find the angle BCD?
2006-12-12 02:51:29 · 4 answers · asked by chaz 1 in Science & Mathematics ➔ Mathematics
AB=8cm
BC=10cm
VA=12cm
2006-12-12 03:34:55 · update #1
area of ABCD=40cm^2
=l*AB where l is perpendicular
to AB
therefore,l=40/8=5cm
sinDCB=l/10=5/10=0.5
arcsin(0.5)=30 degrees
hence,angle BCD=30 degrees
i hope that this helps
2006-12-12 05:58:42 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The vertical height of V above A will not affect this calculation.
Area of a parallelogram = base x perpendicular height
40 = 8 x h
h = 40 / 8
h = 5cm
sin (angle BCD) = opposite / hypotenuse
sin (angle BCD) = 5/10
sin (angle BCD) = 1/2
angle BCD = 30 degrees.
2006-12-12 10:25:40 · answer #2 · answered by Kemmy 6 · 0⤊ 0⤋
I strongly believe there is not enough information given in your question to lead to a particular solution. I sense that the pyramid's base, a parallelogram, is not well-defined simply by knowing its area (40 sq. cm).
Could you re-post your question with additional information?
2006-12-12 03:19:09 · answer #3 · answered by Tim GNO 3 · 0⤊ 0⤋
what
2006-12-12 03:45:28 · answer #4 · answered by Twista-Adzy 2 · 0⤊ 0⤋