 Hello friends welcome again to a new session on problem-solving on trigonometry Again in this session. We are going to take up a problem in heights and distances now unlike previous sessions where We have solved problems related to trigonometry and application of trigonometry in one plane. This problem Involves trigonometry in two different planes. In fact, it is a three-dimensional problem So let us try and understand what this problem says and how we will solve this problem So it is given that a man observes that at a point due south of a certain tower Its angle of elevation is 60 degrees So let us first of all try and draw a representative diagram. So let us say this is a trial and This is my ground. Okay This is the ground and let us say this point is Is P and P in this queue, right? So PQ is the tower So PQ is the tower Okay And let us say a point R is due south as it is mentioned and now don't go by the actual positioning of the point so Usually south will be you know in the in the downward direction But now we are you know for the convenience of or to represent it in three-dimension We are taking R and let us say R is south to Q and Its angle of elevation here is given as 60 degrees Okay, so this is 60 degrees now this guy moves 300 feet due west so if this is let us say if this direction so I'm redrawing the directions So let us say for this purpose this particular question if this is direction of south then west is definitely in this direction Isn't it? So hence he moves. What does he do? He moves 300 feet due west So let us say he is moving now. Let us say this is 300 feet Okay, and to Denote it in 3d. Let me draw it in a slant position. So let us say this is 300 feet now don't get confused now I am drawing this angle with black now this angle is 90 degree now the black line is on the ground Okay, and RP is not on the ground RP is in the vertical plane Okay, now from this point the angle of elevation. So hence I if I have to draw this Let us say now P Q and R and this point is let us say S and now SP Okay, now this angle SP. So SQ please remember SQ and RQ R on the same plane R on the ground Okay, R on the ground and if he moves from this and let us say RS is is equal to 300 feet and it's given That now the angle of elevation now the angle of elevation is How much 30 degrees right? So let us say this angle was beta and this is angle 30 degrees and This is alpha which is 60 degrees. Okay, you have to find out Height of the bar that is what is PQ height. So let us say height is capital H So this is unknown. So we have to find H. Okay. I hope the figure is clear So I will do one more thing. I will just shade this Ground surface so that you can You can you know understand that this is the ground surface. So this triangle is on the ground and SP and SP so this is the triangle on the ground is it is lying on the horizontal ground. Okay, and R Q P and SP Q so triangle triangle R Qp and triangle R sorry not our P Triangle PSQ R in R in vertical plane R in vertical plane Okay, so this is the diagram. So now we have to find out again and there are a few more things you can you can mention So this angle is also 90 degrees so angle PQ R is in 90 degrees and also this angle that is Angle PQ S is also 90 degrees now you can imagine, you know, what you can do is take your notebook and along the Middle crease right the the fold where the or the edge on which all the papers are affixed So you can imagine you can you can you can just you know Open the notebook and put it on the table such that The plane of the notebook is perpendicular to the notebook Okay, so when you see there are two folds to two ends of the notebook or two cover pages of the notebook will be having some angle between Then but both of these cover pages are fixed are attached at one particular hinge and both these pages are in the vertical or both these cover pages are in the vertical plane and And so what I'm trying to say is so this is the this is let us say this is your notebook Where this is one of the pages and the other pages Something like that Okay, so this is the same Same scenario here and now you are joining the diagonals in both the both the Pages both the cover pages So this one is one diagonal and another one is this diagonal and these angles are given So this is similar condition and you have to find out This height right so similar to this. I hope you could understand what I'm trying to say now Let us solve this problem now. So basically what is given RS is given and what is RS RS is Let us say this is X so I'm mentioning this as X and X is equal to 300 feet 300 feet 300 feet. Okay. Now we have to find out what edge so obviously we will be using all the information which is which are known and we will try to use the Triangles as the info and and the related information. So hence guys if you see How can we find out it so you you can find out any of these Sides like RQ or RS or QS rather if you know QS, let us say, okay, and Or RS RQ, let us say then I can use any of the two vertical triangles to find