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DABC is a tetrahedron and ABC is an equilateral triangle and $AB=BC=CA=2a$

Given that $DA=DB=DC$ and the distance to the plane $ABC$ from $D$ is $3a$

Find the angle between $DA$ and $ABC$.

When I was trying to draw a sketch of the tetrahedron arouse a problem,what is the distance mentioned in problem,is it the distance from $D$ to center of $ABC$ triangle? I'll appreciate if someone can explain this.

Michael Rozenberg
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ranmal fernando
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  • Hmm, $AB=BC=CD=2a$ ? Do you mean $AB=BC=CA = 2a$ instead. Otherwise, together with $DA=DB=DC$ and $ABC$ is equilateral will imply all side lengths are $2a$. – achille hui Jun 17 '17 at 04:00
  • @achille hui no change dear..I feel it's impossible to have these values..Isn't it?.. – ranmal fernando Jun 17 '17 at 04:13
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    if $CD = 2a$, then it is impossible for the distance between $D$ and plane $ABC$ to be $3a$ (assume $a > 0$). On the other hand, if it is a typo for $CA = 2a$, then the distance between $D$ and plane $ABC$ indeed equals to the distance between $D$ and centroid of $ABC$ (for this particular case). – achille hui Jun 17 '17 at 04:17

2 Answers2

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Let $DH$ is an altitude of the tetrahedron.

Since $DA=DB=DC$, we get $\Delta DAH\cong\Delta DBH\cong\Delta DCH$,

which says that $AH=BH=CH=\frac{2a}{\sqrt3}$.

Thus, we need to find $\measuredangle DAH$ and we have: $$\measuredangle DAH=\arctan\frac{3a}{\frac{2a}{\sqrt3}}=\arctan\frac{3\sqrt3}{2}.$$

Michael Rozenberg
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Let $D'$ be the point $D$ projected in $ABC$.

$D'$ is in the centre of the equilateral triangle; disects its height in $2:1$ ratio:

  • $\overline{AD'}=2{\sqrt 3 \over 3}a$.

Hence the superior side & base angle is:

  • $\measuredangle DAD'=\text{atan}({3\sqrt 3 \over 2})$.

Let $D''$ be the point $D$ projected on $\overline{AB}$.

$D''$ is the middle point of $\overline{AB}$:

  • $\overline{D''A}=a$,

By Pithagoras:

  • $\overline{DD''}=2\sqrt{7\over 3}$, and,

Finally the superior side & bottom side angle is:

  • $\measuredangle DAD''=\text{atan}(2\sqrt \frac 73)$.
Brethlosze
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