Question

How would I find the angles of the triangle with the given vertices?

Answer 1

Let
`\vec a,\vec b,\vec c` be vectors that point to
`A,B,C`, respectively.

`\vec a = 7\,\vec\imath+8\,\vec\jmath+2\,\vec k`

`\vec b = 2\,\vec\imath + \vec\jmath + 2\,\vec k`

`\vec c = -5\,\vec\imath-6\,\vec\jmath-6\,\vec k`

Then consider the directed line segments
`AB`,
`BC`, and
`CA`, to which we'll assign the vectors

`AB ~:~ \vec b - \vec a = -5\,\vec\imath - 7\,\vec\jmath`

`BC ~:~ \vec c - \vec b = -7\,\vec\imath-7\,\vec\jmath-8\,\vec k`

`CA ~:~ \vec a - \vec c = 12\,\vec\imath+14\,\vec\jmath+8\,\vec k`

whose lengths are

`\|\vec b - \vec a\| = √((-5)^2+(-7)^2) = √(74)`

`\|\vec c - \vec b\| = √((-7)^2+(-7)^2+(-8)^2) = √(162) = 9\sqrt2`

`\|\vec a - \vec c\| = √(12^2+14^2+8^2) = √(404) = 2√(101)`

The angle at vertex
`A` is made by the directed segments
`AB` and
`AC`, corresponding to
`\vec b-\vec a` and
`\vec c-\vec a`. Use the dot product identity to find the measure of
`\angle A`.

`(\vec b - \vec a) \cdot (\vec c - \vec a) = \|\vec b - \vec a\| \|\vec c - \vec a\| \cos\left(m\angle A\right) \\\\ 158 = 2√(7474) \cos\left(m\angle A\right) \\\\ \cos\left(m\angle A\right) = (79)/(√(7474)) \\\\ m\angle A = \cos^(-1)\left((79)/(√(7474))\right) \approx \boxed{23.964^\circ}`

Similarly, the angle at
`B` is made by
`\vec a-\vec b` and
`\vec c-\vec b`[/tex]. Then

`(\vec a-\vec b) \cdot (\vec c - \vec b) = \|\vec a - \vec b\| \|\vec c - \vec b\| \cos\left(m\angle B\right) \\\\ \cos\left(m\angle B\right) = -(14)/(3√(37)) \\\\ m\angle B = \cos^(-1)\left(-(14)/(3√(37))\right) \approx \boxed{140.103^\circ}`

Do the same for
`C`, or simply use the fact that the interior angles in any triangle sum to 180°. You should find

`m\angle C = \cos^(-1)\left((41)/(3√(202))\right)  = 180^\circ - m\angle A - m\angle B \approx \boxed{15.933^\circ}`