Ratio Problem
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Hint:jschwa1 said:
The hypotenuese of the triangle = D * cos(33°) ? where D is the diameter of the circle.
Height = ??
Base = ??
Area = ??
mmm4444bot
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Using Subhotosh's hint, I also get 0.409 (rounded).
I'm hoping that it's not obvious why the cosine of that angle gives the decimal percent of the diameter that the hypotenuse is.
I'm hoping that it's not obvious why the cosine of that angle gives the decimal percent of the diameter that the hypotenuse is.
mmm4444bot
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I asked the guy at Beth's Cafe for some scratch paper; Subhotosh's hint haunted my hamburger.
I thought that I had an insight, but I resorted to my usual 24-step program of analytical geometry, to solve the exercise, instead.
So, the "hint" that I come up with is that the base of the triangle is: 2r cos(33°)^2, where r is the radius.
Wow. I'd like to see Subhotosh's work. I can't believe that's a hint! :wink:
I thought that I had an insight, but I resorted to my usual 24-step program of analytical geometry, to solve the exercise, instead.
So, the "hint" that I come up with is that the base of the triangle is: 2r cos(33°)^2, where r is the radius.
Wow. I'd like to see Subhotosh's work. I can't believe that's a hint! :wink:
Hello, jschwa1!
\(\displaystyle \text{For reference, label the points,}\)
\(\displaystyle \text{The right triangle has: }\;\begin{Bmatrix}A &=& \text{lower-left vertex} \\ B &=& \text{upper vertex} \\ C &=& \text{right angle} \end{Bmatrix} \quad\begin{array}{ccc}D &=& \text{other end of diameter}\)
\(\displaystyle \text{Let }a = AC.\)
\(\displaystyle \text{In right triangle }ABC\!:\;\;\tan33^o \:=\:\frac{BC}{a} \quad\Rightarrow\quad BC \:=\:a\tan33^o\)
\(\displaystyle \text{Area of }\Delta ABC\:=\:\frac{1}{2}(AC)(BC) \:=\:\frac{1}{2}(a)(a\tan33^o) \:=\:\frac{1}{2}a^2\tan33^o\) .[1]
\(\displaystyle \text{In right triangle }ABD\!:\;\;\sec33^o \:=\:\frac{AD}{a\sec33^o} \quad\Rightarrow\quad AD \:=\:a\sec^2\!33^o\;\text{ (diameter)}\)
\(\displaystyle \text{Area of semicircle: }\;\frac{1}{2}\pi\left(\frac{a\sec^2\!33^o}{2}\right)^2 \;=\;\frac{1}{8}\pi a^2\sec^4\!33^o\) .[2]
\(\displaystyle \text{The desired ratio is [1] over [2]: }\;R \;=\;\frac{\frac{1}{2}a^2\tan33^o} {\frac{1}{8}\pi a^2\sec^4\!33^o}\)
. . \(\displaystyle \text{Therefore: }\:R \;=\;\frac{4\tan33^o}{\pi\sec^4\!33^o} \;=\; \frac{4}{\pi}\sin33^o\cos^3\!33^o \;=\;0.409065723\)
\(\displaystyle \text{For reference, label the points,}\)
\(\displaystyle \text{The right triangle has: }\;\begin{Bmatrix}A &=& \text{lower-left vertex} \\ B &=& \text{upper vertex} \\ C &=& \text{right angle} \end{Bmatrix} \quad\begin{array}{ccc}D &=& \text{other end of diameter}\)
\(\displaystyle \text{Let }a = AC.\)
\(\displaystyle \text{In right triangle }ABC\!:\;\;\tan33^o \:=\:\frac{BC}{a} \quad\Rightarrow\quad BC \:=\:a\tan33^o\)
\(\displaystyle \text{Area of }\Delta ABC\:=\:\frac{1}{2}(AC)(BC) \:=\:\frac{1}{2}(a)(a\tan33^o) \:=\:\frac{1}{2}a^2\tan33^o\) .[1]
\(\displaystyle \text{In right triangle }ABD\!:\;\;\sec33^o \:=\:\frac{AD}{a\sec33^o} \quad\Rightarrow\quad AD \:=\:a\sec^2\!33^o\;\text{ (diameter)}\)
\(\displaystyle \text{Area of semicircle: }\;\frac{1}{2}\pi\left(\frac{a\sec^2\!33^o}{2}\right)^2 \;=\;\frac{1}{8}\pi a^2\sec^4\!33^o\) .[2]
\(\displaystyle \text{The desired ratio is [1] over [2]: }\;R \;=\;\frac{\frac{1}{2}a^2\tan33^o} {\frac{1}{8}\pi a^2\sec^4\!33^o}\)
. . \(\displaystyle \text{Therefore: }\:R \;=\;\frac{4\tan33^o}{\pi\sec^4\!33^o} \;=\; \frac{4}{\pi}\sin33^o\cos^3\!33^o \;=\;0.409065723\)
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 .409, but I can't figure out how to get it.
