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So, it's |dw:1353327509163:dw| and |dw:1353327558921:dw| Now, first we'll need to find the length of the hypotenuse and vertex of right angle. That's done by using the tan 45. Now, we'll find the distance from the vertex of the right angle to plane B. Since we have the length, which will be \(4 \tan 45^o\), we'll apply the sine to find that length. It will be |dw:1353327784152:dw| i.e. (4 tan 45)(sin 60)
Do you need further help?
wait.. i'm going to analyze it first
find the distance from the vertex of the right angle to plane B = 4 square root of 3 length of the projection of each leg of the triangle on plane B = 8 is this correct?? regardless of the unit
the first might be 2 sqrt 3 the second..hmm can i see your workings?
aaaawwww.. i was wrong.... what i did was the 30-60-90 angles, but it should be 45-45-90.. right?
why 2 sqrt 3 ?? i got 2 sqrt 2
4 tan 45 = 4 4 sin 60 = 2 sqrt 3?
i don't get it T.T
lol for the first I find the length of d |dw:1353329595834:dw| to do this, i need |dw:1353329616295:dw|
please explain me this one "Plane A which contains an isosceles right triangle forms a dihedral angle of 60 degrees with another plane B." it's kinda confusing
It's the first pic. It means that the plane A is at angle of 60 with B
And on the plane A there is a isoscles right angle triangle.
|dw:1353387385535:dw| is it like this?? sorry for the drawing
Nice drawing. |dw:1353330325620:dw|
is that the right position of the isosceles triangle?
Yup. Since the "the hypotenuse of the triangle lies in plane B"
so where to put the imaginary line for this," from the vertex of the right angle to plane B" and "the length of the projection of each leg of the triangle on plane B."
and where is the 60 degree angle?? sorry if i have so many questions
|dw:1353330869441:dw| lol dun worry about it
where does projection of each leg of the triangle refers to??
I was thinking that they refer to the same line actually...because the triangles projection is really that point.
atlast i got it.. ahhaa.. thanks for the help!