# finding height and angle?

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Feb 18th 2011, 06:54 AM
andyboy179
finding height and angle?
hi, i need to try and find the angle of VAC and find the height of this pyramid!

Attachment 20858

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!
• Feb 18th 2011, 07:52 AM
earboth
Quote:

Originally Posted by andyboy179
hi, i need to try and find the angle of VAC and find the height of this pyramid!

Attachment 20858

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 $\angle(MVC)$:

$\sin(\angle(MVC)) = \dfrac {12.5}{24}~\implies~\angle(MVC)\approx 31.388^\circ$
• Feb 18th 2011, 08:16 AM
andyboy179
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??
• Feb 18th 2011, 12:01 PM
earboth
Quote:

Originally Posted by andyboy179
Height

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

would h= 419.75?

No. That's $h^2$.

Quote:

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.
• Feb 19th 2011, 03:39 AM
andyboy179
Quote:

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?
• Feb 19th 2011, 03:47 AM
HallsofIvy
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.
• Feb 19th 2011, 03:55 AM
andyboy179
i don't understand how to work all of this out though! its really confusing
• Feb 19th 2011, 10:26 AM
earboth
Quote:

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 $\angle(CAV)$.
• Feb 19th 2011, 04:13 PM
scounged
Quote:

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.
• Feb 20th 2011, 03:26 AM
andyboy179
Would VCM=0.3?
• Feb 20th 2011, 04:23 AM
earboth
Quote:

Originally Posted by andyboy179
Would VCM=0.3?

No.

How did you get this result?
• Feb 20th 2011, 07:35 AM
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??
• Feb 20th 2011, 07:52 AM
earboth
Quote:

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 $\angle(CAV)$.

Quote:

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 $\angle(VCM) \approx 58.61^\circ$
• Feb 20th 2011, 08:02 AM
andyboy179
Why does it go from 12.5/24 to 25/48??
• Feb 20th 2011, 09:21 AM
scounged
12.5/24=(12.5*2)/(24*2)=25/48
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last