Q10 Help (1 Viewer)

[UTRazilBay]

Im struggling with the above question, any help is appreciated.

 

[HeroicPandas]

Triangle AOB use trig ratios

 

[HeroicPandas]

Triangle AOB use trig ratios to find the length AO

 

[Timske]

Do you need help with all of it?

 

[UTRazilBay]

the first two

 

[UTRazilBay]

could you show me exactly how to use the trig ratios, I've tried everything!

 

[HeroicPandas]

i mean consider triangle APO, sorry

 

[HeroicPandas]

could you show me exactly how to use the trig ratios, I've tried everything!

Ignore my first 2 posts sorry look at triangle OPB and APO

 

[UTRazilBay]

yeah could you show me your working out, my BO's and AO's don't cancel out

 

[HeroicPandas]

Use the cos ratio (adjacent on hypotenuse) to find PB in triangle OPB

Use the sine ratio (opposite on hypotnuse) to find AP in triangle APO

NOW u'VE GOTTEN the side AB then u do the same thing on the right side (or u say similarly? someone clarify

after u've found the 2 longest slanted sides find the area of ABC, A= 0.5absinC formula

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\\\ S=\frac{1}{2}*AO*BC\\\\ sin\theta=\frac{a}{AO}~\rightarrow AO=\frac{a}{sin\theta} \\\\ tan\theta=\frac{BO}{AO} \rightarrow BO=tan\theta*AO \\\\ AO =\frac{a}{sin\theta} ~and~ BO = \frac{a}{cos\theta}\\\\ BC = 2BO\\\\ \therefore S=\frac{1}{2}*\frac{a}{sin\theta}*\frac{2a}{cos\theta} =\frac{2a^2}{sin2\theta}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\\\ S=\frac{1}{2}*AO*BC\\\\ sin\theta=\frac{a}{AO}~\rightarrow AO=\frac{a}{sin\theta} \\\\ tan\theta=\frac{BO}{AO} \rightarrow BO=tan\theta*AO \\\\ AO =\frac{a}{sin\theta} ~and~ BO = \frac{a}{cos\theta}\\\\ BC = 2BO\\\\ \therefore S=\frac{1}{2}*\frac{a}{sin\theta}*\frac{2a}{cos\theta} =\frac{2a^2}{sin2\theta}" title="\dpi{100} S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\\\ S=\frac{1}{2}*AO*BC\\\\ sin\theta=\frac{a}{AO}~\rightarrow AO=\frac{a}{sin\theta} \\\\ tan\theta=\frac{BO}{AO} \rightarrow BO=tan\theta*AO \\\\ AO =\frac{a}{sin\theta} ~and~ BO = \frac{a}{cos\theta}\\\\ BC = 2BO\\\\ \therefore S=\frac{1}{2}*\frac{a}{sin\theta}*\frac{2a}{cos\theta} =\frac{2a^2}{sin2\theta}" /></a>

 

[HeroicPandas]

<a href="http://www.codecogs.com/eqnedit.php?latex=S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\ S=\frac{1}{2}AO*BC\\\\ sin\theta=\frac{a}{AO} \\\\ tan\theta=\frac{BO}{AO}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\ S=\frac{1}{2}AO*BC\\\\ sin\theta=\frac{a}{AO} \\\\ tan\theta=\frac{BO}{AO}" title="S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\ S=\frac{1}{2}AO*BC\\\\ sin\theta=\frac{a}{AO} \\\\ tan\theta=\frac{BO}{AO}" /></a>

I thought of that first but is AO PERPENDICULAR TO BC?

MY METHOD IS THE LAST RESORT

 

[Timske]

I thought of that first but is AO PERPENDICULAR TO BC?

MY METHOD IS THE LAST RESORT

Yes, because AO bisects the line BC

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} b)i) \frac{dS}{d\theta} > 0\\\\ ~~~~~~~ii)\frac{dS}{d\theta} < 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} b)i) \frac{dS}{d\theta} > 0\\\\ ~~~~~~~ii)\frac{dS}{d\theta} < 0" title="\dpi{100} b)i) \frac{dS}{d\theta} > 0\\\\ ~~~~~~~ii)\frac{dS}{d\theta} < 0" /></a>

 

[UTRazilBay]

oh i see... could you also help with part b)...

For decreasing
(-4a^2 . cos2θ) / (sin2θ)^2 < 0
-4a^2 . cos2θ < 0
cos2θ > 0
therefore θ > pi/4 ??? ( since 0 < θ < pi/2)

Is that right?

 

[HeroicPandas]

oh i see... could you also help with part b)...

For decreasing
(-4a^2 . cos2θ) / (sin2θ)^2 < 0
-4a^2 . cos2θ < 0
cos2θ > 0
therefore θ > pi/4 ??? ( since 0 < θ < pi/2)

Is that right?

and

^ using the unit circle since 0 < θ < pi/2 we take the first range: