# Math Help - Triangle 2

1. ## Triangle 2

I need the method of these... For Q2, no matter how many times I do, I got 9.57 for VB...

1. Each edge of the uniform tetrahedron ABCD has the length 4a. The point P lies on BC such that BP=3a. Find the cosine of the angle APD.

i) show that the perpendicular distance of the corner A from the plane BCD is $\frac{4}{3}a\sqrt{6}$, and hence
ii) show that the angle between the line AP and the plane BCD is $sin^{-1}\sqrt{\frac{32}{39}}$

2. The horizontal base of a pyramid is an equilateral triangle ABC with each side 10m long. The vertex V of the pyramid is at a height of $\frac{10\sqrt{6}}{3}$m above the base. Find VB.

i) If the point P lies on VB such that VP= $\frac{1}{5}VB$, find AP.

1. $\frac{5}{13}$
2. VB=10m, AP= $2\sqrt{21}$m

2. Originally Posted by cloud5
...

2. The horizontal base of a pyramid is an equilateral triangle ABC with each side 10m long. The vertex V of the pyramid is at a height of $\frac{10\sqrt{6}}{3}$m above the base. Find VB.

i) If the point P lies on VB such that VP= $\frac{1}{5}VB$, find AP.

1. $\frac{5}{13}$
2. VB=10m, AP= $2\sqrt{21}$m
1. I assume that the vertex of the pyramid is vertically above the center M of the equilateral triangle. (In an equilateral triangle the orthocenter, the centroid, the common point of the three angle bisectors,... is the same point which I call here the center of the triangle)

2. The point M lies on the median through B. The distance $|\overline{BM}| = \frac23 \cdot (length\ of\ median)$

The length of the median is

$|m| = \sqrt{10^2-5^2}=5 \sqrt{3}$

Thus $|\overline{BM}| = \frac23 \cdot 5 \sqrt{3}$

3. Now use Pythgorean theorem to calculate

$|\overline{VB}| = \sqrt{ \left(\frac23 \cdot 5 \sqrt{3} \right)^2 + \left(\frac{10\sqrt{6}}{3} \right)^2 } = \sqrt{\frac{100}3 + \frac{200}3}$

Since VB = 10 the pyramid is a tetrahedron.

4. I'll leave the rest for you.

3. Originally Posted by earboth

I'll leave the rest for you.
Do you know how to answer Q1?

4. Originally Posted by earboth

2. The point M lies on the median through B. The distance $|\overline{BM}| = \frac23 \cdot (length\ of\ median)$

The length of the median is

$|m| = \sqrt{10^2-5^2}=5 \sqrt{3}$

Thus $|\overline{BM}| = \frac23 \cdot 5 \sqrt{3}$

3. Now use Pythgorean theorem to calculate

$|\overline{VB}| = \sqrt{ \left(\frac23 \cdot 5 \sqrt{3} \right)^2 + \left(\frac{10\sqrt{6}}{3} \right)^2 } = \sqrt{\frac{100}3 + \frac{200}3}$

Since VB = 10 the pyramid is a tetrahedron.

4. I'll leave the rest for you.
Can you sketch the triangle for me? I'm getting more confused... and about this $|\overline{BM}| = \frac23 \cdot (length\ of\ median)$ are you using the vector formula? why there is $\frac{2}{3}$? I'm sorry if I'm asking too much...

5. Originally Posted by cloud5
Can you sketch the triangle for me? <<<<<<<< yes
I'm getting more confused... and about this $|\overline{BM}| = \frac23 \cdot (length\ of\ median)$ are you using the vector formula? <<<<<< no
why there is $\frac{2}{3}$? I'm sorry if I'm asking too much...
1. See attachment.

2. As I've stated previously the point M is simultaneously the point of intersection of the angle bisectors, the heights and the medians.

3. The point M divides each median in triangle in the ratio 2:1 from the vertex. Thus the line segment $\overline{BM}$ must be $\tfrac23$ of the complete median.

4. The sketch was made for another question so the labeling has nothing to do with your problem.

6. Originally Posted by cloud5
I need the method of these... For Q2, no matter how many times I do, I got 9.57 for VB...

1. Each edge of the uniform tetrahedron ABCD has the length 4a. The point P lies on BC such that BP=3a. Find the cosine of the angle APD.

...

...
1. You are dealing with 3 triangles:

$\Delta DPC:$ Calculate the length of $\overline{DP}$ and $\overline{AP}$ using Cosine rule:

$|\overline{DP}| = \sqrt{a^2+16a^2-2 \cdot a \cdot 4a \cdot \frac12} = a\sqrt{13}$

$|\overline{AP}| = \sqrt{16a^2+9a^2-2 \cdot 4a \cdot 3a \cdot \frac12} = a\sqrt{13}$

2. Calculate the Cosine of the angle $\angle(DPA)$:

$\cos(\angle(DPA)) = \dfrac{16a^2-13a^2-13a^2}{-2 \cdot a\sqrt{13} \cdot a\sqrt{13}} = \dfrac{-10a^2}{-26a^2} = \dfrac5{13}$

7. So you are using cosine rule a^2+b^2 -2abCosC, only how did you know C was 1/2? Also where did the 9 come from in the second one?

8. Originally Posted by BugzLooney
So you are using cosine rule a^2+b^2 -2abCosC, only how did you know C was 1/2? Also where did the 9 come from in the second one?
1. The "faces" of a tetrahedron are equilateral triangles. Each interior angle in an equilateral triangle is 60°. And $\cos(60^\circ) = \tfrac12$.

2. The 2nd equation calculates the length of $\overline{AP}$ which belongs to the triangle $\Delta ABP$. The known sides of this triangle are: $|\overline{AB}|=4a$ and $|\overline{BP}|=3a$. And when I square 3a I'll get 9a².

9. Ah I see didn't see the 3a. I see how it is that for part 1. Which law was used to generate the cosine of DPA? Just the law is all I'm looking for so I can work it myself.

10. Originally Posted by BugzLooney
Ah I see didn't see the 3a. I see how it is that for part 1. Which law was used to generate the cosine of DPA? Just the law is all I'm looking for so I can work it myself.
That's the Cosine rule, solved for $\cos(\angle DPA)$.

In general:

$c^2 = a^2+b^2-2\cdot a\cdot b \cdot \cos(\gamma)~\implies~\boxed{\cos(\gamma)=\dfrac{c ^2-a^2-b^2}{-2\cdot a\cdot b}}$