Question

A parallelogram with sides of 6 and 10 has an area of 30. Find the measure of each angle of the parallelogram.

small angle °

large angle °

does anyone know this

Answer 1

isnt the area
\[ab\sin(\theta)\]?

Answer 2

let me look it up but i think that is the formula

Answer 3

For this you need to know the following formulas:
\[\left| \left| A x B \right| \right| = \left| \left| A \right| \right| * \left| \left| B \right| \right| * \left| \sin \theta \right|\]

Answer 4

yes that is it. ok so just plug in the numbers and solve for
\[\theta\]

Answer 5

Once you find that theta you can just subtract it from 180 to find the other angle.

Answer 6

\[60\sin(\theta)=30\]
\[\sin(\theta)=\frac{1}{2}\] etc,
@oreostar this is correct yes?

Answer 7

lolz omg i forgot to put 30square root of 2 sorry

Answer 8

oreostar sent the "even/odd" version. easy enough to understand, if you can keep track of the lines that say "this is even so that is even" etc

Answer 9

well it doesn't matter you get
\[60\sin(\theta)=30\sqrt{2}\]

Answer 10

so
\[\sin(\theta)=\frac{\sqrt{2}}{2}\] making
\[\theta=\frac{\pi}{4}\]

Answer 11

or if you are working in degrees, 45 degrees

Answer 12

solution to previous question has been completed

Answer 13

well first u calculate the height: area = height * side
therefore, height = area/side = 30/10 = 3 (if u choose the other side you will get the other hide which is 30/6 = 5) now, you know that one of the corners let's call it θ1 has sin: sinθ1 = height /side = 5/10 = 1/2 therefore θ1 = 60 degrees. Then the opposite corner of θ1 let's call it θ2 will be equal to θ1 and 60 degrees and since the other two corners are equal and have sum 360 all together θ3 = θ4 = (360 - 2*60) / 2 =120 degrees each so: 60,60,120,120