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Regarding the tetrahedron OABC in the picture, with sides $BC=10$, $AC=8$, $OA=4$, $\sin(\angle ACB)=\frac{3}{4}$ and $\triangle ABC \equiv \triangle OBC$. With this, you find that the area of $\triangle ABC=30$.

Moreover, if $AH$ denotes the perpendicular line drawn from point $A$ to side $BC$, you can find its value is $6$.

Now, to the questions:

Since $\triangle ABC$ and $\triangle OBC$ have a common side, all their sides have the same values, right?

Let $\theta$ denote the angle formed by the plane $ABC$ and the plane $OBC$. How do I find $\cos \theta$ and $\sin\theta$?

And at last, how do I find the the volume?

I need help on how to visualize these concepts.

Tetrahedron

Renato
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enter image description here

As you know that $AH=6$ and $\triangle ABC\equiv\triangle OBC$, so you have $OH=6$.

Now, apply Cosine Rule in $\triangle AOH$ given that $AO=4$ since the angle between the planes $ABC$ and $OBC$ is $\angle OHA$.

SarGe
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  • But shouldn't AH be perpendicular? How do I find the volume after that? – Renato Aug 05 '20 at 11:58
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    @Renato, $AH$ **is** perpendicular to $BC$, I didn't mention as it is obvious. For volume drop perpendicular $O$ on $AH$ and find its length knowing $\angle OHA$. Now, you have base area and height of the tetrahedron. – SarGe Aug 05 '20 at 12:04
  • I managed to do it! thanks for your help! – Renato Aug 05 '20 at 12:35