finding height and angle?

**andyboy179** · February 18th 2011, 06:54 AM

hi, i need to try and find the angle of VAC and find the height of this pyramid!

to find the height of it would i do:
1/3x20x15= 100??

as for finding the angle of VAC i don't know what to do!

if you can help please let me know how to do this please as i will be very grateful!!

thanks!

**earboth** · February 18th 2011, 07:52 AM

Originally Posted by andyboy179

hi, i need to try and find the angle of VAC and find the height of this pyramid!

Click image for larger version.

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to find the height of it would i do:
1/3x20x15= 100??

as for finding the angle of VAC i don't know what to do!

if you can help please let me know how to do this please as i will be very grateful!!

thanks!

As far as I understand your sketch ...

1. The vertex of the pyramid (Obviously you labeled it V) is perpendicularly above the intersection M of the two diagonals of the base area.
The diagonal has a length of 25.

2. The height h, a half diagonal and an edge form a right triangle:

$24^2 = 12.5^2+h^2$

Solve for h.

3. The angle $\angle(MVC)$ is half as large as the angle $\angle(AVC)$ . Use the right traingle MCV to determine the angle $\angle(MVC)$ :

$\sin(\angle(MVC)) = \dfrac {12.5}{24}~\implies~\angle(MVC)\approx 31.388^\circ$

**andyboy179** · February 18th 2011, 08:16 AM

Height

you said 24^2= 12.5^2 + H^2

would h= 419.75?

angle of VAC

i don't really understand where MVC came from??

**earboth** · February 18th 2011, 12:01 PM

Originally Posted by andyboy179

Height

you said 24^2= 12.5^2 + H^2

would h= 419.75?

No. That's $h^2$ .

angle of VAC

i don't really understand where MVC came from??

I've attached a new sketch with all the necessary labels so you can see which angles are meant.

**andyboy179** · February 19th 2011, 03:39 AM

Originally Posted by earboth

No. That's $h^2$ .

I've attached a new sketch with all the necessary labels so you can see which angles are meant.

to get the height would i square root 419.75= 20.49??

i know what the angles are now but how would this help me find the angle of VAC?

**HallsofIvy** · February 19th 2011, 03:47 AM

Once you know angle VAC, use the fact that the angles in a triangle sum to 180 degrees (or $\pi$ radians) and that, since traingle AVC is an isosceleles triangle, that angles VAC and VCA are equal.

**andyboy179** · February 19th 2011, 03:55 AM

i don't understand how to work all of this out though! its really confusing

**earboth** · February 19th 2011, 10:26 AM

Originally Posted by andyboy179

i don't understand how to work all of this out though! its really confusing

1. You should have seen that $|\angle(CAV)| = |\angle(VCM)|$ .

2. Let d denote the length of one diagonal of the base area and e the length of one edge, then you see that

$\cos(\angle(VCM)) = \dfrac{\frac12 d}{e}$ .

3. Calculate the value of $\angle(VCM)$ and subsequently the value of $\angle(CAV)$ .

**scounged** · February 19th 2011, 04:13 PM

Originally Posted by andyboy179

i don't understand how to work all of this out though! its really confusing

Remember that the law of sines is your friend. I'm sure you can use it in the picture earboth has drawn in order to first calculate the angle MCV, as the angle VMC is 90 degrees. After that you don't need to do much more, as the angle VAC is equal to the angle MCV.

**andyboy179** · February 20th 2011, 03:26 AM

Would VCM=0.3?

**earboth** · February 20th 2011, 04:23 AM

Originally Posted by andyboy179

Would VCM=0.3?

No.

How did you get this result?

**andyboy179** · February 20th 2011, 07:35 AM

Sorry, misread my calculator. My answer was 74.91, to get that I did 12.5x1/2 /24=0.26 inverse cos=74.91??

**earboth** · February 20th 2011, 07:52 AM

Originally Posted by earboth

1. You should have seen that $|\angle(CAV)| = |\angle(VCM)|$ .

2. Let d denote the length of one diagonal of the base area and e the length of one edge, then you see that

$\cos(\angle(VCM)) = \dfrac{\frac12 d}{e}$ .

3. Calculate the value of $\angle(VCM)$ and subsequently the value of $\angle(CAV)$ .

Originally Posted by andyboy179

Sorry, misread my calculator. My answer was 74.91, to get that I did 12.5x1/2 /24=0.26 inverse cos=74.91??

As I've suggested in the post #8 you'll get the Cosine of $\angle(VCM)$ by calculating
$\cos(\angle(VCM)) = \dfrac{\frac12 d}{e}$
$\cos(\angle(VCM)) = \dfrac{12.5}{24} = \dfrac{25}{48}$

And now use the inverse Cosine function of your calculator.

The final result should be $\angle(VCM) \approx 58.61^\circ$

**andyboy179** · February 20th 2011, 08:02 AM

Why does it go from 12.5/24 to 25/48??

**scounged** · February 20th 2011, 09:21 AM

12.5/24=(12.5*2)/(24*2)=25/48

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