 Hi, I'm Zor. Welcome to Inizor Education. I would like to solve a few trigonometric problems related to geometry, basically continuation of the topic. Obviously, as usually, I do recommend you to try to solve these problems yourself first, because that's the whole purpose. So you will be actually solving the problems, and only then you'll listen to this lecture. So if you didn't do it yet, just press the pause button and try to engage yourself into thinking about these problems. So, okay, so let's not do anything, and let's go straight to the problems. Okay, problem number one, if you have a triangle, vertices are A, B, C, angles are alpha, beta, gamma, and three A, and three altitudes. The altitudes are HA, because it falls on the side A, which is this one. This is B, opposite to the vertex B, and this is HB, and this is HC falls to the side C. So, any triangle, and I would like to prove the following property. HA times sine of alpha equals HB sine beta equals HC sine gamma. Well, all problems today will be relatively simple. Now, looking at that, I don't know, it reminds me immediately the law of sines, so something should be related to the law of sines. Now, what is the law of sines? It's A over sine alpha equals B over sine beta equals C over sine gamma, right? At the same time, we know that A times the altitude, which is falling to A. It's the same as BHB, the same as CHC. This is double area of the triangle, right? The base times altitude divided by two, or the base times altitude divided by two, or this base times altitude divided by two. All these are errors, so side times the altitude towards this side is double area. Now, from these two, I can very easily derive this one. How? Well, for instance, HA equals 2S divided by A, HB is equal to 2S divided by B, and HC is equal to 2S divided by C, and if I will substitute in this, I will have exactly the law of sines with the multiplier S everywhere, right? So, since HA is 2SS divided by A, I have this, right? Now, this is 2S over A sine B, and this is 2S over B sine gamma. Now, these are equal, right? Because S is just a multiplier everywhere, and this is the reverted, or reversed, but they're inverted law of sines. Law of sines is A divided by sine of alpha this is B, I'm sorry, and this is C. So, this is inverted law of sines, basically, because law of sines is A divided by sine of alpha, B divided by sine of beta, and C divided by sine of gamma, and this is just inverted. So, these are equal, then inverted are equal as well. So, this is basically the proof. We start from the law of sines, we can invert it, multiply by 2S, and we got this one. So, everything is, like, irreversible in both directions. Okay, that's it for number one. Number two, given a parallelogram with a cute angle alpha, we have to prove that 4 times A B times A D times cosine alpha equals to A C square minus B D square. Now, A C square is a diagonal, B D is another diagonal. So, the difference of squares of diagonals, the bigger minus smaller. It's 4 times A B times A D and cosine. Well, whenever you see something like product of two sides and the cosine between them, or the angle between them, that actually reminds the law of cosines, right? Now, what is the law of cosines? The applicable to triangle A B D law of cosines is, well, let me just call this thing, I will call, let's say, P, this thing I will call Q, just to make it easier, right? So, A B is P and this is Q, alright? This is X and this is Y, X, Y. And the triangle A B D, Y square equals P square plus Q square minus 2 P Q cosine alpha, right? That's the law of cosines. Now, from triangle A B C, X square equals P square. Now, this is parallelogram, which means this is also Q, right? So, P square plus Q square minus 2 P Q cosine. And what is this angle? Well, this angle is pi minus alpha, right? 180 degree minus alpha. So, this is pi minus alpha. Cosine of pi minus alpha equals 2 watt. Cosine is abscissa, right? So, if this is the unit circle, this is alpha, this is pi minus alpha. Abscissa is the same by absolute variable but with a minus sign. So, it's minus cosine alpha. So, that's why this is equal to P square plus Q square plus 2 P Q cosine of alpha, right? Now, what is AC square minus BD square, X square minus Y square? If I will subtract from this this, P square will be reduced, Q square will be reduced. This is plus 2 P Q cosine. This is minus but I'm subtracting so it will be plus. So, it will be 2 P Q cosine minus minus 2 P Q which is 4 P Q cosine of alpha. That's exactly what this is. That's a very simple problem. By the way, for the previous two cases, two previous problems, the first one contains signs and it kind of implied that the law of signs should be applicable and the second one contains the product of two sides by the cosine of the angle between them and again, this prompts for law of cosines and indeed, in both cases, these two theorems did help. Okay, now, let's say you have any triangle. So, this is A, this is B and this is angle