Video: Proof of the Law of Sines | Nagwa Video: Proof of the Law of Sines | Nagwa

# Video: Proof of the Law of Sines

Work through a proof of the law of sines by finding equivalent expressions for the perpendicular height of a triangle using the sine ratio and equating and rearranging them into the format we are familiar with.

05:46

### Video Transcript

In this video weβre going to look at one method of proving the law of sines. First of all, a reminder of what the law of sines is. So I have here a triangle, in which Iβve labelled the three angles as π΄, π΅, and πΆ using capital letters and then Iβve labelled the sides as π, π, and π using lowercase letters. And you can see that side π is opposite angle π΄, side π is opposite angle π΅, and side π is opposite angle πΆ.

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.

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 thatβs also equal to π over sin πΆ.

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 πΆ over π, but itβs this version of the law of sines that weβre going to look at a proof for.

So the first stage in this proof, Iβm going to draw in a perpendicular height from angle πΆ down to the base of this triangle π΄π΅. So thereβs this height here which Iβve labelled as β and you can see that it divides the triangle up into two smaller right-angled triangles which Iβve labelled as triangles one and two.

Now those triangles arenβt necessarily congruent to each other unless the original triangle was isosceles, so weβre assuming that theyβre not the same as each other here. So as Iβve now got right-angled triangles, I can perform standard trigonometry using sine, cosine, or tangent.

And what Iβm going to do is Iβm gonna recall the definition of just the sine ratio in a right-angled triangle. Remember then that itβs defined like this: sine of an angle π is equal to the opposite divided by the hypotenuse.

So what Iβm going to do then Iβm going to work in triangle one first of all, and my first step is Iβm going to label the three sides of this right-angled triangle in relation to angle π΄.

So I have the opposite, the adjacent, and the hypotenuse, and now Iβm going to write down the sine ratio for angle π΄, so itβs opposite divided by hypotenuse, and thatβs going to be β divided by lowercase π using the notation in this diagram.

So I have sin π΄ is equal to β over π. Iβm gonna do one rearrangement of this. Iβm going to multiply both sides of this equation by π. And Iβve 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΄.

Iβm now going to move over to triangle two and do the same thing in relation to angle π΅. So that side β is still the opposite, but Iβm going to fill in the hypotenuse and the adjacent as well. And now Iβm going to write down the sine ratio for angle π΅. So opposite divided by hypotenuse, it will be β divided by π.

I have then that sine of the angle π΅ is equal to β over π. As I did before, Iβm gonna do one rearrangement to this where I multiply both sides of this equation by π. This tells me then that β is equal to π sin π΅.

Now perhaps you can see what the next stepβs going to be. I have got two expressions for β: one is π sin π΄ and the other is π sin π΅. And if theyβre both equal to β, then I can set them equal to each other.

So I have then that π sin π΄ is equal to π sin π΅. Remember when Iβm saying π and π here, thereβs a difference between capital π΄ representing an angle and lowercase π representing a side, and the same thing for π.

Now perhaps you can see this is looking close to the law of sines, but it isnβt quite there yet. 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΅.

If I just divide by sin π΄ first of all, then youβll see that it cancels out the fact of sin π΄ on the left-hand side and appears in the denominator on the right. 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.

So what you see is that I have the first two parts of the law of sines. Theyβre 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΄.

Now thatβs 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.

So this gives you then a fairly concise proof of the law of sines. We started off with a triangle, drew in the perpendicular height from one of the vertices. 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.

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