Can we find the lengths and angles in this pyramid?

Can we find the lengths and angles in this pyramid?

Question

An equilateral triangle $ABC$ has side length $6$ units. The three altitudes of the triangle meet at $N$. Show that $AN=2\sqrt{3}$ units.

This triangle is the base of a pyramid whose apex $V$ lies on the line through $N$ perpendicular to the plane $ABC$. Given that $VN=2$ units, prove that $\widehat{VAN}=30^\circ.$

The perpendicular from $A$ to the edge $VC$ meets $CV$ produced at $R$. Prove that $AR=\frac{3}{2}\sqrt{7}$ units, and find the exact value of $\cos \widehat{ARB}$.

Suggestion

Could we draw a picture of $ABC$ with the altitudes and $N$ marked? Can you spot any similar triangles? Or right-angled triangles?

Could we draw another picture? In three dimensions this time?

The phrase “$CV$ produced” means, “the line segment $CV$ extended”.

Could we draw a picture of $AVC$, marking in $R$? Which lengths in this triangle do we already know?

Solution

To find the length $AN$, let’s call the midpoint of $AC, M$, and the midpoint of $BC, P$.

Then the triangles $ANM$ and $ACP$ are similar, since they are both right-angled, and they also share another angle.

Using Pythagoras, we find that $AP=\sqrt{6^2-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}$. We know that

$$\frac{AN}{3}=\frac{AN}{AM}=\frac{AC}{AP}=\frac{6}{3\sqrt{3}},$$

since the triangles are similar, and so $AN=\dfrac{6}{\sqrt{3}}=6\dfrac{\sqrt{3}}{3}=2\sqrt{3}$ units.

Alternatively, we could have recognised that $ANM$ is a standard $30$-$60$-$90$ triangle with edges in the ratio $1:2:\sqrt{3}$.

We know $AN$, and $VN$ is given in the question giving $\tan\widehat{VAN}=\dfrac{2}{2\sqrt 3}=\dfrac{1}{\sqrt 3}$ which we should recognize as $\tan 30^\circ$ giving $\widehat{VAN}=30^\circ.$

Alternatively we find $AV = 4$, by Pythagoras.

We now have that $\sin\widehat{VAN}=\frac{2}{4}=\frac{1}{2},$ Therefore $\widehat{VAN}=30^\circ.$

The perpendicular from $A$ to the edge $VC$ meets $CV$ produced at $R$. Prove that $AR=\dfrac{3}{2}\sqrt{7}$ units, and find the exact value of $\cos \widehat{ARB}$.

The triangle $AVC$

By symmetry, we know that $AV=VC$. We’ll use the cosine rule to find $\widehat{ACV}$.

Applying the cosine rule to

$ACV$

, we have that

$$\begin{align*} 4^2&=6^2+4^2-2\times 4\times 6\times \cos\widehat{ACV}, \\ 48\cos\widehat{ACV}&=36, \\ \cos\widehat{ACV}&=\frac{3}{4}. \end{align*}$$

Now by right-angled trigonometry, we have that $\sin\widehat{ACV}=\frac{AR}{6},$ and so $AR=6\sin\widehat{ACV}=6\sqrt{1-\cos^2\widehat{ACV}}=6\sqrt{\frac{7}{16}}=\frac{6}{4}\sqrt{7}=\frac{3}{2}\sqrt{7}.$

Alternatively, we could say $RV = x$, and $AR = y$, and then use Pythagoras twice to create simultaneous equations that enable us to find $y$.

We need now to compute $\cos\widehat{ARB}$.

By symmetry, we have that $BR=AR$. So we can apply the cosine rule to $ARB$.

The triangle $ARB$

We have that

$$\begin{align*} 6^2&=2\left(\frac{3\sqrt{7}}{2}\right)^2-2\left(\frac{3\sqrt{7}}{2}\right)\left(\frac{3\sqrt{7}}{2}\right)\cos\widehat{ARB}, \\ \implies 36&=\frac{9\times 7}{2}-\frac{9\times 7}{2}\cos\widehat{ARB} \\ \implies 63\cos\widehat{ARB}&=63-72,\\ \implies \cos\widehat{ARB}&=-\frac{9}{63}=-\frac{1}{7}. \end{align*}$$