
# 16.2: Pythagorean Theorem


Here is an analog of the Pythagorean theorems (Theorem 6.2.1 and Theorem 13..1) in spherical geometry.

Theorem $$\PageIndex{1}$$ Spherical Pythagorean Theorem

Let $$\triangle_sABC$$ be a spherical triangle with a right angle at $$C$$. Set $$a=BC_s$$, $$b=CA_s$$, and $$c=AB_s$$. Then

$$\cos c=\cos a \cdot \cos b.$$

Proof

Since the angle at $$C$$ is right, we can choose the coordinates in $$\mathbb{R}^3$$ so that $$v_C\z=(0,0,1)$$, $$v_A$$ lies in the $$xz$$-plane, so $$v_A=(x_A,0,z_A)$$, and $$v_B$$ lies in $$yz$$-plane, so $$v_B=(0,y_B,z_B)$$.

Applying, 16.2.3, we get that

\begin{aligned} z_A&=\langle v_C,v_A\rangle =\cos b, \\ z_B&=\langle v_C,v_B\rangle =\cos a.\end{aligned}

Applying, 16.2.1 and 16.2.3, we get that

\begin{aligned} \cos c &=\langle v_A,v_B\rangle= \\ &=x_A\cdot 0+0\cdot y_B+z_A\cdot z_B= \\ &=\cos b\cdot\cos a.\end{aligned}

In the proof, we will use the notion of the scalar product which we are about to discuss.

Let $$v_A=(x_A,y_A,z_A)$$ and $$v_B=(x_B,y_B,z_B)$$ denote the position vectors of points $$A$$ and $$B$$. The scalar product of the two vectors $$v_A$$ and $$v_B$$ in $$\mathbb{R}^3$$ is defined as

$\langle v_A,v_B\rangle := x_A\cdot x_B+y_A\cdot y_B+z_A\cdot z_B.$

Assume both vectors $$v_A$$ and $$v_B$$ are nonzero; suppose that $$\phi$$ denotes the angle measure between them. Then the scalar product can be expressed the following way:

$\langle v_A,v_B\rangle=|v_A|\cdot|v_B|\cdot\cos\phi,$

where

\begin{aligned} |v_A|&=\sqrt{x_A^2+y_A^2+z_A^2}, & |v_B|&=\sqrt{x_B^2+y_B^2+z_B^2}.\end{aligned}

Now, assume that the points $$A$$ and $$B$$ lie on the unit sphere $$\Sigma$$ in $$\mathbb{R}^3$$ centered at the origin. In this case $$|v_A|=|v_B|=1$$. By 16.2.2 we get that

$\cos AB_s=\langle v_A,v_B\rangle.$

Exercise $$\PageIndex{1}$$

AShow that if $$\triangle_s ABC$$ is a spherical triangle with a right angle at $$C$$, and $$AC_s=BC_s=\dfrac{\pi}{4}$$, then $$AB_s=\dfrac{\pi}{3}$$.

Hint

Applying the Pythagorean theorem, we get that

$$\cos AB_s = \cos AC_s \cdot \cos BC_s = \dfrac{1}{2}.$$

Therefore, $$AB_s = \dfrac{\pi}{3}.$$

Alternatively, look at the tessellation of a hemisphere on the picture; it is made from 12 copies of $$\triangle_s ABC$$ and yet 4 equilateral spherical triangles. From the symmetry of this tessellation, it follows that $$[AB]_s$$ occupies $$\dfrac{1}{6}$$ of the equator; that is, $$AB_s = \dfrac{\pi}{3}$$.