bisector. So, these two angles are the same and each one is equal to gamma over 2, right? Since it's a bisector, this is angle gamma and this is LC, right? The bisector. Now, we have to prove that LC times A plus B equals 2AB cosine gamma over 2. So, bisector times sum of the sides in between them, this bisector actually goes, is equal to 2AB and cosine. All right, now, actually, I'm sure there are more in one way to prove this particular equality. The one which I have chosen is let's just have these two perpendicular as altitudes of triangles ADC and BDC. And what I will do is I will compare some of the areas of these two triangles with the area of an entire triangle. Now, what is the area of ADC in terms of the base and an altitude? Well, the base is B. Altitude from this triangle DCM. It's the right triangle. Hypotenuse is LC. The acute angle is gamma over 2. So, I know the DM, the altitude in this triangle. And it's a cadetus in the right triangle GMC. It's equal to hypotenuse times sine of gamma over 2. So, my area of triangle ACG equals ACG equals B times altitude, which is LC times sine of gamma over 2 and divided by 2. Right? Now, area of triangle BCG equals, again, one-half of the base, which is A times LC times DM is exactly the same. It's also LC times sine of gamma over 2. On the other hand, the area of triangle ABC, the entire triangle is, let's think. Well, if I will take this particular altitude, so it would be B one-half B and this particular altitude is A times sine of the gamma, right? From the triangle BPC, BP is a cadetus, BC is hypotenuse, lengths of A and the angle is gamma, so the opposite cadetus. Now, sum of these two is equal to this, right? Well, therefore, well, let's just forget about one-half. If I will sum these two, I will get LC times A plus B sine of gamma over 2, right? If I will sum these two and I drop one-half because it's reduced, it will be reduced anyway. And on the right side, I have this, well, we are close to this one, but just need a little bit more. Now, obviously, sine of gamma, it's 2 of gamma over 2, right? So it's gamma over 2 plus gamma over 2, double gamma over 2. So it's 2AB sine gamma over 2 and cosine gamma over 2, right? That's what sine of gamma is. It's 2 sine gamma over 2, cosine gamma over 2. I still remember the formula for the sine and the cosine of the sum of two angles, right? This is gamma over 2 plus gamma over 2, that's where it is. And from here, obviously, we can reduce by sine of gamma over 2 and we get exactly this. That's it. Now, what was interesting about this particular problem? I was using additional construction. I put all the altitudes and the concept of area. I don't know what exactly prompted me to do this. However, we do have to include this bisector, LC, into some kind of an equality, right? From which we can derive the equality between it. And, you know, sum of two areas of the small triangle is equal to the area of the big triangle is really an equality. And I'm sure there are some other components of this triangle that can be connected together with some kind of an equation. Well, again, the problem can have many different solutions. This is just one of them and it seems to me it's relatively simple because all I need to know is that the catch-up is equal to hypotenuse times the sine of the angle or something. All right. Next. Next, we have a nice socialist triangle, ABC, and we have altitudes. One altitude, another altitude, and another altitude. So these are right angles. Now, this angle is given. Let's call it gamma. So gamma is given. Now, what I have to find out is now H is the intersection of these three altitudes. It's called orthocenter, by the way, of the triangle. I have to find out in what ratio point H divides the altitude from C to F. So basically I have to find out the ratio HF divided by CH. All right? It's actually quite easy. First of all, this is a socialist triangle, right? So this altitude is everything. It's altitude, it's the angle bisector, it's median of this side, and it's basically the axis of symmetry, right? The whole picture is completely symmetrical because this is perpendicular, and CD is equal to dB, so point C and point B are symmetrical relatively to this one. So everything is symmetrical. So it's very easy to prove, for instance, that EH and HF are exactly the same. Well, why? For instance, you can consider right triangles AEH and AFH. They have the common hypotenuse and the angles are the same because the altitude is a bisector, an angle bisector as well, right? So everything is symmetrical. But now let's think about it. Angle HC. Consider angle EHC. Well, this line is perpendicular to this. This line is perpendicular to this. So basically this angle is the same as this angle, right? Because it's two angles which are formed by mutually perpendicular lines. Now from this we can obviously and this is equal to this is alpha, so this is alpha. So from this we can see that angle ECH ECH, right? is equal basically pi over 2 