Two Questions (1 Viewer)

[kaz1]

How would you do part (ii)?

and

I have no clue for this one.

 

[Timothy.Siu]

the second question u can move the absolute stuff up, because its always positive, so no change of sign is required.

so u get

|x-1|>1

x-1>1 or x-1<-1
x>2 or x<0

 

[kaz1]

the second question u can move the absolute stuff up, because its always positive, so no change of sign is required.

so u get

|x-1|>1

x-1>1 or x-1<-1
x>2 or x<0

Thanks, didn't think of that.

 

[Timothy.Siu]

width of path using area of triangle?

is the width root3/14 x h^2 ?

 

[kaz1]

width of path using area of triangle?

is the width root3/14 x h^2 ?

How did you use area of the triangle? If it helps I got the height to be rt(432/7).

btw I don't have the answers (It's the 2007 Independent Schools Trial Exam).

 

[Timothy.Siu]

How did you use area of the triangle? If it helps I got the height to be rt(432/7).

btw I don't have the answers (It's the 2007 Independent Schools Trial Exam).

lol i got h=root(21) but i'm not sure....

umm, well the area of the triangle is 1/2 xbxh or 1/2 AB sin C, so i made them equal each other.

 

[kaz1]

I think I got the answer to (ii) after using Timothy's method. The answer I got is (216√3)/7 but correct me if I'm wrong.

 

[Timothy.Siu]

someone else can do it, i got 3root3/2 because my h is different.

 

[jet]

Are you assuming BC is perpendicular to AB, because i dont think it is.

 

[Timothy.Siu]

Are you assuming BC is perpendicular to AB, because i dont think it is.

nope, i found the area of that triangle 2 times using the 2 different formulas

 

[jet]

I got the same as Timothy Siu.
i) I used the cos rule. h=sqrt(21)
ii) did the thing with the areas of the triangles. h=3sqrt(3)/2

i) BC = h/tan60, AC = h/tan30

Using the cos rule, AB^2 = BC^2 + AC^2 -2AC x BC x cos60
Therefore (h^2/tan^2(60)) + (h^2/tan^2(30)) - ((2h^2 cos(60))/tan60 tan30)=49
Multiplying up the denominators,
h^2 tan^2(30) + h^2 tan^2(60) - 2h^2(cos(60)tan(60)tan(30))=49tan^2(60)tan^2(30)
h^2(tan^2(60) + tan^2 (30) - 2cos(30)tan(60)tan(30) = 49
h^2(3 + (1/3)-1) = 49
(7h^2)/3=49
h=sqrt(21)

ii)First, Area = 1/2 (base) (perp. height=H)
=(7H/2)

Second, Area = 1/2 BC x AC x sin(60)
=1/2 x (sqrt(3)/2) x sqrt(21/3) x sqrt(21) x sqrt(3)
=1/4 x 21 x sqrt(3)
=21sqrt(3)/4

Hence,
7H/2 = 21sqrt(3)/4
7H = 21sqrt(3)/2
H=3sqrt(3)/2

 

[azureus88]

u can also draw a perpendicular from B to a point D on line AB produced, calculate angle A and length AC using sine rule, so BD=ACsin(angleA)

 

[Timothy.Siu]

u can also draw a perpendicular from B to a point D on line AB produced, calculate angle A and length AC using sine rule, so BD=ACsin(angleA)

much easier lol