# AEA Vectors - 2003Watch

#1
Can someone please show me how to do the attached question. Thanks.
0
12 years ago
#2
|OA| = |AB| = 1 unit

COSINE RULE
(1/rt2)^2 + (1+1/rt2)^2 = 1^2 + 1^2 - 2(1)(1)cosOAB
=> 1/2 + 1 + 2/rt2 + 1/2 = 2 - 2cosOAB
=> 2 + rt2 = 2-2cosOAB
=> 2cosOAB = -rt2
=> cosOAB = -rt2/2
=> angle OAB = 135 degrees = 3pi/4 radians

Therefore cos(3pi/4) = -0.5rt2
cos2A = 2cos^2A - 1
=> cos^2A = (cos2A+1)/2
=> cos^2(3pi/8) = (-0.5rt2+1)/2

1+tan^2(3pi/8) = sec^2(3pi/8)
=> tan(3pi/8) = (1/cos^2(3pi/8) - 1)^0.5
=> tan(3pi/8) = (2/(1-0.5rt2)-1)^0.5
tan(3pi/8) = [(2+rt2)/0.5 - 1)^0.5
tan(3pi/8) = [4+2rt2-1]^0.5
tan(3pi/8) = [(rt2+1)^2]^0.5
tan(3pi/8) = 1+rt2
0
#3
Thanks Widowmaker - really appreciate it.
0
12 years ago
#4
Notice that |OA|=|AB|, so OAB is isosceles. Draw a line from A to the x-axis, call the intersection point X. Notice that |OX|=|AX|, so angle OAX=arctan(1)=pi/4. Therefore OAB=3pi/4, and since the triangle is isosceles, BOA=pi/8. Therefore BOX=3pi/8. Therefore tan(3pi/8)=BX/OX which gives the right answer.
0
#5
(Original post by Kernel)
Notice that |OA|=|AB|, so OAB is isosceles. Draw a line from A to the x-axis, call the intersection point X. Notice that |OX|=|AX|, so angle OAX=arctan(1)=pi/4. Therefore OAB=3pi/4, and since the triangle is isosceles, BOA=pi/8. Therefore BOX=3pi/8. Therefore tan(3pi/8)=BX/OX which gives the right answer.
How do you notice that OX = AX?
0
12 years ago
#6
OA=OX+AX
OX has a j component of 0, so call it [x,0] while AX has an i component of 0, so call it [0,y]
OA=[x,0] + [0,y]=[rt(2)/2,rt(2)/2], so x and y=rt(2)/2.
So: OX=[rt(2)/2,0] AX= [0,rt(2/2] so their moduluses are equal.
0
12 years ago
#7
(Original post by joe8232)
How do you notice that OX = AX?
i-component of OX = 1/rt2
length of AX = j-component of OX = 1/rt2

Therefore OX = AX
0
#8
Thanks again - it's so obvious once you see it.
0
X

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