Two opposite sides of a parallelogram each have a length of #12 #. If one corner of the parallelogram has an angle of #(5 pi)/6 # and the parallelogram's area is #48 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram each have a length of #12    #.  If one corner of the parallelogram has an angle of #(5 pi)/6       # and the parallelogram's area is #48      #, how long are the other two sides?

Violet Aldrich · Accepted Answer

#b=2# Area of Parallelogram #=48=ab sintheta# where #a=12# and #b=?# and #theta=(5pi)/6# or #48=12timesbtimessin((5pi)/6)# or #b=48/12times(0.5)# or #b=4times(0.5)# or #b=2#

Avery Allen · Answer

undefined Given that two opposite sides of the parallelogram have a length of 12, and the area of the parallelogram is 48, we can use the formula for the area of a parallelogram:

$$ \text{Area} = \text{base} \times \text{height} $$

Since the height corresponds to the length of the other two sides, we can let $h$ represent the length of these sides.

Using the given angle of $ \frac{5\pi}{6} $, we can find the height using the formula:

$$ \text{Area} = \text{base} \times \text{height} = 12 \times h \sin(\frac{5\pi}{6}) $$

Solving for $h$:

$$ h = \frac{\text{Area}}{12 \sin(\frac{5\pi}{6})} $$

$$ h = \frac{48}{12 \sin(\frac{5\pi}{6})} $$

$$ h = \frac{48}{12 \times \frac{1}{2}} $$

$$ h = \frac{48}{6} = 8 $$

So, the length of the other two sides of the parallelogram is 8.