Sine, Cosine and Projection Rules

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Imagine a triangle with sides five, seven and eight, and no right angle anywhere. Can you find one of its angles exactly, without a protractor? Pythagoras will not help here, right? It needs a right angle. So today we learn the rules connecting the sides and angles of any triangle: the sine, cosine and projection rules. Triangle questions come up often in JEE, and many of them start right here. So, let's begin.

0:33Naming a triangle, and the sine rule

So, first the names. In triangle capital ABC, capital A, B and C are the angles at the three corners. Small a is the side opposite angle capital A, small b is opposite capital B, and small c is opposite capital C, okay? Now the sine rule. Small a over sine A, equals b over sine B, equals c over sine C, and each of these equals two R, twice the capital R on screen. So every side, divided by the sine of its opposite angle, gives the same number. Right? Here, capital R is the circumradius: the radius of the circle that passes through all three corners. Clear? So tell me, if small a over sine A equals two R, how can you write small a by itself? Yes, exactly, just multiply across! Small a equals two R sine A, b equals two R sine B, and c equals two R sine C, as on screen. This line is the key to today's trick.

1:28The cosine rule: Pythagoras, corrected

Got it? Next, the cosine rule. Small a squared equals b squared plus c squared, minus two times b times c times cos A, okay? The little two at the top means squared. Now look at that last term. Suppose angle capital A is ninety degrees; the small circle on screen means degrees. What happens to this formula then? Very good, cos ninety degrees is zero, so the last term vanishes, and we get small a squared equals b squared plus c squared, which is Pythagoras! So the cosine rule is simply Pythagoras, corrected for any angle. With me? Now, to find an angle from three sides, we rearrange it. Cos A equals b squared plus c squared, minus small a squared, all over two times b times c. Okay? Now let me ask you this: in the formula for cos A, which side's square gets the minus sign, and why that one? It is always the side opposite the angle! For cos A, we subtract small a squared, since small a faces angle capital A, directly across from it. So cos B equals small a squared plus c squared, minus b squared, over two times small a times c; and cos C equals small a squared plus b squared, minus c squared, over two times small a times b. Subtracting the wrong side is a common mistake, so be careful. Right?

2:53The projection rule

Now the third rule, the projection rule. Small a equals b cos C, plus c cos B, and look at the pattern. Angle capital C sits between side small a and side b, and capital B between side small a and side c. So b cos C and c cos B are the projections of sides b and c on side small a, and together they give small a; hence the name. Okay? Now you try the pattern. What should side b equal? Very good, side b equals c cos A, plus small a cos C, the same pattern again. Clear? And side c equals small a cos B, plus b cos A, so all three lines follow one pattern. Got it?

3:32Napier's analogy, and the topper's trick

One more result: Napier's analogy. Tan of, B minus C over two, equals b minus c, all over b plus c, times cot of A over two. It is handy when you know two sides and the angle between them, okay? Now the topper's trick. Suppose a question gives you a condition in the sides, but your identities are all about angles. So how do you switch from sides to angles? Yes, exactly, with the sine rule! Replace small a, b and c by two R sine A, two R sine B, and two R sine C, as on screen. Now everything is in angles, and the conditional identities of chapter two take over. Right? And to go the other way, from angles to sides, use the cosine rule: write each cosine in terms of the sides, small a, b and c. With me?

4:22Example 1: find angle B when a = 5, b = 7, c = 8

Now, remember our puzzle from the start? Let's solve it. In triangle capital ABC, small a is five, b is seven and c is eight. We want angle capital B, the angle opposite side b. Pause the video and try it first. Done? Okay, let's check. Three sides are given and we want an angle, so we use the cosine rule. But which square gets the minus sign this time? It is b squared, because side b is opposite capital B, the angle we want. So cos B equals small a squared plus c squared, minus b squared, over two times small a times c. Right? Small a squared is twenty five, c squared is sixty four, and b squared is forty nine. Twenty five plus sixty four is eighty nine, and eighty nine minus forty nine gives forty. Below, two times five times eight is eighty; the dots on screen mean times. So cos B is forty over eighty, which is one half. Okay? In a triangle, the angle whose cosine is one half is sixty degrees, so capital B is exactly sixty degrees, no protractor needed. That was our puzzle! Clear?

5:38Example 2: a cos A = b cos B, isosceles or right-angled

Now a harder one, JEE style. In a triangle, small a times cos A equals small b times cos B, as on screen. What can you say about the triangle? Hint: use the topper's trick. Pause the video and try it first. Done? Let's see. The condition is in the sides, so we switch to angles with the sine rule. Small a becomes two R sine A, and small b becomes two R sine B, so two R sine A cos A equals two R sine B cos B, okay? Now, two sine A cos A is sine two A, the double angle formula, and the same on the right. Capital R is not zero, so it cancels. We get sine two A equals sine two B; the double arrow means implies. Clear? So sine two A equals sine two B, which means two A equals two B, so the triangle is isosceles, and we are done, right? No, no, wait, this is exactly the trap! Two different angles can have the same sine, because sine of pi minus x equals sine x, where pi, the Greek letter, stands for one hundred and eighty degrees. So either two A equals two B, or two A equals pi minus two B, got it? In the first case, capital A equals capital B, so the triangle is isosceles. In the second case, capital A plus capital B equals pi over two, so capital C is pi minus pi over two, which is ninety degrees: a right-angled triangle. So the triangle is either isosceles, or right-angled. Students who cancel too fast lose the second case. Okay?

7:10Quick revision: sine, cosine and projection rules

So, quick revision. Sine rule: each side over the sine of its opposite angle gives the same value, two R, twice the circumradius. Cosine rule: to find an angle, subtract the square of the side opposite it. Projection rule: small a equals b cos C, plus c cos B, and to turn sides into angles, put two R sine A in place of small a. Now go and try the practice problems on this page. All the best!