Ratio Problem

BigGlenntheHeavy said:
\(\displaystyle Afterthought: \ Why \ is \ triangle \ ABD \ a \ right \ triangle?\)
Because your teacher told you so, as a fait accompli !
 
Denis said:
Quite easy if we let the hypotenuse = 1 :

Ratio = SIN(33)SIN(57) / (PI x^2) where x = [SIN(57) + SIN^2(33) / SIN(57)] / 2
That's .409065723....
a = SIN(33), b = SIN(57), p = PI

Ratio = ab / (px^2) where x = (b + a^2 / b) / 2

I'm depressed: nobody seems impressed with my beautiful SINful solution :cry:
 

Hello, everyone!

I found a "geometric" approach . . . well, sort of.
I still had to use Trigonometry at the very end.



I'm trying to find the ratio of the area of this right triangle to the area of the semicircle.
The solution is supposed to be 0.409.

\(\displaystyle \text{For reference, label the points and segments.}\)

\(\displaystyle \text{The right triangle has: }\;\begin{Bmatrix}A &=& \text{lower-left vertex} \\ B &=& \text{upper vertex} \\ C &=& \text{right angle} \end{Bmatrix} \quad\begin{array}{ccc}D &=& \text{other end of diameter}\)

\(\displaystyle \text{Let: }\;\begin{Bmatrix} a &=& BC & \text{altitude} \\ b &=& AC & \text{base} \\ c &=& CD \end{Bmatrix}\)

\(\displaystyle \text{Hence: }\;\text{Area of }\Delta ABC \:=\:\tfrac{1}{2}ab\;\;[1]\)



\(\displaystyle \text{Note that }\Delta ABD\text{ is a right triangle}\)
. . \(\displaystyle \text{and }a\text{ is the altitude to the hypotenuse.}\)

. . \(\displaystyle \text{We have: }\:\frac{b}{a} \:=\:\frac{a}{c} \quad\Rightarrow\quad c \:=\:\frac{a^2}{b}\)
\(\displaystyle \text{The radius is: }\:r \:=\:\tfrac{1}{2}(b + c) \:=\:\tfrac{1}{2}\left(b + \frac{a^2}{b}}\right) \:=\:\frac{a^2+b^2}{2b}\)
\(\displaystyle \text{Hence: area of the semicircle }\:=\;\tfrac{1}{2}\pi r^2 \:=\:\tfrac{1}{2}\pi\left(\frac{a^2+b^2}{2b}\right)^2 \:=\:\frac{\pi(a^2+b^2)^2}{8b^2}\;\;[2]\)


\(\displaystyle \text{The ratio is: }\;R \;=\;\frac{\frac{1}{2}ab}{\dfrac{\pi(a^2+b^2)^2}{8b^2}} \;=\;\frac{4ab^3}{\pi(a^2+b^2)^2}\)
\(\displaystyle \text{Divide top and bottom by }b^4\!:\;\;R \;=\;\frac{\dfrac{4ab^3}{b^4}} {\pi\dfrac{(a^2+b^2)^2}{(b^2)^2}} \;=\;\frac{4\left(\dfrac{a}{b}\right)}{\pi\left[\dfrac{a^2+b^2}{b^2}\right]^2} \;=\;\frac{4\left(\dfrac{a}{b}\right)}{\pi\left[\left(\dfrac{a}{b}\right)^2 + 1\right]^2}\)


\(\displaystyle \text{Since }\,\frac{a}{b}\,=\,\tan33^o\)

. . \(\displaystyle \text{we have: }\;R \;=\;\frac{4\tan33^o}{\pi\left(\tan^2\!33^o + 1\right)^2} \;=\;\frac{4\tan33^o}{\pi\sec^4\!33^o} \;=\;\frac{4}{\pi}\sin33^o\cos^3\!33^o\)

 
Denis said:
Denis said:
Quite easy if we let the hypotenuse = 1 :

Ratio = SIN(33)SIN(57) / (PI x^2) where x = [SIN(57) + SIN^2(33) / SIN(57)] / 2
That's .409065723....
a = SIN(33), b = SIN(57), p = PI

Ratio = ab / (px^2) where x = (b + a^2 / b) / 2

I'm depressed: nobody seems impressed with my beautiful SINful solution :cry:

No matter - how elegant - Who would ever want to join a hockey-puck chasing SINner??
 
H..ALL !!!
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°)...!!!
" TNX & GOD BLESS "


SPAM
 
Subhotosh Khan said:
No matter - how elegant - Who would ever want to join a hockey-puck chasing SINner??
Well, sure more fun chasing pucks than getting dressed like a surgical doctor and bowling cricket balls:
the players have to pay people to come watch them :)
 
markjay50 said:
H..ALL !!!
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°)...!!!
" TNX & GOD BLESS "


SPAM

It has been explained in the postings above - please read those and ask question without your "SPAM".
 
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