# Finding angles in spherical triangle using law of cosines

Problem: Assume that the earth is a sphere of radius $5280$ miles, find the length of the sides, the measure of the angles and the area of the spherical triangle with vertices $A(70°N,10°E)$,$B(10°S,100°E)$ and $C(50°S,80°W)$. The earth's radius will be used as the unit of length. The spherical coordinates $(r,v,u)$ of the three vertices are $(1,10,20)$, $(1,100,100)$ and $(1,−80,140).$

Their Cartesian coordinates are:

$$(\sin(20°)\cos(10°),\sin(20°)\sin(10°),\cos(20°))=(0.3368,0.0594,0.9397)$$ $$(\sin(100°)\cos(100°),\sin(100°)\sin(100°),\cos(100°))=(−0.1710,0.9698,−.1736)$$ $$(\sin(140°)\cos(−80°),\sin(140°)\sin(−80°),\cos(140°))=(0.1116,−0.6330,−.7660)$$

The cosines of the angles between radii $OA,$ $OB$ and $OC$ are equal to the dot product of the corresponding position vectors. Thus:

$$∠AOB=\cos^{−1}(-0.1631)=1.7347 \ \mathrm{radians}$$

$$∠BOC=\cos^{−1}(-0.5000)=2.0944 \ \mathrm{radians}$$

$$∠COA=\cos^{−1}(-0.7198)=2.3743 \ \mathrm{radians}$$

Since the length of the arc of a circle is the product of its radius by the radian measure of its central angle, it follows that the length of the sides of the spherical triangle $ABC$ are:

$9,159$ miles, $11,058$ miles, and $12,536$ miles.

The angle of this triangle are derived by means of the appropriate Law of cosines. Thus:

!This is where I begin to have problem understanding the question!!

$$\cos A = \frac{\cos(2.0944) - \cos(1.7346) \cos(2.3743)}{\sin(1.7346) \sin(2.3743)} = -0.9013$$

So $A=154°=2.6941$ radians

Similarly,

$B=160°=2.7874$ radians

$C=150°=2.6261$ radians

So, my question is how did they calculate $B$ and $C$ to be $160°$ and $150°$?

The book says:

$$\cos A = \frac{\cos(a)-\cos(b) \cos(c)}{\sin(b) \sin(c)}.$$

Therefore, I know how they calculated A. So i'm guessing they did $B$ and $C$ the same way but I dont get the same answer.

SO $$a=2.0944$$

$$b=1.7346$$

$$c=2.3743$$

So if I wanted to calculate $B$:

$$\cos B = \frac{\cos(b)-\cos(a) \cos(c)}{\sin(a) \sin(c)}.$$

$$\cos B = \frac{\cos(1.7346)-\cos(2.0944) \cos(2.3743)}{\sin(2.0944)\sin(2.3743)}.$$

and I get $B=45°$ not $160°$

I'm not sure where I went wrong.

• All the cosines on the right side of your equation for $\cos B$ are negative, and the sines are positive. So we have negative minus positive gives a negative, then divide by positive and the result is still negative. Arc cosine then is greater than $90$ degrees. You may have an arithmetic error somewhere but it's hard to guess what without seeing more of the steps after the equation for $\cos B$. Commented Feb 11, 2015 at 21:03
• $$\cos B = \frac{0.00106}{0.0015}=0.7066$$ Either way, I get positive in top and bottom.
– mika
Commented Feb 11, 2015 at 21:08
• It looks like you are taking the sines and cosines of $1.7346$ degrees, $2.0944$ degrees, and so forth. Make sure your calculator is set to use radians. Commented Feb 11, 2015 at 21:22