minus alpha because triangle CH E is the right triangle. Now, I just mentioned that EH is equal to HF. This is equal to this because these triangles are congruent, right? So basically if you consider right now triangle ECH it's the right triangle. The angle alpha is something which we definitely know. So what is the ratio of HF to CH? It's the same as EH to CH, right? And this is by definition a cosine of alpha, right? That's basically that's basically EH over CH which is the same as HF over CH. So this is what we actually need. The fact that this angle is perpendicular and this angle is equal to this one that's sufficient to basically say that the ratio between segments into which point H divides this altitude is the same as cosine of alpha. So as long as one of the angles is given to us if this angle is given then you can obviously calculate what this one is because this is an isosceles triangle. So as soon as we know one of the angles of the isosceles triangle we know this angle on the top and the cosine of this angle on the top is the ratio which we need. Well, that's it. Next, give it a rhombus. This is point M and it's chosen in such a way that area of triangle A D M this triangle is equal to one third of the area of an entire parallelogram A, B, C, D. What's also known is that the cosine of alpha is equal to one fourth. What we have to prove is that A M is divided by D M equals 2. This would have to be proved. So knowing that the area of this triangle is one third of the area of an entire parallelogram and knowing that the cosine of this angle is one fourth we have to prove that the A M is twice as big as D M. So what I'm going to do is I will put some abbreviations. Let's say this is A this is X this is Y All right? Now, what I will also do since we are dealing with areas I will need altitudes. So this is one altitude and this is another altitude BQ Let's say this is equal to U and this is equal to B right? So Mp equals to U and BQ equals to B. All right, now let's do the calculation of the areas, right? Now the area of the parallelogram is base which is A times height which is B. Here I have A times B or if you wish this is a rhombus, right? So all sides are A and knowing the cosine of this angle I can very easily find the altitude BQ right? So this is A times sine of alpha but let's just leave it for later. Now the area of a triangle is equal to base which is also A times altitude. So this is equal to A times U but it's a triangle so we have to divide it by 2, right? Let me wipe out this so it doesn't feel bother us and we know this. Now this is angle alpha and this is angle alpha, right? Because this is a rhombus which is therefore parallelogram so these are parallel. Now knowing X I can calculate the altitude as well. So this particular equation would look like this A times instead of U I will put X sine alpha X sine alpha divided by 2 right? And on the right I have one third A times and instead of V I will put A times sine alpha. So from here I immediately see that X is equal to two-thirds of A. That's interesting, right? Such a complicated calculations and conditions related to area etc etc and it looks like X is equal to two-thirds of A even without the angle alpha. It's independent, so for any angle alpha that would be the case. So if I draw the line which cuts off the triangle which area is one-third of the area of the rhombus then this particular piece would be two-thirds of the side. Alright but we don't need really this we need the ratio between AM and MD, right? So in this case since I already know this X I can find out AM which is Y using the law of cosines, right? So law of cosines says that Y-square is equal to A-square plus X-square minus 2AX and the cosine now what is this angle? This angle is pi minus alpha, right? or A-square plus X-square is four-ninths of A-square minus 2A X is two-thirds of A, so I put square here Now the cosine of pi minus alpha we already spoke in during this lecture it's negative cosine of alpha so I'll put this plus and cosine of alpha but likely we know what is the cosine of alpha it's one-quarter, right? So this would be A-square plus four-ninths A-square plus this is one-quarter, this is two times two which is four so it will be reduced so I have A-square divided by three X is equal to two-thirds of A that's what we have found and this is what the common denominator is nine so this is nine plus four-thirteen plus three which is sixteen so it's equal to sixteen-ninths A-square from which Y is equal to four-thirds of A Now look at this and look at this obviously Y divided by X is equal to two and that's exactly what's necessary to prove This wasn't actually too difficult to come up with a solution because again we were dealing with areas and basically all I did was use the definition of the area and you know the formulas for rhombus or parallelogram and the triangle so that wasn't really difficult to prove Now the last problem well in notes on Unisor.com I'm using triangle in this case