees or we can say angle pad is equal to 30 degrees clearly we can see this is horizontal line this is line of sight and line of sight is below horizontal so this is angle of depression. So we get angle pad is equal to 30 degrees you also know that from key idea that angle pad is equal to angle adc. We know these two angles are alternate angles so angle adc is equal to angle pad so we can write pad is parallel to pad and ad is transversal this implies angle pad is equal to angle adb. We also know that angle of depression of the ship at point C from point A is equal to 45 degrees or we can say angle PAC is equal to 45 degrees this is the horizontal line this is the line of sight line of sight is below the horizontal so angle PAC is a angle of depression and from key idea we know angle PAC is equal to angle ACB here clearly we can see CB is parallel to PA and AC is transversal so angle PAC is equal to angle ACB so we can write PA is parallel to CB and AC is transversal so angle PAC is equal to angle ACB is equal to angle 45 degrees since they are alternate angles now here also angle pad is equal to angle adb is equal to 30 degrees now in this problem we have to find CD we have to find distance between the two ships so distance between the two ships is CD now first of all let us consider triangle ACB so we can write in right triangle ACB tan 45 degrees is equal to AB upon BC we know tan theta is equal to perpendicular upon base in this triangle AB is perpendicular and BC is base so tan 45 degrees is equal to AB upon BC now we know AB is equal to 75 meters and tan 45 degrees is equal to 1 so substituting corresponding values of tan 45 degrees and AB in this expression we get 1 is equal to 75 upon BC now multiplying both the sides by BC we get BC is equal to 75 meters now let us consider triangle ABD in right triangle ABD we know AB upon BD is equal to tan 30 degrees here we can draw this triangle separately and clearly we can see in right triangle ABD tan 30 degrees is equal to AB upon BD we know tan theta is equal to perpendicular upon base here AB is the perpendicular and DB is the base now we know AB is equal to 75 meters and tan 30 degrees is equal to 1 upon root 3 so substituting corresponding values of tan 30 degrees and AB in this expression we get 1 upon root 3 is equal to 75 upon BD now multiplying both the sides by BD we get BD upon root 3 is equal to 75 now multiplying both the sides by root 3 we get BD is equal to 75 root 3 meters now we have to find CD we can write CD is equal to BD minus BC here we have shown that BC is equal to 75 meters and BD is equal to 75 root 3 meters now substituting corresponding values of BD and BC in this expression we get CD is equal to 75 root 3 minus 75 meters now taking 75 common on right hand side we get 75 multiplied by root 3 minus 1 meters is equal to CD so required distance between the two ships is equal to 75 multiplied by root 3 minus 1 meters so this is our required answer this completes the session hope you understood the solution take care and have a nice day