Last one today (I hope).

[The Preacher]

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

 

[stapel]

Draw a parallelogram; the scale of the drawing isn't important. (I drew mine with the parallelogram "leaning" to the right.)

Label the base as "10"; label the slanty sides as "6".

The area is 30sqrt[2]. The area is also A = bh. So what is the height?

Draw a dashed line for the height "h". Label with the "height" value you just found. This forms a right triangle with height "h" and hypotenuse "6". The angle opposite the height can be related to these two labelled sides using a [which trig function?] ratio.

Take the inverse of that to find the angle, or just simplify the ratio and recall the basic reference angle that has that value.

Eliz.

 

[pka]

If a and b are the lengths of the sides of a parallelogram then its area is:
\(\displaystyle \L
ab\sin (\theta )\) .

 

[The Preacher]

If a and b are the lengths of the sides of a parallelogram then its area is:
\(\displaystyle \L
ab\sin (\theta )\) .

Thanks, pka.

By the way, what's a "\(\displaystyle \theta\)"? I've done sine ratios, but I have no idea what that stands for.

Draw a dashed line for the height "h". Label with the "height" value you just found. This forms a right triangle with height "h" and hypotenuse "6".

Um, I don't remember finding a height value. :? Just the little newbie formula for the angles. I've labeled it with h, but... for some reason I can't find the height. I forget so much of the stuff I go over in math.

Here's what I've got so far:

(Sorry about the poor Paint skills).

That's basically all the stuff you told me to draw, except I can't figure out the height. I'm guessing it has to do with using the area and the sides or something, though.

 

[Gene]

As Staple said, the area = base*height so
30sqrt(2) = 10h
h=3sqrt(2)
\(\displaystyle \Theta\) is the greek letter theta which is commonly use as an angle variable. You can use any variable name.
Pka said A=a*b*sin(\(\displaystyle \Theta\))
10*3sqrt(2)=10*6*sin(\(\displaystyle \Theta\))

 

[The Preacher]

So... this is the sine, right?

\(\displaystyle \[
\frac{{3\sqrt 2 }}{6}
\]\)

That means I need to divide \(\displaystyle 3\sqrt 2\) by 6, right?

...I don't know how to divide radicals.


EDIT:

Nevermind, I was able to figure it out. Thanks a lot for helping me guys, especially stapel!

God bless y'all,
-The Preacher