ARITHMETIC ALGEBRA GEOMETRY QUADRATIC function STRAIGHT LINES
RIGHT ANGLED TRIANGLE & COORDINATE GEOMETRYAbove we have the standard right angled triangles which you are already familiar with in Lesson 1 and Lesson 2. Now we will put this right angled triangle in the Coordinate Geometry space of the x and y axes.
Coordinate Geometry is nothing but using the x axis and the y axis to correctly show the location of each and every point in space. The area is divided into the four quadrants and any point is shown as (a,b) where a is the measure in the horizontal direction and b is the measure in the vertical direction. So, look at the diagram below which places the right angled triangle amidst the coordinate axes The triangle above must be very familiar to you by now. The hypotenuse is called r here (but you can call it by any other name too). And the two sides are thus rCos x and rSin x and so the coordinates of the point A is simply (a,b) or we can also write it as A (rCos x, rSin x). Do you see how simple it is. And do you see how trigonometry and geometry are so smoothly connected with Coordinate Geometry. Two further things here is that, tan x is simply b divided by a tan x = b / a and we also have the Pythagoras Theorem r^2 = a^2 + b^2 as shown in the picture above I will give you a task now. In the Pythagoras Theorem, replace a by rCos x and b by rSin x and see what you find.
Now the above diagram looks complicated. But the trick to deal with complex matter is to break it down into simpler elements. And what simple element can you see? Yes, Right angled triangle, which you know so well now.
Do you see that there are THREE small right angled triangles inside the above picture with hypotenuse p, q & r. Now, as you know, as soon as you know the hypotenuse and one of the angles, you can immediately find out the two sides. Here in all the three triangles, the angle is same: x. Why? This comes from the parallel line theorem which you learnt in junior classes. So, the hypotenuse being p,q & r, the sides of the triangles are known e.g. for the first triangle the sides are pCos x and pSin x and so on and so forth. And that gives the coordinates of A as (pCos x, pSin x). Now, how to find the coordinates of B and C. For that you must notice that there are two more larger triangles in the picture, one with the hypotenuse as p+q and the largest triangle with hypotenuse as p+q+r. Do you see? So, that is how, as shown in the diagram, you can find the coordinates of B and C immediately. Do you see how the complex has become simple? We will have another look at the same diagram below in a slightly different way that will give you more insight and control. |
Finally, we look at another way to look at the same diagram. In the diagram above, B has been given as (a,b). Now, again what you should do is to locate a right angled triangle that you can work with. Since B is given, let us find the triangle which has B in it.
Which is the triangle that has a and b as its sides? And what is its hypotenuse.
Do you see that the answer is p+q. And as you know by now very well, you can immediately write
a=(p+q)Cos x and b=(p+q)Sin x
Now I want you to do another thing. Come down from B to A. Although the route from B to A is on the slanted line, as per Coordinate Geometry, you are vertically coming down as well as horizontally sliding to the left, is it not? And how do you calculate that. Yes, by right angled triangle with side q, because as you climb down from B to A you traverse a slanted length of q. And thus, by right angled triangle with hypotenuse q, you come down vertically by qSin x and slide leftwards by qCos x, right? Do you get it?
Thus, the coordinates of A is (a – q Cos x, b – q Sin x)
Now your task is to prove that the coordinates of C will be (a + r Cos x, b + r Sin x)
Good to go? Right.