# Last one today (I hope).

#### The Preacher

##### Junior Member
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
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
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
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. :? 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
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
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