out The height edge because the angle of elevations are given now How to approach this problem? So clearly what all information is given? we have been given only one X that is RS and We know that triangle SRQ is a right is a right triangle is a right triangle. Why because SR is perpendicular to RQ because if you remember the guy moved due west from south So due west from south if you see this is this is how it would be So let us say if this was south and he moved due west from south So hence he will make an angle of 90 degrees is it so hence SR is perpendicular to RQ Now and SR is given now Let us try and express SQ in terms of H and RQ in terms of H and then we can use Pythagoras theorem or the information in the triangle RSQ, okay So consider a triangle PQS. Can I write in this PQS? H upon SQ So can I write H upon SQ is equal to tan of beta Isn't it? Tan of beta so hence SQ can be written as H upon tan beta is equal to H upon tan beta is equal to H cot beta I can write that similarly in Triangle see which one PRQ PRQ PRQ again angle Q is 90 degrees. So hence I can write Tan of alpha this angle this angle tan of alpha is nothing but opposite opposite to tan alpha is PQ Opposite to alpha. Sorry is PQ and adjacent is RQ and We know PQ is nothing but H and let us say RQ. So can I not find out RQ from here? So RQ is H upon tan alpha Is equal to H cot Alpha why because one upon tan alpha is cot alpha now in the triangle SQ R what do I know in triangle triangle SQ R? What do I know? Angle R is 90 degrees. Isn't it? Yes, then what can I say SQ square? That is hypotenuse square. So please take care of this SQ square is SR square So I'm writing it here and then I'll take it there SQ square is equal to SR square plus RQ square, right? Why because this angle R is 90 degrees. So SQ square. So SQ square is equal to I'm writing again SR square Plus RQ square Now what is SQ in terms of H? We just found out it is nothing but H cot beta So hence I can write H square cot square beta because it's a square term is equal to SR What is SR? SR is if you see SR in the figure is X So I can write X square plus RQ square. What is RQ? So if you see RQ is H cot alpha So H cot alpha Okay, so hence now. Sorry. This is a square term, right? So hence what do you do rearrange H square and Then it is writing. We are writing cot square beta minus cot square alpha You take this term H square alpha on the left-hand side and take H square is common is equal to X square. So what is H square guys? So H square will be nothing but X square upon cot square beta minus cot square Alpha, is it it? So what is H then? So H will be nothing but under root X square by cot square beta minus cot square alpha, right? What is The value round. Let us calculate X was 300 feet. Isn't it? Let us see. Yes, 300 feet alpha was 60 and beta was 30 So I will write 300 square divided by cot square beta was 30 so 30 degrees minus cot square 60 degrees Okay, and then It is nothing but 300 Will come out of the root and cot 30 is nothing but root 3 So root 3 square minus cot 60 is 1 by root 3. So This so hence answer is 300 divided by 3 minus 1 by 3 Hence it is nothing but 300 by under root 9 minus 1 by 3 which is equal to 300 times root of 3 by 8 if you simplify you'll get 300 into root of 3 by 8 and this if you calculate the value comes out to be 183.71 feet Okay, so this is what is the Height of the tower edge, right? And it was also the second part of the question asked us to find out what the original distance from it Original distance of the man from it that is RQ. This was the original distance, isn't it? So let us find out RQ So what is RQ guys RQ if you see if you see we have calculated RQ is H cot alpha from here if you see this is H cot alpha So H H we just found out H is 183 0.71 cot of alpha is cot of 60 degrees Isn't it? It's and if you calculate this 183.71 into cot 60 is root. Sorry 1 by root 3 1 by root 3. Yes So hence the value will come out to be 106.06 feet This is the calculation now in this question one thing is noteworthy And that is as you are moving away from the point from our point towards the west, right your angle of inclination of the point P is going down, right? So the at point R the angle of inclination was 60 degrees at point S which is let us say some distance away from R in the west direction the Value gets decreased so as you move away So let us say if you keep on walking on this west direction in this west direction Your angle of elevation will keep on dropping dropping right? This is an important learning of on in this problem Okay, so I hope you understood the solution. So hence this problem was a little different because it involved trigonometry in two planes two planes and and but then if you are in command of the figure and if you run in if you have understood the question properly, you will be able to find out the The desired result always keep in mind that the desired result has to be in terms of known so try to You know eliminate all the unknowns and hence eventually you will get the desired result