# Trigonometry - Calculating the pyramid volume - where am I going wrong?

There seems to be a mistake in my solution to a problem, which I cannot find.

The problem:

There be the points $A(-2, -2, 0)$, $B(2, 6, 0)$, $C(8, 4, 0)$ and $D(0, 0, 9)$. Calculate the volume of the pyramid, whose base area is the triangle $ABC$ and whose apex is $D$.

Seems simple enough.

I calculated the three ground sides, which are $AB = \sqrt{80}$, $BC = \sqrt{52}$ and $CA = \sqrt{136}$

Then I got the height $h$ of $AB$ by finding the cosinus of the angle enclosed by $AB$ and $CA$, then using $\arccos$ to get the angle in radians $\Big(\dfrac{41\sqrt{170}}{680}\Big)$, whose sinus I multiplied with the side $CA$. Result: approximately $8.2523$.

The height of the base gets multiplied with the side it is perpendicular to (which in this case is $AB$), divided through $2$, which becomes the base area: approximately $36.9055$.

Lastly, the height of the pyramid is $9$, which can be determined by the z-coordinate of point $D$. Thus, according to the formula $V = \dfrac{A * h}{3}$, the volume is approximately $110.7165$.

But well, my solution does not match with the one presented in my book. So, where did I go wrong?

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Why did you compute the arcsin of something just take the sine of it again? – Qiaochu Yuan May 28 '12 at 17:52
Ups, it should be $\arccos$ instead. – Miroslav Cetojevic May 28 '12 at 17:56
Well, your computation of the base area is wrong; it's $28$. So you either computed the length of the altitude incorrectly or the length of $AB$ incorrectly. – Qiaochu Yuan May 28 '12 at 18:00
By the way, $BC=\sqrt{40}$. – André Nicolas May 28 '12 at 18:00

An idea: if you already have the lengths of $\,\underline{u}:=\overrightarrow{AB}=(4,8,0)\,\,,\,\,\underline{v}:=\overrightarrow{AC}=(10,6,0)$ , why don't you use the trigonometric formula for a triangle's area? First, the angle between $\,\underline{u}\,,\,\underline{v}\,$:$$\cos\theta:=\frac{\underline{u}\cdot\underline{v}}{||\underline{u}||\,||\underline{v}||}=\frac{11}{\sqrt{170}}\Longrightarrow \theta=32.1192^\circ$$ $$S_\Delta=\frac{||\underline{u}||\,||\underline{v}||\sin\theta}{2}=28$$Since the pyramid's height is clearly $\,9\,$ , the volume is $\,\,\displaystyle{V=\frac{S_\Delta\cdot 9}{3}=84}$
There is an alternative approach. Volume of a pyramid is $\frac{1}{6}$ of the volume of a parallelepiped built on $AB$, $AC$ and $AD$. Therefore $$V=\frac{1}{6}\left(\vec{AC} \times \vec{AB}\right)\cdot \vec{AD}$$ In components $$V=\frac{1}{6}\left|\begin{array}{ccc} 10 & 4 & 2\\ 6 & 8 & 2\\ 0 & 0 & 9 \end{array}\right|=\frac{1}{6}9\left(80-24\right)=84$$