Here’s the first calculation using the law of cosines – we can see that the 3rd side does indeed equal roughly 6.67

We should be able to get the other two angles from the law of sines, but what happens when we look for the value of the angle opposite the side with length 11?

Since the law of sines only tells us the sine of the angle, we actually have two choices – which one is right? Let’s try 97 first:

Seems close enough – what happens when we check 83?

Dang, that doesn’t work at all. In fact, the side that has length 8 would have to have a length of about 9.6 to satisfy the law of sines.

But, are the two triangles related in some way? One cool relationship comes from the “extended law of sines” that says for any triangle:

A / Sin(a) = B / Sin(b) = C / Sin(c) = 2R, where R is the radius of the circumcircle.

We can check on Wolfram Alpha that the circumcircle for each of the two triangles has the same radius:

I have a soft spot for the extended law of sines because talking about it was one of the first videos that I ever put on youtube:

Here the extended law of sines tells us that although the two triangles have some different sides and angles, they can be inscribed in the same circle. Pretty neat 🙂

I think this would be a really fun problem to share with a trig class.

4 thoughts on “A neat question to share with a trig class from Mary Bourassa”

Ok (deep breath) I would usually try to work through this and see where I am making a mistake in my thinking, but I can’t get past first step where you do, c^2 = 11^2 + 8^2 . It seems to me that c in the triangle shown is not the hypotenuse, so c = 7.549. I thought maybe that the triangle is not drawn to scale, so I drew it out, and there is no way c is the hypotenuse. Like I said, I didn’t work through this, so if you addresses this in later steps, oops.

Your white equation is clear enough, I can see that now; I’m not fluent enough in trig to have made that connection. I just stopped in my tracks using Pythagorean. and getting
a different measurement than 6.67. Oh, but now I’m remembering, the Pythagorean is for 90 degree triangles…blush.

Another good line of questioning for the trig class is how we could tell that angle B has to be larger than 90 degrees?

In addition to Mike’s test, two other ways to reason start with noticing
(1) 11 cos(37deg) > 8
(2) Once you calculate c, sqrt(121-64) > c

Ok (deep breath) I would usually try to work through this and see where I am making a mistake in my thinking, but I can’t get past first step where you do, c^2 = 11^2 + 8^2 . It seems to me that c in the triangle shown is not the hypotenuse, so c = 7.549. I thought maybe that the triangle is not drawn to scale, so I drew it out, and there is no way c is the hypotenuse. Like I said, I didn’t work through this, so if you addresses this in later steps, oops.

Sorry, I should have been more clear. That step is the law of cosines which says:

C^2 = A^2 + B^2 – 2*A*B*cos(c)

So, the second line of the formula in the whiteboard is part of the calculation, too.

Your white equation is clear enough, I can see that now; I’m not fluent enough in trig to have made that connection. I just stopped in my tracks using Pythagorean. and getting

a different measurement than 6.67. Oh, but now I’m remembering, the Pythagorean is for 90 degree triangles…blush.

Another good line of questioning for the trig class is how we could tell that angle B has to be larger than 90 degrees?

In addition to Mike’s test, two other ways to reason start with noticing

(1) 11 cos(37deg) > 8

(2) Once you calculate c, sqrt(121-64) > c