The Preacher
Junior Member
- Joined
- Sep 13, 2005
- Messages
- 53
Surprising as it may be, I solved a few problems on my own, too. Haha. This one's a word problem:
A parallelogram with sides of 6 and 10 has an area of \(\displaystyle \
30\sqrt 2
\\). Find the measure of each angle of the parallelogram.
Small angle _____\(\displaystyle \
^\circ
\\)
Large angle _____\(\displaystyle \
^\circ
\\)
Just about all I can think of is that the opposite angles will be equal (duh). If I can figure out one angle, then I can double it. Then I can take that sum and subtract it from 360 \(\displaystyle ^\circ\), and divide that difference by two to get my other angle.
Let x be the measure of angle 1, and y be the measure of angle 2.
360-\(\displaystyle \
x^2
\\)=\(\displaystyle y^2\)
If I can figure out x or y, I can figure out the other. Can anyone give me a hint in the right direction?
I hope the equation I just came up with isn't common knowledge or something I'm supposed to know already. Haha. Thanks a lot.
-The Preacher
A parallelogram with sides of 6 and 10 has an area of \(\displaystyle \
30\sqrt 2
\\). Find the measure of each angle of the parallelogram.
Small angle _____\(\displaystyle \
^\circ
\\)
Large angle _____\(\displaystyle \
^\circ
\\)
Just about all I can think of is that the opposite angles will be equal (duh). If I can figure out one angle, then I can double it. Then I can take that sum and subtract it from 360 \(\displaystyle ^\circ\), and divide that difference by two to get my other angle.
Let x be the measure of angle 1, and y be the measure of angle 2.
360-\(\displaystyle \
x^2
\\)=\(\displaystyle y^2\)
If I can figure out x or y, I can figure out the other. Can anyone give me a hint in the right direction?
I hope the equation I just came up with isn't common knowledge or something I'm supposed to know already. Haha. Thanks a lot.
-The Preacher