Trail HSc questions (1 Viewer)

[FOB all the way]

hi

i am in year 11 but doing the mathematics hsc this year
there are a fue quaestions i have been unable to do all holidays could some one please help

Question 1:
The 5 angles of a pentagon are in arithmetic progression. Given the size of the largest angle is three times the size of the smallest angle, find the size of all the angles.

Question2:
ABC is a triangle in which AB=AC=x metres and AB+BC+CA=1 metre and angle ABC= @ (thertra) radians. D is the mid point of BC

i) show that the altitude AD= Square root(4x-1)/2 metres

ii) Hence or otherwise show that sin@= square root(x-1/4)/x

Question 3:
f(x)=log3(1/x^2)




Thank you sooooooooooooooooooooooooooooooooooooooooooooooo much

 

[iorlas]

okay i can only be bothered doing question two.. don't know what you want with question three...

okay, you got your triangle...

AC + AB + BC = 1
therfore

BC = 1 - AB - AC
BC = 1 - 2x

then BD = BC/2

therefore BD = (1-2x)/2

then by pythagoras

AD^2 = x^2 - ((1-2x)/2)^2

simplify that and you get

AD^2 = (4x - 1)/4
take the square root...

then for part two, in your triangle ABD, use sine = o/h

o = what you got above, and h = x

simplify that, take a four out of the square root (leaving you with 1/4 inside the root) and then the 4, becomes a 2 (sqrt of 4) and that cancels with the 2 on the bottom...

 

[ellen.louise]

Question 3:
f(x)=log3(1/x^2)

and? what do you want from this?

 

[FOB all the way]

thanks soooooooooooooooooooo much

and by Question three (sorry typo)

f(x)log3(1/x^2) find f'(x)

 

[ellen.louise]

k im doing it now, then i'll try q1

 

[jannny]

you mind telling us the answer for question 1?

I got a = 54 degrees

d = 27

therefore the angles
are : 54, 81, 108, 135, 162

edit

 

[ellen.louise]

man my notes suck.

I only just found the changing the base law.
I should really know that one

 

[FOB all the way]

hi sorry i dont have any answers yet i will get them after the holidays

 

[FOB all the way]

would you mind telling me how you worked out question 1

 

[jannny]

alright im prety sure with my answer

a = 54
d= 27

for question 1.
Let a = smallest angle
Let 3a = biggest angle

Then find the sum of all angles of a polygon
S = 2n -4 x 90 degrees
S= (2x5) -4 x 90 degrees
= 540

therefore the sum of all the angles in the polygon is 540 degrees.

Then you find the Nth equation for the arithmetic progress

Tn = a + (n-1)d
T5 = 3a, a = the first term
3a = a + ( 5-1) d
3a - a = 4d
2a = 4d
therefore d = 1/2 a

Find the Sum of AP equation

Sn = n/2 (2a + (n-1)d)
Sn = 540, n=5 d= 1/2a

540 = 5/2 (2a + (5-1) * 1/2a)
1080 = 5(2a+ 4*1/2 a)
1080 = 20a
a = 54

sub this back in to

1/2a = d
1/2 * 54 = d

d= 27
and a = 54

therefore
the series is a, a +d, a+ 2d, a+3d, a+4d
which makes it 54+ 81 +108 +135+162
which also proves 3a = 162
a = 54

which adds to 540 degrees =]

 

[ellen.louise]

3:

f(x)=log3(x^-2)

f(x)=ln(x^-2)/ln3
use y = lnx
y'= f '(x)/f(x)
f '(x)=1/ln3 . (-2x^-3)/(x^-2)
=1/ln3 . (-2x^2)/x^3
=1/ln3 . (-2/x)
=-2/(xln3)

I hope this is right.

 

[FOB all the way]

thanks so much
sorry to be such a pain and i guess the last think you want to be doing on the holidays is someone elles home work but can i ask one more question

if sin(A+B)=sinAcosB+cosAsinB and Cos (a+B)=cosAcosB-sinAsinB

show that tan(A+B) = tanA+tanB / 1-tanAtanB

and hence find a similar expression for tan(A-B)

thanks so much
its a bit of a worry that i have soo many questions looks like i will be doing it again next year

 

[jannny]

for question 2.. Why would you use pythagoras for i) Unless there's a diagram coz you can only use pythagoras if the triangle is a right angle right?

Did you get a diagram with question 2 OP?

 

[jannny]

thanks so much
sorry to be such a pain and i guess the last think you want to be doing on the holidays is someone elles home work but can i ask one more question

if sin(A+B)=sinAcosB+cosAsinB and Cos (a+B)=cosAcosB-sinAsinB

show that tan(A+B) = tanA+tanB / 1-tanAtanB

and hence find a similar expression for tan(A-B)

thanks so much
its a bit of a worry that i have soo many questions looks like i will be doing it again next year

I dont recall this being in 2 unit, u sure this is a 2 unit q?

 

[FOB all the way]

no there was no diagram and yes i am sure it is a 2u question it is in a 2006 trail mathematics paper

 

[iorlas]

for question 2.. Why would you use pythagoras for i) Unless there's a diagram coz you can only use pythagoras if the triangle is a right angle right?

Did you get a diagram with question 2 OP?

what do you mean 'unless there's a diagram'? always, ALWAYS draw yourself a diagram if they give you information like that...

once you put AD in you have two right angled triangles...

 

[ellen.louise]

for question 2.. Why would you use pythagoras for i) Unless there's a diagram coz you can only use pythagoras if the triangle is a right angle right?

Did you get a diagram with question 2 OP?

You're finding the altitude of AD.

as ABC is an isosceles triangle, and D is the midpoint of CB, AD is automatically perpendicular to CB. Thus you use pythagoras

 

[FOB all the way]

so no one doing 2u can do the last question

 

[jannny]

Ah k thanks, I forgot the fact that its an isosceles cheers =]

 

[ellen.louise]

so no one doing 2u can do the last question

I think technically they should be able to.
They have all the work for it.
I still want to know if I got it right cause I'll feel pretty lame if I didn't.

Post the answer up when you get it, K?