Two opposite sides of a parallelogram each have a length of #6 #. If one corner of the parallelogram has an angle of #(7 pi)/8 # and the parallelogram's area is #5 #, how long are the other two sides?

To find the lengths of the other two sides of the parallelogram, we first need to determine the length of the diagonal formed by the two given sides and the angle between them. We can use the Law of Cosines to find this diagonal length.

Given:

• Length of one side of the parallelogram = 6
• Angle between the two sides = ( \frac{7\pi}{8} )

Using the Law of Cosines: [ c^2 = a^2 + b^2 - 2ab \cos(C) ]

Where:

• ( c ) is the length of the diagonal
• ( a ) and ( b ) are the lengths of the given sides (both equal to 6)
• ( C ) is the angle between the sides

Plugging in the values: [ c^2 = 6^2 + 6^2 - 2(6)(6) \cos\left(\frac{7\pi}{8}\right) ]

Evaluate ( \cos\left(\frac{7\pi}{8}\right) ) using trigonometric identities or a calculator. Then, solve for ( c ).

Once you have the length of the diagonal, you can divide the parallelogram into two congruent triangles and find their areas using the formula: [ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]

Since the given area of the parallelogram is 5, you can set up an equation involving the areas of the triangles and solve for the height. Once you have the height, you can use it to find the lengths of the other two sides of the parallelogram.