Last one today (I hope).

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
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
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

Elite Member
Joined
Jan 29, 2005
Messages
7,823
If a and b are the lengths of the sides of a parallelogram then its area is:
\(\displaystyle \L
ab\sin (\theta )\) .
 

The Preacher

Junior Member
Joined
Sep 13, 2005
Messages
53
pka said:
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.

stapel said:
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. :? :oops: 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

Senior Member
Joined
Oct 8, 2003
Messages
1,904
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

Junior Member
Joined
Sep 13, 2005
Messages
53
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. :oops:


EDIT:

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

God bless y'all,
-The Preacher
 
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