I am struggling with the following four questions (look at image attached). Any help would be much appreciated:

http://farm3.static.flickr.com/2536/...777d3a53_b.jpg

Printable View

- Oct 6th 2009, 12:56 PMyeah:)Hard Questions - PLEASE HELP!!
I am struggling with the following four questions (look at image attached). Any help would be much appreciated:

http://farm3.static.flickr.com/2536/...777d3a53_b.jpg - Oct 6th 2009, 09:53 PMGrandad
Hello yeah:)

The way you find the angle between two planes is this:

- Look at the line where the two planes intersect
- Find two lines perpendicular to this line of intersection, one in each plane
- Find the angle between these two lines

So, in question 1, the planes intersect along the line BD. This line BD has two important properties, each relating to one of the planes:

- It is the base of the isosceles triangle BGD. (Can you see why it's isosceles?)
- It is the diagonal of the square ABCD.

So, can you see that if we take M as the mid-point of BD, and draw the lines MG and MC, these lines are perpendicular to BD? So it's the angle between these two lines, $\displaystyle \angle GMC$, that you need to calculate. You can do this by finding the length of MC, and then looking at the triangle MCG.

Do question 2 in a similar way: the planes intersect along the line AC. Mark its mid-point, M, and draw the lines MS and MD.

In question 3, you need to mark in the mid-point of BC, and question 4 is very similar, except that you're given the vertical height of the pyramid instead of the height of one of the slant-edges.

Have another go at these questions. Get back to us if you need more help.

Grandad

- Oct 7th 2009, 02:32 PMyeah:)
Thank you very much for that. Unfortunately, I could do the first two, but I still struggle with numbers 3 and 4. Could you please show me how to do them (i.e. full answer), so that I can see where I am going wrong, because these questions are already getting on the nerves!!!

- Oct 7th 2009, 10:48 PMGrandad
Hello yeah:)

Here's what you need - but you can finish off!

3) If $\displaystyle M$ is the mid-point of $\displaystyle BC$, then the angle we want is $\displaystyle \angle EMN$.

$\displaystyle MN=\tfrac12AB=1.5$

$\displaystyle EN^2=5^2-NC^2$ (using Pythagoras on $\displaystyle \triangle ENC$)

and $\displaystyle NC=\tfrac12AC=\tfrac12\sqrt{3^2+3^2}$ (from $\displaystyle \triangle ABC$)

You now have enough to calculate $\displaystyle EN$. Then use $\displaystyle \tan(\angle EMN)=\frac{EN}{MN}$ to find the angle you need.

4) This is easier, because we already have the vertical height. Using the same lettering as in #3, then, the angle you want is the same, and you can simply say:

$\displaystyle EN = \text{vertical height} =145$

$\displaystyle MN = \tfrac12AB=115$

So $\displaystyle \tan(\angle EMN)=\frac{EN}{MN}=...$

You can finish it now.

Grandad