Just another thought.
Let the triangle be ABC, A left vertex, B right vertex and C upper vertex.
Let Let AB = x, BC = y and AC = z.
Let r = the radius of the semicircle.
Then, z = 2r(cos(33)), x = 2r(cos^2(33)) and y = 2r(sin(33))(cos(33))
Triangle area = [2r(sin(33))cos(33)2r(cos^2(33))]/2 = 2r^2(sin(33))(cos(33))cos^2(33))
Semicircle area = Pi(r^2)/2
Tarea/SCarea = 4r^2(sin(33))(cos(33))(cos^2(33))/Pi(r^2)
With r = 1, Tarea/SCarea = .409
TchrWill - please check your algebra - r[sup:1zd6w3oz]2[/sup:1zd6w3oz] should cancel out from the numerator and the denominator (making the assumption r=1 unneccessary)
Thanks for the acute observation.
Sorry for the tardiness.
Just another thought.
Let the triangle be ABC, A left vertex, B right vertex and C upper vertex.
Let Let AB = x, BC = y and AC = z.
Let r = the radius of the semicircle.
Then, z = 2r(cos(33)), x = 2r(cos^2(33)) and y = 2r(sin(33))(cos(33))
Triangle area = [2r(sin(33))cos(33)2r(cos^2(33))]/2 = 2r^2(sin(33))(cos(33))cos^2(33))
Semicircle area = Pi(r^2)/2
Tarea/SCarea = 4r^2(sin(33))(cos(33))(cos^2(33))/Pi(r^2)
With r = 1, Tarea/SCarea = .409
TchrWill - please check your algebra - r[sup:1zd6w3oz]2[/sup:1zd6w3oz] should cancel out from the numerator and the denominator (making the assumption r=1 unneccessary)
Thanks for the acute observation.
Sorry for the tardiness.
mmm4444bot
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Denis said:
Seriously, yes, that's an option. The exercise implies that the ratio is the same for all diameters, so we could start by assigning any value.
(Choosing D = 1 would make D "disappear" from the calculations even sooner.)
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soroban said:
Working backwards, I can resolve this into the sine ratio BD/AD, but it takes several steps.
Will you show me how you determined it, Soroban?
D
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But why a$$-u-me something - that we really don't need to????
After finishing the calculation (specially at high-school level) - the student will have nagging suspicion that the results are true only for r = 1 or D=1
After finishing the calculation (specially at high-school level) - the student will have nagging suspicion that the results are true only for r = 1 or D=1
mmm4444bot
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Subhotosh Khan said:
I think Denis meant "assign D = 1" versus "assume the diameter = 1".
In other words, a numerical approach (boring!) versus an algebraic approach.
Subhotosh, how did you arrive at your hint? I had to go through Pythagorean Theorem, trig identity, and radical simplification. Did you do these, too?
In the figure, if you create another triangle that is inscribed in the circle by drawing a line segment from the current triangle's top vertex to the point on the right where the diameter intersects the circle, you create another right triangle and can use Subhotosh's hint.
D
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Re:
hypotenuese = D*cos(33°)
Base = D*cos(33°) * cos(33°) = D*cos[sup:15afxeuc]2[/sup:15afxeuc](33°)
height = D*cos(33°) * sin(33°)
Area ? = 1/2 * D*cos(33°) * sin(33°) * D*cos[sup:15afxeuc]2[/sup:15afxeuc](33°) = 1/2 * D[sup:15afxeuc]2[/sup:15afxeuc] *cos[sup:15afxeuc]3[/sup:15afxeuc](33°)* sin(33°)
Area [½?] = ?/8 * D[sup:15afxeuc]2[/sup:15afxeuc]
?/[½?] = 4/? *cos[sup:15afxeuc]3[/sup:15afxeuc](33°)* sin(33°) = 0.409065723
.
mmm4444bot said:I asked the guy at Beth's Cafe for some scratch paper; Subhotosh's hint haunted my hamburger.
I thought that I had an insight, but I resorted to my usual 24-step program of analytical geometry, to solve the exercise, instead.
So, the "hint" that I come up with is that the base of the triangle is: 2r cos(33°)^2, where r is the radius.
