# Thought I did this right but apparently got it wrong

• May 10th 2012, 09:18 AM
brokaliv
Thought I did this right but apparently got it wrong
Attachment 23836
Two strips of cloth with width 1 overlap at an angle a as shown. Find the area of the overlap.

I swear I am missing something on this.

Rhombus Area is base x height
Height is 1
Let base = x

unknown angle a

So all I could come up with was:

$\displaystyle x = 1/sine (a)$

Which gave me the Area = A

$\displaystyle A = 1 * (1/Sine (a))$

I don't recall ever figuring out an area with so little information given before.
• May 10th 2012, 11:05 AM
emakarov
Re: Thought I did this right but apparently got it wrong
I agree with your solution.
• May 13th 2012, 06:24 AM
Soroban
Re: Thought I did this right but apparently got it wrong
Hello, brokaliv!

If you submitted the answer $\displaystyle A = 1 * (1/Sine (a))$ to an online exam,
. . it probably did not recognize the expression.

. . First of all, sine would be entered/written $\displaystyle \sin$.

. . Second, $\displaystyle \frac{1}{\sin a}$ is equal to $\displaystyle \csc a.$

. . Third, we never EVER submit an answer like $\displaystyle x \,=\,1\cdot3$ .**

If you submitted that answer to a human teacher/professor,
. . he/she would deduct points for the very same reasons.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Consider this insulting example . . .

. . $\displaystyle \begin{array}{cccccc}\text{Solve for }x\!: & 2x + 3 &=& 9 \\ \\[-3mm] \text{Subtract 3:} & 2x &=& 9 - 3 \\ \\[-4mm] \text{Divide by 2:} & x &=& \dfrac{9-3}{2} \end{array}$

Would you submit $\displaystyle x \,=\,\frac{9-3}{2}\;?$

You are saying, "Hey, I did all the algebra!"
. . followed by, "Oh, did you want me to simplify, too?"