which has an inscribed circle but actually I would like to in this lecture I would like to address a general case what if you have an inscribed circle into any convex polygon now how to express the area of polygon in terms of radius of inscribed circle this one and angles so let's say we have all angles and the radius well that's actually sufficient to calculate the area now how do we do it well obviously since it's inscribed circle these are all perpendicular to sides right these are all perpendicular to sides now if you will connect the center with vertices you basically subdivide the area of the entire polygon into many triangles now what do we know about each triangles about each triangle well let's consider for instance O A B C D E M N P Q R what do we know about triangle AOM triangle AOM now what's the area of this triangle well it's the right triangle we know one calculus which is R and we know this angle which is basically alpha divided by 2 therefore we can calculate our casualties and the area will be equal to 1 half R and this one since A M divided by O M is cotangent of this angle A M would be equal to this calculus which is R times cotangent so R square of alpha over 2 right times R by cotangent since A M divided by R is equal to cotangent of alpha over 2 A M is equal to R times cotangent that's fine that's right so that's the formula for area of AOM now for triangle EOM EOM this one it's symmetrical it's exactly the same thing the same triangle right oh no sorry not EOM NOM N okay this one because these are equal tangential lines we have proven this many times before so area A M O and A M O are exactly the same because these triangles are congruent both are right triangle common hypotenuse and these two pieces of tangential lines are the same this is R as well so these two triangles are equal and that's why the area is equal to exactly the same cotangent of alpha over 2 so therefore the area of A N O M sum of these it's some kind of a quadrilateral A N O M is equal to R square cotangent of alpha over 2 now absolutely similarly the area of quadrilateral the next one B P O N this one sum of these two triangles again triangles are exactly the same because they have the common hypotenuse and these two these two are are radiuses so they are equal so these this is equal to this so everything is the same congruent triangles and the area is R square cotangent of in this case if this angle is beta it would be beta over 2 and I can continue this thing for all these quadrilaterals this one, this one and this one so basically what I have is the total area if I summarize all of them together would be equal to R square sum of cotangents cotangents of alpha over 2 plus cotangents of beta over 2 plus etc all angles are supposed to be taken care of so if I know the angles and I know the the radius of the inscribed circle I can calculate the area of the entire convex polygon into which this circle is inscribed using something like this now obviously the very last angle I don't have to use I don't have to know basically the value of the last angle because I know that the sum of angles of any convex polygon can be expressed as basically the number of sides and the previous angle so sum of all the angles is dependent on the number of sides actually so if I know how many sides this polygon has then I know the sum of the angles and that's why I don't need the last one I can calculate it based on the previous ones and the number of sides but that's besides the point basically so knowing angles and the radius of an inscribed circle if the reason inscribed circle because not into every convex polygon you can inscribe the circle so if the resets a circle then it's reduced and the angles are sufficient to calculate the area now obviously in case of a triangle it would be only three angles and the third one is pi minus sum of the previous one so we can just basically replace the cotangent of gamma with the cotangent of some expression based on alpha and beta well that's it these are all the problems I wanted to present today I do recommend you to go through these problems again and solve them accurately I do suggest in writing actually I mean if you accurately write your solution on the paper or in the computer basically and try to be as specific as possible that's very very useful exercise so no statement is left without some kind of reasoning behind it so every step should be based on some logical reasoning so I do recommend you to go through these problems again and solve them again and well don't forget that Unizor has a lot of other things and if you sign in as a student you not only can just go yourself through all these lectures and problems etc it will also present you exams and it will give you some score an exam you can take any time you want and as many times as you want to improve your score the goal is for every exam to get the maximum score that's very important try to think about this that's it for today thank you very much and good luck