Wow. I'd like to see Subhotosh's work. I can't believe that's a hint! :wink:
hypotenuese = D*cos(33°)
Base = D*cos(33°) * cos(33°) = D*cos[sup:15afxeuc]2[/sup:15afxeuc](33°)
height = D*cos(33°) * sin(33°)
Area ? = 1/2 * D*cos(33°) * sin(33°) * D*cos[sup:15afxeuc]2[/sup:15afxeuc](33°) = 1/2 * D[sup:15afxeuc]2[/sup:15afxeuc] *cos[sup:15afxeuc]3[/sup:15afxeuc](33°)* sin(33°)
Area [½?] = ?/8 * D[sup:15afxeuc]2[/sup:15afxeuc]
?/[½?] = 4/? *cos[sup:15afxeuc]3[/sup:15afxeuc](33°)* sin(33°) = 0.409065723
.
mmm4444bot
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Subhotosh, I might have misworded my request.
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°).
My hint for the base expression took too much effort, to be called a hint.
(I'm concerned that I'm missing something obvious.)
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°).
My hint for the base expression took too much effort, to be called a hint.
(I'm concerned that I'm missing something obvious.)
D
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Re:
jschwa1 explained it above -
I used the fact the angle contained by a semicircle is 90°.
mmm4444bot said:Subhotosh, I might have misworded my request.
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°).
My hint for the base expression took too much effort, to be called a hint.
(I'm concerned that I'm missing something obvious.)
jschwa1 explained it above -
I used the fact the angle contained by a semicircle is 90°.
mmm4444bot
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Subhotosh Khan said:
Rats. I did not even see jschwa1's third post! Missed it, in the flurry of activity.
Thanks, I'll recheck.
mmm4444bot
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Oh, DUH!!
Note: I solved the exercise, using Subhotosh's hint at face value; I never employed an additional triangle, in my calculations.
(Now, I wait to be embarrased again, by Soroban.)
Note: I solved the exercise, using Subhotosh's hint at face value; I never employed an additional triangle, in my calculations.
(Now, I wait to be embarrased again, by Soroban.)
BigGlenntheHeavy
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\(\displaystyle mmm4444bot, \ to \ steal \ soroban's \ thunder, \ he \ has \ tan(33^0) \ = \ \frac{BC}{a}.\)
\(\displaystyle Hence, \ BC \ = \ atan(33^0), \ I \ trust \ no \ problem \ there.\)
\(\displaystyle Therefore, \ Area \ of \ given \ right \ triangle \ ABC \ = \ \frac{1}{2}a^2tan(33^0).\)
\(\displaystyle Now, \ right \ triangle \ ABD. \ sec(33^0) \ = \ \frac{AD}{AB}.\)
\(\displaystyle Now \ here's \ the \ cute \ part \ of \ soroban's \ analysis.\)
\(\displaystyle (AB)^2 \ = \ a^2+a^2tan^2(33^0) \ = \ a^2[1+tan^2(33^0)] \ = \ a^2sec^2(33^0).\)
\(\displaystyle Ergo, \ AB \ = \ asec(33^0), \ hence \ sec(33^0) \ = \ \frac{AD}{asec(33^0)}, \ implies \ AD \ = \ asec^2(33^0).\)
\(\displaystyle Afterthought: \ Why \ is \ triangle \ ABD \ a \ right \ triangle?\)
\(\displaystyle The \ rest \ should \ be \ self-explanatory.\)
\(\displaystyle See \ graph.\)
[attachment=0:35o0wkbe]aza.JPG[/attachment:35o0wkbe]
\(\displaystyle Hence, \ BC \ = \ atan(33^0), \ I \ trust \ no \ problem \ there.\)
\(\displaystyle Therefore, \ Area \ of \ given \ right \ triangle \ ABC \ = \ \frac{1}{2}a^2tan(33^0).\)
\(\displaystyle Now, \ right \ triangle \ ABD. \ sec(33^0) \ = \ \frac{AD}{AB}.\)
\(\displaystyle Now \ here's \ the \ cute \ part \ of \ soroban's \ analysis.\)
\(\displaystyle (AB)^2 \ = \ a^2+a^2tan^2(33^0) \ = \ a^2[1+tan^2(33^0)] \ = \ a^2sec^2(33^0).\)
\(\displaystyle Ergo, \ AB \ = \ asec(33^0), \ hence \ sec(33^0) \ = \ \frac{AD}{asec(33^0)}, \ implies \ AD \ = \ asec^2(33^0).\)
\(\displaystyle Afterthought: \ Why \ is \ triangle \ ABD \ a \ right \ triangle?\)
\(\displaystyle The \ rest \ should \ be \ self-explanatory.\)
\(\displaystyle See \ graph.\)
[attachment=0:35o0wkbe]aza.JPG[/attachment:35o0wkbe]
