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Law of Cosines

An algebra student posted this question on

How would you express the length n of a chord of a circle with radius 10cm as a function of the central angle x? And what is a chord is that important to know when writing out the equation and how can you tell what trig function to use?

The answer is:

join the endpoints of the chord and the center of the circle, that'll give you a isosceles triangle. and since the radius of the circle is 10, thus the lengths of this isosceles triangles are 10,10,n.  Now we apply the cosine law:

n�= 10�+10�- 2*10*10*cosx = 200 -200 cos x

hence, n=sqrt(200-200cosx)

Here's how that works:  The objective is given the lengths of two legs of an isosceles triangle and the angle between them, find the length of its base.

Draw an isosceles triangle so that its base is horizontal, and the angle, x, between its equal legs is at the top. Let the length of each leg equal r units. Now draw a line perpendicular to one of the legs that intersects the opposite corner of the triangle. This makes two right triangles inside the isosceles triangle. The upper right triangle has hypotenuse r and legs r sin x and r cos x. The lower right triangle, then has a leg in common with the upper right triangle, whose length is r sin x. The other leg of the lower right triangle is r - r cos x, which you can see from the diagram you're drawing. The lower right triangle's hypotenuse is the 3rd side of the isosceles triangle, which is the length, n, we're looking for.

n� = (r-r cos x)� + r�sin�x
n� = r� - 2r�cos x + r�cos�x + r�sin�x

By the Pythagorean Theorem, sin�x+cos�x=1, so:

n� = r� - 2r�cos x + r�
n� = 2r� - 2r�cos x
n� = r�(2-2 cos x)
n = r sqrt(2-2 cos x)

To generalize this to any triangle, imagine the original triangle has sides r, n, and q.  That is, the left side of the triangle in the diagram, above, is replaced by a side of length q.

Now, instead of r-r cos x as the second leg of the lower right triangle, it's q-r cos x.  Now the calculations look like this:

n� = (q-r cos x)� + r�sin�x
n� = q� - 2qr cos x + r�cos�x + r�sin�x

By the Pythagorean Theorem, sin�x+cos�x=1, so:

n� = q� - 2qr cos x + r�
n� = q� + r� - 2qr cos x

This is the Law of Cosines, which gives the length of the third side, n, of a triangle, when you know two sides, p and q, and the angle between them, x.  You'll note that when x=pi/2, a right angle, the cosine term goes to zero, and you're left with the Pythagorean equation, n = sqrt(q� + r�).

Here's something else that's interesting: From first diagram of the isosceles triangle, above, you can see that n is also equal to 2r sin (x/2). To show this, draw the axis of symmetry of the isosceles triangle. This makes a right triangle whose hypotenuse is r, and whose legs are sin (x/2) and cos (x/2). The first leg is half the third side of the isosceles triangle.

r sqrt(2-2 cos x) = 2r sin(x/2)

So we have the equality,

sqrt(2-2 cos x) = 2 sin(x/2)

Related pages in this website:

The Law of Cosines -- a quick and easy proof

Triangles -- formulas for area and circumscribed and inscribed circles

Trig Equivalences -- cos(x) = 1 - 2sinĀ²(x/2) = 2cosĀ²(x/2) - 1 

Heron's formula for the area of a triangle, if all you know is the lengths of its sides.


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