Preliminary applied trig, Help please! (1 Viewer)

crimp · May 25, 2015

Hey, I was wondering if anyone could help me with question 10.
Thanks!

http://imgur.com/S4Flh94

InteGrand · May 25, 2015

crimp said:
Hey, I was wondering if anyone could help me with question 10.
Thanks!

http://imgur.com/S4Flh94

$Consider $\triangle OAC$. Since we are given $OG:OC=2:3$, we can let $OG=2$ and $GC=1$. We are also given that $\angle AOC = \alpha$ and $\angle OAG=\beta$. Label all these on a diagram of the triangle to make things clear. Also note that $\angle C$ is a right angle.$

$Now, $\angle AGC = \alpha + \beta$ (external angle of $\triangle OAG$). Hence by right-angled trigonometry in $\triangle AGC$, $\tan (\alpha +\beta)=\frac{AC}{1}\Rightarrow AC=\tan(\alpha+\beta)$. Then by right-angled trigonometry in $\triangle OAC$, we see that $\tan \alpha=\frac{AC}{OC} = \frac{\tan(\alpha+\beta)}{3}\Rightarrow \tan(\alpha+\beta) = 3\tan \alpha$.$

$Now we just expand the above equation's LHS: $\frac{\tan \alpha + \tan \beta}{1+\tan \alpha \tan \beta}=3\tan \alpha$, by the compound angle formula for $\tan$. Now we just need to solve for $\tan \beta$ (which is simple to do) and we're done.$

$We write $A\equiv \tan \alpha$ and $B\equiv \tan \beta$ for simplicity, so $\frac{A+B}{1+AB}=3A$, and we want to solve for $B:$$

$\Rightarrow A+B=3A(1+AB) = 3A + 3A^2 B$

$\Rightarrow B(1-3A^2) = 2A$

$\Rightarrow B= \frac{2A}{1-3A^2}$

$i.e. $\tan \beta =\frac{2\tan \alpha}{1-3\tan^2 \alpha}.$ \blacksquare$

braintic · May 25, 2015

crimp said:
Hey, I was wondering if anyone could help me with question 10.
Thanks!

http://imgur.com/S4Flh94

Where did you get this question from?

Not that it affects the maths required for the question, but I would have thought that the question writer would have tried to get the physics right.

Assuming it has uniform density, the centre of gravity of a cone is 3/4 of the way from the apex to the base, not 2/3.

Edit: Actually, since it is described as a 'vessel', it is probably hollow. In that case, it is definitely not 3/4. Not sure if it's 2/3 though.

crimp · May 26, 2015

braintic said:
Where did you get this question from?

Not that it affects the maths required for the question, but I would have thought that the question writer would have tried to get the physics right.

Assuming it has uniform density, the centre of gravity of a cone is 3/4 of the way from the apex to the base, not 2/3.

Edit: Actually, since it is described as a 'vessel', it is probably hollow. In that case, it is definitely not 3/4. Not sure if it's 2/3 though.

I got it from the "New Senior Mathematics" Three unit course for year 11 and 12 textbook.

crimp · May 26, 2015

InteGrand said:
$Consider $\triangle OAC$. Since we are given $OG:OC=2:3$, we can let $OG=2$ and $GC=1$. We are also given that $\angle AOC = \alpha$ and $\angle OAG=\beta$. Label all these on a diagram of the triangle to make things clear. Also note that $\angle C$ is a right angle.$

$Now, $\angle AGC = \alpha + \beta$ (external angle of $\triangle OAG$). Hence by right-angled trigonometry in $\triangle AGC$, $\tan (\alpha +\beta)=\frac{AC}{1}\Rightarrow AC=\tan(\alpha+\beta)$. Then by right-angled trigonometry in $\triangle OAC$, we see that $\tan \alpha=\frac{AC}{OC} = \frac{\tan(\alpha+\beta)}{3}\Rightarrow \tan(\alpha+\beta) = 3\tan \alpha$.$

$Now we just expand the above equation's LHS: $\frac{\tan \alpha + \tan \beta}{1+\tan \alpha \tan \beta}=3\tan \alpha$, by the compound angle formula for $\tan$. Now we just need to solve for $\tan \beta$ (which is simple to do) and we're done.$

$We write $A\equiv \tan \alpha$ and $B\equiv \tan \beta$ for simplicity, so $\frac{A+B}{1+AB}=3A$, and we want to solve for $B:$$

$\Rightarrow A+B=3A(1+AB) = 3A + 3A^2 B$

$\Rightarrow B(1-3A^2) = 2A$

$\Rightarrow B= \frac{2A}{1-3A^2}$

$i.e. $\tan \beta =\frac{2\tan \alpha}{1-3\tan^2 \alpha}.$ \blacksquare$

Thanks man! You're like a math god @_@

Preliminary applied trig, Help please! (1 Viewer)

crimp

New Member

InteGrand

Well-Known Member

braintic

Well-Known Member

crimp

New Member

crimp

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)