Hello! I need to be reminded how to find a angle between two planes! So, I have a problem:

Via isosceles triangle BC goes a plane a. The distance from A to plana a (I have labelled it as AM) is square root of 10. PROBLEM: find a angle between triangle and plane a of BC=18 amd AB=AC=11 (look at the picture below!)

Question

Hello! I need to be reminded how to find a angle between two planes! So, I have a problem:

Via isosceles triangle BC goes a plane a. The distance from A to plana a (I have labelled it as AM) is square root of 10. PROBLEM: find a angle between triangle and plane a of BC=18 amd AB=AC=11 (look at the picture below!)

kamille · Answer

attachment

kamille · Answer

@phi

kamille · Answer

@jhonyy9 any help!?

kamille · Answer

Maybe you don't understand anything? I can try to reword it :o

phi · Answer

You are saying the base BC lies in the plane  and A is above the plane

kamille · Answer

There are two planes : ABC (triangle) and plane called "a". this two planes interesects . and yes, A is "above" plane "a"

kamille · Answer

Still don't get the problem?

phi · Answer

I get the problem, it's the answer that is slowing me down

kamille · Answer

Well, firstly, can you explain me how to get an angle between these two planes?

phi · Answer

I just noticed BC=18.  so now we can do this

we can find the distance from C to M using pythagoras

phi · Answer

call the mid point between B and C   D
CD is 1/2 of 18= 9
now use pythagoras using CD  and CM to find DM

phi · Answer

finally, the tan(angle)  = sqrt(10)/DM

kamille · Answer

oh, thank you. can you draw me exatly where the angle between these two planes are? Or I can draw and you can say if it is correct?

phi · Answer

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