one question (1 Viewer)

[super.muppy]

can somebody giv me an answer for question 84 all of it plz
http://www.boardofstudies.nsw.edu.au/schoolcertificate/2007/mathematics-sc-test-07.pdf
cant think anymore or mayb its jst bcuz im stupid

 

[Naminé]

ForA, use Pythagoras:

the triangle with AB is just the other triangle flipped, and its base is half of the entire line given, as the line given is the width of the two put together. so it's 6.2/2 = 3.1

a^2 + b^2 = c^2

1.4^2 + 3.1^2 = c^2

take the square root of the answer, which is c, and it should be 3.4

As for C, I'm so worn out from studying, 'bout to head to bed, but you could use trig. ratios, I guess.

AB = 3.4, remember
AX = 3.1
BX = 1.4

As for B, those two triangles put together make a square... so the metal needed is 3.1m by 1.4m.

 

[rolror88]

B) would be just adding up the sides.

C would be tan ^-1 1.2/3.1 = 24 (nearest degree)

I'm taking a bit of a wild stab at D.. I'm too tired to concentrate properly either.

From C, you know that Angle BAC =Angle BCA = 24 degrees

180 - 24 - 24 = 132.

I'm not sure the angle at B is bisected, if it is, then Angle YBX = YXB = 66 degrees

So, 180 - 66 - 66 = 48 degrees. (Angle BYX)

Meh, I'm probably wrong.

____

If you just use your answer and iscoceles properties and the exterior angle rule (is that what it's called), you can also get Angle BYX = 48 degrees.

 

[super.muppy]

OMG
dats fucking da most easiest questions alive and i asked for help
(not saying dat i am smart or anything but OMG)

oh yea thanx

 

[Naminé]

It's alright. You're probably exhausted like me and finding it hard to think even when the answer is really obvious. Don't feel bad =)

 

[tommykins]

Soln's up to 70, the other thread got deleted for some odd reason.

40. D)
56. D)
57. B)
58. D)

http://img87.imageshack.us/my.php?image=scmaths1fy4.jpg
http://img255.imageshack.us/my.php?image=scmaths2qt7.jpg

 

[ratcher0071]

Question 84

a)
AB=BC (given on diagram)

XC= 6.2 / 2
XC= 3.1 m

BC² = BX² + XC²
BC² = (1.4)² + (3.1)²
BC = sqrt (11.57)
BC=3.4 m (1 dec. pl.)
therefore AB= 3.4 m

b)
Total length of metal = AB + BC + AC + XY + ZY + BX
= 3.4 + 3.4 + 6.2 + 1.7 + 1.7 + 1.4
=17.8 m

c) Tan A = 1.4 / 3.1
A= 24^o

d)
Angle ABX = 180 - (90+24) (angle sum of triangle)
Angle ABX = 66^o
therefore Angle BXY = 66^o (base angles of isosceles triangle are equal)
therefore Angle BYX = 180 - (66+66) (angle sum of triangle)
Angle BYX = 48^o