Solving a Triangle

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two students must draw a triangle with one side eight, another side five, and an angle of thirty degrees opposite the side five. Both follow the data exactly, and they draw two different triangles! So, is one of them wrong? Today we will see that two sides and an angle that is not between them can give no triangle, one, or even two. This is called the ambiguous case, and JEE questions on triangles do test it. So, let's begin!

0:32What is given, and what we find

So, what does solving a triangle mean? It means finding every side and angle from the ones given. In triangle ABC on screen, small a, b and c are the sides opposite the angles capital A, B and C; ignore the two circles and their labels today. Okay? Here, we are given two sides, small a and small b, and the angle capital A, which is acute, meaning less than ninety degrees. Got it? Now look at the figure and tell me. Is the angle capital A between the two given sides, small a and small b? No, it is not between them. Small a is the side B C, opposite the corner A, and it does not even touch that corner. That is exactly where the trouble starts. Right? And what do we have to find? The third side, small c, and the other two angles, capital B and capital C, okay?

1:26One quadratic tells the whole story

Let us not just memorise a table; let us understand it. The unknown is small c. So tell me, which rule connects small a, small b, small c and the angle capital A? The cosine rule, yes, exactly! Small a squared equals b squared plus c squared, minus two times b times c times cos A, from the earlier section. With me? Now bring everything to one side. c squared, minus two times b times c times cos A, plus b squared minus a squared, equals zero. Small a, small b and the angle are known. So, as an equation in c, what kind of equation is it? It is a quadratic in c, absolutely! And each positive root c gives one possible triangle. So we can get no triangle, one or two, and that is why this case is called ambiguous. Clear? Call the two roots c one and c two; the small one and two below are subscripts, just labels. From the quadratic, their sum is two times b times cos A, and their product is b squared minus a squared. Right? And for their difference, the two vertical bars mean the modulus, the size without the sign. Mod of c one minus c two equals two times the square root of, a squared minus b squared times sine squared A, okay? So the roots are real only when what is under the root is not negative.

2:47How many triangles? (angle A acute)

Now let us count, with the angle A acute, with the three facts about the roots on top. First case: if small a is less than b times sine A, the quantity under the root is negative, there is no real root, so no triangle at all. Second case: if small a is exactly equal to b times sine A, the difference of the roots is zero, so the two roots are equal. We get just one triangle, and it is right-angled at the corner B, okay? Now the interesting one: small a is bigger than b times sine A, but smaller than b. The roots are real and different. Their product, b squared minus a squared, is positive. And their sum, two times b times cos A, is also positive, because the angle is acute. So what can you say about the signs of the two roots? Both roots are positive, exactly! Two different positive values of c, so two different triangles. This is the true ambiguous case. With me? And last, small a greater than or equal to b, the sign with a line under it. Now the product, b squared minus a squared, is zero or negative, so exactly one root is positive: one triangle. Got it?

3:57Example 1: a = 3, b = 6, A = 30 degrees, one triangle

Alright, the first example, from this section. Small a is three, small b is six, and the angle A is thirty degrees; the small circle means degrees. How many triangles are possible? Pause the video and try it first. Done? Okay, let's check. Many students see that small a is less than small b, and say two triangles straight away. Is that enough? No, no, wait, that is only half the condition! For two triangles, small a must also be bigger than b times sine A, so always find b times sine A first. Six times sine thirty degrees, which is one half, gives three; the dot means times. Okay? So small a, which is three, is exactly equal to b times sine A, and we are on the boundary, not inside the two triangle range. Right? And on the boundary we get exactly one triangle, right-angled at the corner B, as our table says. So the answer is one, not two. Got it?

4:59Example 2: a = 5, b = 8, A = 30 degrees, find c and B

Now a harder one, and it is our puzzle from the start! Small a is five, small b is eight, and the angle A is thirty degrees. How many triangles are possible? And find the possible values of small c, and of the angle B, too. Pause the video and try it first. Done? Okay, let's check. First, b times sine A: eight times one half gives four. Four is less than five, which is less than eight, so this is the two triangle case. Both students in our puzzle were right! Okay? Now the quadratic for c. Two times eight times cos thirty degrees, which is root three over two, gives eight root three. And sixty four minus twenty five gives thirty nine. So c squared, minus eight root three times c, plus thirty nine, equals zero. Right? Use the quadratic formula; the plus or minus sign gives the two roots. Eight root three, squared, is one hundred and ninety two, and four times thirty nine is one hundred and fifty six. Their difference is thirty six, whose square root is six. So c equals eight root three plus or minus six, all over two, which is four root three plus or minus three. Clear? Now the angle B, by the sine rule. Sine B equals b times sine A, over small a: eight times one half, over five, which is four over five. With me? So the angle B is sine inverse four over five; the minus one up there is not a power, it means sine inverse. Is that the only answer? No, no, careful, that is the classic trap: do not stop after one! Sine inverse only gives angles from minus pi over two to pi over two, so it gives just the acute angle. But pi minus that angle has the same sine; pi here is one hundred and eighty degrees. So the angle B is sine inverse four over five, or pi minus sine inverse four over five: two triangles, two answers. Well done!

6:57Quick revision: none, one or two triangles via b sin A

So, quick revision. Given small a, small b and an acute angle A, the third side comes from one quadratic in c. If small a is less than b times sine A, there is no triangle, and if it is equal, there is one, right-angled at the corner B, okay? Between b times sine A and b, there are two triangles, and if small a is at least b, just one. And when there are two triangles, find both values of c and of the angle B, never just one. Now go and try the practice problems on this page. All the best!