 Welcome to another session on trigonometry friends. So this in this session we are going to discuss another problem. I know heights and distances is one of the topics where kids struggle a lot. So the only way to deal with it is getting exposed to multiple types of problems and trying on your own. So even if you're watching this session I would recommend request you to pause the video and see the question try to solve it and I am very sure you'll be able to solve it but just in case if you're not able to you can then go through this session. So here is the solution to this question. The question says from the top of a cliff 200 feet high so the the height of the cliff is 200 feet feet high the angles of depression of the top and bottom of a tower obviously now if you see there's a 200 feet high and it is talking about the angles of depression of the top and bottom of a tower that means some there is a cliff so you have let us say this is a very steep cliff okay and you are here standing over here let us say you are standing on the top of the cliff and it is given that the cliff height from the ground is how much 100 100 sorry 200 feet high now so let us say this is the ground level so this is the ground level and and there is a tower somewhere tower somewhere let us say this tower is this here now obviously the tower will be having a height less than 200 feet why because you are seeing both the the top and the bottom as a depression isn't it so if you see that it has been given that the angle of depression so what is angle of depression you draw a horizontal line from wherever you are watching observing and now try to connect the point which you are observing so let us say this is the point this is the top of the tower let me say this tower is PQ okay this tower is PQ now you are trying to you are trying to see or look at P from let us say this is the height of the tower what is the height of the sorry cliff let the cliff be RS okay so this is RS is the cliff and and PQ is the tower just for you not to be confused I am deleting the diagram of the cliff which I drew earlier right so now this RS represents the cliff now it's given that the angle of depression of the tower's height the top point is 30 degrees isn't it so let us call this as alpha and alpha is 30 degrees and it is also given that the angle of depression of the bottom of the bottom of the tower is how much beta so this angle of depression is beta and this is equal to 60 degrees right now it is it is asked to find the height of the tower let us say and let me also rename RSS capital H and it is given as 200 feet okay so now you will you we will apply the basic trigonometric ratios to find out the height H now what to do first of all if you observe if you observe closely this angle is also nothing but beta why because this is alternate interior angle if this is let's say RT so if you see RT is parallel to QS isn't it so hence angle TRQ is equal to angle RQS and S is definitely 90 degrees right it is perpendicular cliff is always perpendicular to the ground TRQ is RQS and hence but we get we get this is equal to beta is equal to 60 degrees this is one of the first step and now what do you do now you drop a perpendicular on to RS from P let us say this is you this point is you right just to differentiate it from H let me write it let me write it with black so this is P you and H was given to be this is H is 200 feet which is RS okay so hence let me mention that here as well so H is RS is equal to 100 sorry 200 feet okay now why did you why did I drop a perpendicular P you from P on to RS is simply this that one of the right angle triangles I could RQS yeah but I have to use this alpha 30 degree somewhere right now either you could have dropped this perpendicular that is perpendicular from here on to RT or you could have dropped P you perpendicular to RS both would fetch you the same result now what happens if I drop P you as a perpendicular to RS then automatically by the same logic which we adopted for beta this angle is 30 degree alternate interior angle because P you is also parallel to TR so this line is parallel to this line so hence this angle will be equal to this angle what will be this height are you are we are you will be nothing but capital H minus small H yes or no right so hence I can say are you is equal to capital H minus H and why am I doing this because I am interested in triangle are you P because I can see a 90 degree I know an angle alpha so hence I can you and I can use this information and specially when I am when I am you know and I have to find out small H and capital H is given so I would adopt that ratio what ratio should I adopt so I will go for tan of alpha why tan and not sign in cause is simply this because if you use sign in cause then PR this this this length will be used but it is unknown to you isn't it what is known and what is common to both the right angle triangles if you notice there are two right angle triangles triangle P you are and triangle are QS and in both one thing is common that is the length this length QS let it be why so if this is why then P is also why and why is this because PQ SU is a rectangle is it it's a rectangle so this is why so this one will also be why so hence we will exploit this information so tan alpha is nothing but H minus small H upon why simply this right and hence the let us say this is equation number one equation number two is for the next triangle and beta will be nothing but you can guess this is H upon why so at this end divide by opposite is it it so hence this is equation number two now let us see how many unknowns are here so we know H alpha and beta but what are unknown is this small H and this small why so we have two equations two variables so we can eliminate one so what should we eliminate definitely why because that's not needed so hence from one we can say y is equal to H minus H upon tan alpha and from second equation you can say why is equal to capital H upon tan beta isn't it so hence you can equate these two new equations now and you can say which two so now you can equate these two why because both are equal to why so H minus H upon tan alpha will be equal to capital H upon tan beta and what information do I need I need H small H right so hence this is H by tan alpha minus H by tan alpha so I separated the fraction this is equal to H upon tan beta and why am I doing this because I need this H isolated at one place so let us say H upon tan alpha will be given by H upon tan alpha capital H upon tan alpha minus capital H upon tan of beta is it so hence small H will be equal to tan alpha times H upon tan alpha rather you can take H as common as well from both of these so hence you can say H is taken common so H tan alpha one minus tan alpha divide minus one minus one by tan alpha minus one by tan beta sorry this will be beta beta now we know all the given quantity so what is H so now let us deploy a value so if you notice I am deploying the values only in the last step this will eliminate as many calculation mistakes as possible so I don't need to and other advantages if you make a mistake in any one of the steps earlier it will not be translated down under right so it will be it is good enough if you are doing it in terms of variable so tan so H is 200 tan alpha let us see what is alpha value so alpha was 30 and beta was 60 isn't it it's it's given alpha is 30 and beta is 60 so now it is nothing but tan of 30 degrees into one upon tan of 30 degrees minus one upon tan of 60 degrees okay so hence what is the value 200 and tan of 30 is one upon root 3 please remember the values it will help you solve the problems quickly one upon root 3 minus one upon root 3 so hence it is 200 into one upon root 3 and then it is root 3 minus one upon root 3 so hence if you see this is 200 into one upon root 3 then this is if you multiply this will become root 3 square take LCM minus 1 by root 3 isn't it so hence the answer is something like this 1 by root 3 and root 3 square is 3 3 minus 1 is 2 and there is a root 3 so this will be 200 into 2 by 3 so the answer would be 133.33 feet right so quick recap what did we do we drew a diagram representative diagram and deployed or showed all the information in the diagram itself then we tried to find out two right angle triangles where we could utilize the given data so and then we use the concepts of geometry so that you know the put two parallel lines and hence alternate interior angles all these were found out and then finally equations basis the trigonometric ratios were found out and the equations were solved unknown was eliminated and finally the known quantity was arrived at