# I have three points with coordinates: A(5,−1,0),B(2,4,10), and C(6,−1,4). I have the following vectors vec(CA)=(−1,0,−4) and vec(CB)=(−4,5,6). I think they have used the trig identity cos2(θ)+sin2(θ)=1 to find the value of sin(theta) rather than arccos(theta) to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using arccos and the other using the trig identity.

I have three points with coordinates: $A\left(5,-1,0\right),B\left(2,4,10\right)$ and $C\left(6,-1,4\right)$
I have the following vectors $\stackrel{\to }{CA}=\left(-1,0,-4\right)$ and $\stackrel{\to }{CB}=\left(-4,5,6\right)$
To find the area of the triangle I used the dot product between these vectors to get the angle and then applied the formula $A=0.5ab\mathrm{sin}C$ to find the area of the triangle which gave me 15.07(2dp).
However in the given solutions the answer is given as $\left(3\ast \sqrt{\left(}102\right)\right)/2$
I think they have used the trig identity ${\mathrm{cos}}^{2}\left(\theta \right)+{\mathrm{sin}}^{2}\left(\theta \right)=1$ to find the value of $\mathrm{sin}\left(\theta \right)$ rather than $\mathrm{arccos}\left(\theta \right)$ to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using $\mathrm{arccos}$ and the other using the trig identity.
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tal1ell0s
Note that $a=\sqrt{17}$,$b=\sqrt{77}$ and $\mathrm{cos}C=-\frac{20}{ab}$, which yields the area
$A=\frac{1}{2}ab\sqrt{1-{\mathrm{cos}}^{2}C}=\frac{1}{2}\sqrt{17\cdot 77-400}=\frac{3\sqrt{101}}{2}$

Zack Chase
If you calculate the angle, you introduce numerical errors. The most elegant way to calculate the area is to use the cross product. You will not need any trigonometric functions:
$A=\frac{1}{2}|\stackrel{\to }{CA}×\stackrel{\to }{CB}|$
So all you need will be some multiplications, additions(subtractions), and a square root. By the way, the answer should be $\frac{3}{2}\sqrt{101}$, if I did the math right, which is closer to your calculations