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In this video weβre going to look at one method of proving the law of sines.
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First of all, a reminder of what the law of sines is.
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So I have here a triangle, in which Iβve labelled the three angles as π΄, π΅, and πΆ using capital letters and then Iβve labelled the sides as π, π, and π using lowercase letters.
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And you can see that side π is opposite angle π΄, side π is opposite angle π΅, and side π is opposite angle πΆ.
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The law of sines is written at the top in red here, and what it tells us is that the ratio between a side and the sine of its opposite angle is constant throughout the triangle.
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So for any of these pairs, if I take the side length and divide it by sine of the opposite angle, then I get the same result, which is why I have π over sin π΄ is equal to π over sin π΅ and thatβs also equal to π over sin πΆ.
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We can also specify the law of sines in an alternative form where we invert each of these fractions, so we have sin π΄ over π is equal to sin π΅ over π and equal to sin πΆ over π, but itβs this version of the law of sines that weβre going to look at a proof for.
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So the first stage in this proof, Iβm going to draw in a perpendicular height from angle πΆ down to the base of this triangle π΄π΅.
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So thereβs this height here which Iβve labelled as β and you can see that it divides the triangle up into two smaller right-angled triangles which Iβve labelled as triangles one and two.
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Now those triangles arenβt necessarily congruent to each other unless the original triangle was isosceles, so weβre assuming that theyβre not the same as each other here.
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So as Iβve now got right-angled triangles, I can perform standard trigonometry using sine, cosine, or tangent.
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And what Iβm going to do is Iβm gonna recall the definition of just the sine ratio in a right-angled triangle.
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Remember then that itβs defined like this: sine of an angle π is equal to the opposite divided by the hypotenuse.
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So what Iβm going to do then Iβm going to work in triangle one first of all, and my first step is Iβm going to label the three sides of this right-angled triangle in relation to angle π΄.
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So I have the opposite, the adjacent, and the hypotenuse, and now Iβm going to write down the sine ratio for angle π΄, so itβs opposite divided by hypotenuse, and thatβs going to be β divided by lowercase π using the notation in this diagram.
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So I have sin π΄ is equal to β over π.
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Iβm gonna do one rearrangement of this.
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Iβm going to multiply both sides of this equation by π.
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And Iβve swapped the order of the two sides here, but it tells me that β is equal to π multiplied by sin π΄, π sin π΄.
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Iβm now going to move over to triangle two and do the same thing in relation to angle π΅.
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So that side β is still the opposite, but Iβm going to fill in the hypotenuse and the adjacent as well.
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And now Iβm going to write down the sine ratio for angle π΅.
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So opposite divided by hypotenuse, it will be β divided by π.
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I have then that sine of the angle π΅ is equal to β over π.
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As I did before, Iβm gonna do one rearrangement to this where I multiply both sides of this equation by π.
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This tells me then that β is equal to π sin π΅.
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Now perhaps you can see what the next stepβs going to be.
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I have got two expressions for β: one is π sin π΄ and the other is π sin π΅.
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And if theyβre both equal to β, then I can set them equal to each other.
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So I have then that π sin π΄ is equal to π sin π΅.
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Remember when Iβm saying π and π here, thereβs a difference between capital π΄ representing an angle and lowercase π representing a side, and the same thing for π.
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Now perhaps you can see this is looking close to the law of sines, but it isnβt quite there yet.
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In order to get this to the law of sines, I need to divide both sides of this equation by sin π΄ and by sin π΅.
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If I just divide by sin π΄ first of all, then youβll see that it cancels out the fact of sin π΄ on the left-hand side and appears in the denominator on the right.
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Now if I divide both sides by sin π΅, well then it cancels out the fact of sin π΅ in the numerator on the right and it now appears in the denominator on the left.
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So what you see is that I have the first two parts of the law of sines.
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Theyβre written the opposite way round how they appear at the top, but I have that π divided by sin π΅ is equal to π divided by sin π΄.
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Now thatβs only demonstrated that two parts of this ratio are equal, but you could demonstrate it for the other pairs by drawing a perpendicular from π to the opposite side for example.
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So this gives you then a fairly concise proof of the law of sines.
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We started off with a triangle, drew in the perpendicular height from one of the vertices.
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We then used the sine ratio in each of the right-angled triangles in turn to get an expression for this perpendicular height β, equated the two expressions, and then with a little bit of rearranging we had our proof of the law of sines.