Question

Find, correct to the nearest degree, the three angles of the triangle with the given vertices.

A(1, 0, -1) B(2, -3, 0) C(1, 4, 4)

angle CAB
angle ABC
angle BCA

Answer 1

In order to find the angles of the triangle with the specified vertices, we must first find the length of each side using the distance formula, and then we may use the law of cosines to find every other angle.

\[SideAB = \sqrt{(1-2)^2+(0-(-3))^2+(-1-0)^2}\]
\[SideAB = \sqrt{1+9+1}= \sqrt{11}\]
\[SideAC = \sqrt{(1-1)^2+(0-4)^2+(-1-4)^2}\]
\[SideAC = \sqrt{0+16+25}= \sqrt{41}\]
\[SideBC = \sqrt{(2-1)^2+(-3-4)^2+(0-4)^2}\]
\[SideBC = \sqrt{1+49+16}= \sqrt{66}\]

Now to find angle CAB we use the law of cosines. \[BC^2 = AB^2 +AC^2-2AB(AC)cosa\] where a = the angle CAB\[{\sqrt{66}}^2= {\sqrt{41}}^2+{\sqrt{11}}^2-2(\sqrt{11*41}\cos(a)\]\[66=52-2\sqrt{451}\cos(a)=52-42.48\cos(a)\]\[\cos^{-1}(-14/42.48)=109.2 degrees = a\] Now to find BAC we repeat the process for BAC. \[AC^2 = BC^2 +AB^2-2BC(AB)cosb\] where b = the angle BCA\[{\sqrt{11}}^2= {\sqrt{66}}^2+{\sqrt{41}}^2-2(\sqrt{66*41}\cos(b)\]\[11=107-2\sqrt{2706}\cos(b)=107-104.04\cos(b)\]\[\cos^{-1}(96/104.04)=22.67 degrees = b\] To find the third angle ABC, we use the polygon interior angle sum theorem, so \[180-(109.2+22.67)=48.13\]

So we have our angles as being:
CAB = 109.2 degrees
BCA = 22.67 degrees
ABC = 48.13 degrees

you can check these answers by using the law of sines\[\frac{ \sin(109.2) }{ \sqrt{66} }= \frac{ \sin(22.67) }{ \sqrt{11} }=\frac{ \sin(48.13)}{ \sqrt{41}}=.116\]